Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals

International Journal of Arts Humanities and Social Sciences Studies Volume 6 Issue 12 ǁ December 2021 ISSN: 2582-1601 www.ijahss.com

Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals Teodor Dumitru Vălcan Abstract: According to the curricular documents, learning Mathematics in school aims to raise awareness of the nature of Mathematics as a problem-solving activity, based on a body of knowledge and procedures, but also as a dynamic discipline, closely linked to society through its relevance in everyday life and its role. in Natural Sciences, in Technologies and in Social Sciences. The major meaning of the current references in teaching learning Mathematics is shifting the emphasis from teaching knowledge that has essentially remained the same from the old programs, on capacity building, where "using different strategies in solving problems" occupies a special place. This paper falls into this context; continuing the steps begun some time ago, a way to calculate the limits for four categories of strings is presented here, with the particularities and related examples.

Keywords: string, convergence, limit, function, integral.

I.

Putting the problem

In a recent paper (Vălcan, 2016), titled very suggestively From Mathematics Spectacle. Examples of Good Practice for Pupils, Students and Teachers, we have proposed, obtaining remarkable limits of sequences, at the upper level of the high school, so only using mathematical knowledge learned in school, but in a comprehensive approach. So, first we used Moivre's formula and Newton's binomial in solving three higherlevel algebraic equations, after which, using Viête's relations, we got some inequalities that helped us calculate the limits of remarkable sequences. Finally, we applied the results obtained when calculating the limits of less known sequences.

In the paper From limits of sequences to definite integrals - (Vălcan, 2017) we have approached from the inductive but also deductive perspective the problem of calculating the limits of some special classes of sequences of real numbers, using the define integral, and in the paper From integrable functions to limits of sequences - (Vălcan, (I), 2018), I started the reverse: from certain definite integrals we have determined two classes of convergent sequences and their limits.

Finally, in the paper (Vălcan, (II), 2018) - entitled The spectacles of Mathematics ... continues, we continued the ideas presented in the papers mentioned above and we calculated the limits of other categories of sequences of real numbers.

In this paper, as mentioned above, we will continue this approach and present ways to calculate the limits of other sequences of real numbers.

II.

Main results, particular cases and examples

This paper is a continuation of the paper (Vălcan, 2017) and, implicitly, of the paper (Vălcan, 2016), where I have shown that the following equalities hold: 1

1 1  (1  x p )  x q 1  =     dx , with p, qN and pq. (q  k )  (q  k  p) p  1  x q  k 1 0 n

A) lim



B) lim

 (1  x r  p )  x p 1  1 1 =     dx , with p, q, rN and prq. (q  k  p)  (q  k  r ) r  p  1  xq  k 0 0

n 



1

n

n 





 [( r  p)  r  x p  p  x r ]  x q 1  1 1 =     dx , n  (q  k )  (q  k  p)  (q  k  r ) p  r  (r  p)  1  xq  k 1 0 with p, q, rN and prq.

1

n

C) lim

 n

D) lim

n 



 (q  k  p)  (q  k  r)  (q  k  s) = (r  p)  (s  p)  (s  r)  1

1

k 0

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals  [(s  r )  (s  p)  x r  p  (r  p)  x s  p ]  x p 1      dx , with p, q, r, sN and prsq. q 1  x    0 Also in (Vălcan, 2017) we calculated or proposed for calculation 16 limits for type A sequences, 24 limits for type B sequences, 24 limits for type C sequences and 16 limits for type D sequences.

Therefore, in the following, we will respect the ideas and numbering made in (Vălcan, 2016) and (Vălcan, 2017).

We start presenting the new results: 1



E) Theorem E: If p, q, r, sN and prs, then: n

lim

n 



1 =· lim n  ( q  k )  ( q  k  p )  ( q  k  r )  (q  k  s ) k 1

1

 E( x )   dx , q  0

 1  x

(E)

where:

E(x)=(1-2xp+3xr-4xs)xq-1, 1 = , p  r  s  (r  p)  (s  p)  (s  r ) 1=(r-p)(s-p)(s-r), 2=rs(s-r), Proof: For every kN, 1 = q  k  (q  k  p)  (q  k  r )  (q  k  s)

3=ps(s-p),

4=pr(r-p).

 (r  p)  (s  p)  (s  r ) r  s  (s  r ) s  p  (s  p) r  p  (r  p)     =·   qk qk  p qk  r qk s   1



=· (1  x qk 1   2  x qk p1   3  x qk r 1   4  x qk s1 )  dx 0 1



=· (1   2  x p   3  x r   4  x s )  x q  k 1  dx , 0

where:

1 , p  r  s  (r  p)  (s  p)  (s  r ) 1=(r-p)(s-p)(s-r), =

2=rs(s-r),

3=ps(s-p),

4=pr(r-p).

So, n

lim

n 

 ( q  k )  ( q  k  p )  ( q  k  r )  (q  k  s ) = 1

k 1

n 1

=· lim

n 

  [(

1

k 1 0

1 n

=· lim

n 

=· lim

n 

  2  x p  3  x r   4  x s )  x qk 1 ]  dx

 [(

1

0 k 1 1

  2  x p  3  x r   4  x s )  x qk 1 ]  dx

n   p r s (     x    x    x )  x qk 1   dx  1 2 3 4  k 1  0





n q   p r s q 1 1  x (1   2  x  3  x   4  x )  x    dx 1  x q   0

1

=· lim

n 

 1

=· lim

n 

 (1   2  x p   3  x r   4  x s )  x q 1     dx 1  xq   0



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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals  (1   2  x p  3  x r   4  x s )  x ( n 1)q 1     dx 1  xq   0

1

=· lim

n 



1

 (   2  x p   3  x r   4  x s )  x q 1  =·  1   dx , q 1  x    0



because, according to (Vălcan, 2016, Proposition),

 (1   2  x p  3  x r   4  x s )  x ( n 1)q 1     dx =0. q 1  x    0

1

lim

n 



The following remark is required here: Remark E0: A simple calculation shows us that: 1-2+3-4=0, which tells us that none of the above integrals is not improper.

Corollary E1: If qN, then: 1

1 1  x q1  (1  x ) 3  =     dx . n  (q  k )  (q  k  1)  (q  k  2)  (q  k  3) 6  1  x q  k 1 0 Proof: Equality (E1) is obtained from Theorem (E), for: p=1, r=2 and s=3, or by direct calculation. Thus, for every kN, 1 3 3 1  1  1  = ·     q  k  (q  k  1)  (q  k  2)  (q  k  3) 6  q  k q  k  1 q  k  2 q  k  3  n

lim





1

1

=

=

(E1)





0 1

0

1 1 · ( x q k 1  3  x q k  3  x q  k 1  x q  k  2 )  dx = · [(1  3  x  3  x 2  x 3 )  x q k 1 ]  dx 6 6



1 · [(1  x )3  x q  k 1 ]  dx .

6 0

So, n



1 1 lim = · lim n  (q  k )  (q  k  1)  (q  k  2)  (q  k  3) 6 n  k 1 1 n

1 · lim 6 n 



=

1 · lim 6 n 

n q   3 q 1 1  x (1  x )  x    dx 1  x q   0

=

1 · lim 6 n 

=

[(1  x )3  x q  k 1 ]  dx =

0 k 1 1

1 · lim 6 n 

n 1

  [(1  x)

3

 x q  k 1 ]  dx

k 1 0

n   3 x q k 1   dx (1  x )   k 1  0

1







 x q 1  (1  x )3  1    dx - · lim q 6 n   1 x  0

1



1

 x ( n 1)q 1  (1  x )3     dx 1  xq   0



1

 x q 1  (1  x )3  1 ·    dx , 6 1  xq   0 because, according to (Vălcan, 2016, Proposition),

=

 1

lim

n 

 x ( n 1)q 1  (1  x )3     dx =0.

1  xq   0



Example E1.1: The following equality holds: n

lim

n 

 k  (k  1)  (k  2)  (k  3) = 18 .

1

1

(E1.1)

k 1

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals Proof: Equality (E1.1) is obtained from equality (E1), for: q=1, or by direct calculation. Thus, for every kN, 1 3 3 1  1 1 = ·     k  (k  1)  (k  2)  (k  3) 6  k k  1 k  2 k  3  1

1





0 1

0

1 1 = · ( x k 1  3  x k  3  x k 1  x k  2 )  dx = · [(1  3  x  3  x 2  x 3 )  x k 1 ]  dx 6 6 =



1 · [(1  x )3  x k 1 ]  dx .

6 0

So, n

lim

n 



1 1 = · lim 6 n  k  ( k  1 )  ( k  2 )  ( k  3 ) k 1 1 n

1 = · lim 6 n 



1 = · lim 6 n 

n  1 3 1 x (1  x )    dx = · lim 6 n  1  x   0

[(1  x )3  x k 1 ]  dx =

0 k 1 1



1

=

1 · lim 6 n 



1 1 · (1  x ) 2  dx - · lim 6 6 n  0

n 1

  [(1  x)

 x k 1 ]  dx

k 1 0

n   3 ( 1  x )  x k 1   dx   k 1  0

1



 1

 [(1  x)

2

 (1  x n )]  dx

0

1



3

1

[(1  x ) 2  x n ]  dx =

0



1 · (1  2  x  x 2 )  dx 6 0

x 2 x 3  1 1  1 1 1 1   = · x  2 = · 1  2    = , 6  2 3  18 2 3  0 6  since: 1

lim

n 

 [(1  x)

2

 x n ]  dx =0.

0

Example E1.2: The following equality holds: n

lim

n 

 (2  k)  (2  k  1)  (2  k  2)  (2  k  3) = 6 ·  6  4  ln 2  .

1

1

17

(E1.2)

k 1

Proof: Equality (E1.2) is obtained from equality (E1), for: q=2, or by direct calculation. Thus, for every kN, 1 3 3 1  1  1    = ·  2  k  (2  k  1)  (2  k  2)  (2  k  3) 6  2  k 2  k  1 2  k  2 2  k  3  1

1





0 1

0

1 1 = · ( x 2k 1  3  x 2k  3  x 2k 1  x 2k  2 )  dx = · [(1  3  x  3  x 2  x 3 )  x 2k 1 ]  dx 6 6 =



1 · [(1  x )3  x 2k 1 ]  dx .

6 0

So, n

lim

n 

=



1 1 = · lim (2  k )  (2  k  1)  (2  k  2)  (2  k  3) 6 n   k 1

1 · lim 6 n 

1 n

 0 k 1

[(1  x )3  x 2k 1 ]  dx =

1 · lim 6 n 

n 1

  [(1  x)

3

 x 2k 1 ]  dx

k 1 0

n   3 ( 1  x )  x 2k 1   dx   k 1  0

1





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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals =

1 · lim 6 n 

 1  x 2 n  1 3  dx = · lim (1  x )  x  2  6 n  1  x   0

1



1

(1  x )  x 1  dx - lim = · n  1 x 6 2

 0

=

1 6

1

 0

(1  x )  x 1 x 2

n 1

1

(1  x ) 2  x  (1  x n )  dx 1 x

 0

1

4  1   dx = ·  x 2  3  x  4    dx x 1  6  0



  1 1  17 x 2 x3  ·  4  x  3   4  ln(x  1)  = ·   4  ln 2  ,   0 2 3 6  6   

since, 1

lim

n 

 0

(1  x ) 2  x n 1  dx =0.

1 x

Example E1.3: The following equality holds: n

lim

n 

 (3  k)  (3  k  1)  (3  k  2)  (3  k  3) = 18 ·  2  1

11

1

k 1

 3   . 

(E1.3)

Proof: Equality (E1.3) is obtained from equality (E1), for: q=3, or by direct calculation. Thus, for every kN, 3 3 1  1  1 1 = ·     3  k  (3  k  1)  (3  k  2)  (3  k  3) 6  3  k 3  k  1 3  k  2 3  k  3  1

1

=

=





0 1

0

1 1 · ( x 3k 1  3  x 3k  3  x 3k 1  x 3k  2 )  dx = · [(1  3  x  3  x 2  x 3 )  x 3k 1 ]  dx 6 6



1 · [(1  x )3  x 3k 1 ]  dx .

6 0

So, n

lim

n 

=

=



1 1 = · lim ( 3  k )  ( 3  k  1 )  ( 3  k  2 )  ( 3  k  3 ) 6 n  k 1



1 · lim 6 n 

3n   1 3 2 1 x  dx = · lim (1  x )  x  3  6 n  1  x   0

[(1  x )3  x 3k 1 ]  dx =

0 k 1 1

(1  x )  x 1 1  dx - · lim · 2 6 6 n  1 x  x 0 2

1



(1  x )  x

0

1

2

1

3 1 1   ·  x2  3 x  3  2   dx = 6 6 x  x 1   0

=

1 6





 x 3k 1 ]  dx

k 1 0

(1  x ) 2  x 2  (1  x 3n ) 1 x  x2

0

3n  2

1 x  x2

=

3







2



1 · lim 6 n 

  [(1  x)

n   3 x 3k 1   dx (1  x )   k 1  0

1

1 · lim 6 n 

1

=

1 n

n 1

 dx

 dx

  2  x 1 1 x 2 x3     2  3  arctg  · 3 x  3   2 3 3   0 

 11 3    ·  , 6 3  

since, 1

lim

n 



(1  x ) 2  x 3n  2

0

1 x  x2

 dx =0.

Example E1.4: The following equality holds: n

lim

n 



 (4  k)  (4  k  1)  (4  k  2)  (4  k  3) = 12 ·  3  3  ln 2  2  .

1

1

11

(E1.4)

k 1

Proof: Equality (E1.4) is obtained from equality (E1), for:

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1





0 1

0

1 1 = · ( x 4k 1  3  x 4k  3  x 4k 1  x 4k  2 )  dx = · [(1  3  x  3  x 2  x 3 )  x 4k 1 ]  dx 6 6 =



1 · [(1  x )3  x 4k 1 ]  dx .

6 0

So, n

lim

n 

=

=



1 1 = · lim ( 4  k )  ( 4  k  1 )  ( 4  k  2 )  ( 4  k  3 ) 6 n  k 1 1 n

 [(1  x)

1 · lim 6 n 

4 n   1 3 3 1 x  dx = · lim (1  x )  x  4  6 n  1  x   0

 x 4k 1 ]  dx =

0 k 1 1

(1  x ) 2  x 3 1 1  dx - · lim · 2 3 6 6 n  1 x  x  x 0



1

0

1

 x 4k 1 ]  dx

k 1 0



(1  x ) 2  x 3  (1  x 4n )

2

1  x  x 2  x3

0

(1  x ) 2  x 4n  3

 1 x  x

3







1

=

1 · lim 6 n 

  [(1  x)

n   3 x 4k 1   dx (1  x )   k 1  0

1

1 · lim 6 n 

3

n 1

 x3

 dx

 dx

1

=

1 1 2 x 1  1    2  2   dx ·  x2  3 x  3  2  x 1 2 x 1 x 1  6  0

=

1 6

=

3 1 1   1  11  1  ·  3    2  ln 2   ln 2   = ·   3  ln 2   2 3 2 4  12  3 2 6 



 1 x 2 x3 1 · 3 x  3   2  ln(x  1)   ln(x 2  1)  arctgx   0 2 3 2  

since, 1

lim

n 

(1  x ) 2  x 4n  3

 1 x  x

2

0

 x3

 dx =0.

Corollary E2: If qN, then: 1  x q 1  (1  x )3  ( x  3)  1 1 =     dx . n  (q  k )  (q  k  1)  (q  k  2)  (q  k  4) 24  1 xq  k 1 0 Proof: Equality (E2) is obtained from Theorem (E), for: p=1, r=2 and s=4, or by direct calculation. Thus, for every kN, 1  3 1 8 6 1   = ·    q  k  (q  k  1)  (q  k  2)  (q  k  4) 24  q  k q  k  1 q  k  2 q  k  4 

n

lim





1

=

(E2)

1

1 1 · (3  x qk 1  8  x qk  6  x qk 1  x qk  3 )  dx = · [(3  8  x  6  x 2  x 4 )  x q k 1 ]  dx .

24 24





0

0

So, n

lim

n 

=



1 1 = · lim n  ( q  k )  ( q  k  1 )  ( q  k  2 )  ( q  k  4 ) 24 k 1

1 · lim 24 n  

1 n

 [(3  8  x  6  x 0 k 1

2

 x 4 )  x qk 1 ]  dx =

n 1

 [(3  8  x  6  x

2

 x 4 )  x qk 1 ]  dx

k 1 0 1

1 · lim 24 n  

n   2 4 x qk 1   dx (3  8  x  6  x  x )   k 1  0





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1 · lim 24 n  

n q   2 4 q 1 1  x  dx (3  8  x  6  x  x )  x  q  1  x     0

1



1  x q1  (3  8  x  6  x 2  x 4 )  1 lim = ·   dx 24 n  0  1 xq 

1 · lim 24 n  

 x ( n 1)q 1  (3  8  x  6  x 2  x 4 )     dx 1  xq   0

1



 x q 1  (3  8  x  6  x 2  x 4 )   x q 1  (1  x )3  ( x  3)  1 1 ·  = ·  dx     dx , 24 24 1 xq 1 xq     0 0 because, according to (Vălcan, 2016, Proposition), 1

1



=



 x ( n 1)q 1  (3  8  x  6  x 2  x 4 )     dx =0. n  1 xq   0 Example E2.1: The following equality holds: 1

lim



n

lim

n 

 k  (k  1)  (k  2)  (k  4) = 288 .

1

13

(E2.1)

k 1

Proof: Equality (E2.1) is obtained from equality (E2), for: q=1, or by direct calculation. Thus, for every kN, 1 3 8 6 1  1   = ·   k  (k  1)  (k  2)  (k  4) 24  k k  1 k  2 k  4  1

1

=

=

1 1 · (3  x k 1  8  x k  6  x k 1  x k  3 )  dx = · [(3  8  x  6  x 2  x 4 )  x k 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  ( x  3)  x k 1 ]  dx .

24

 0

So, n



1 1 lim = · lim n  k  (k  1)  (k  2)  (k  4) 24 n   k 1 =

=

1 n

=

1 · lim 24 n  

 1 xn  1 3 · lim (1  x )  ( x  3)    dx = n  1  x 24     0

0 k 1 1

1 · lim 24 n  



 [(1  x)

0 1

0



1 1 · ( x 3  x 2  5  x  3)  dx = 24 24

 ( x  3)  x k 1 ]  dx

n   3 x k 1   dx (1  x )  ( x  3)   k 1  0

1



 1

[(1  x)

2

 ( x  3)  (1  x n )]  dx

0

1



1 1 · [(1  x ) 2  ( x  3)]  dx · lim 24 24 n  

3

k 1 0



[(1  x )3  ( x  3)  x k 1 ]  dx =

0

=

  [(1  x)

1 · lim 24 n  

1

=

n 1

2

 ( x  3)  x n ]  dx

 x 2 x 3 x 4  1   · 3 x  5  2 3 4  0 

1  1 1 1  1 13 13 · 3  5    = · = , 2 3 4  24 12 288 24 

since: 1

lim

n 

 [(1  x)

2

 ( x  3)  x n ]  dx .

0

Example E2.2: The following equality holds:

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lim

n 

 (2  k)  (2  k  1)  (2  k  2)  (2  k  4) = 24 ·  4  8  ln 2  .

1

1

23

(E2.2)

k 1

Proof: Equality (E2.2) is obtained from equality (E2), for: q=2, or by direct calculation. Thus, for every kN, 1  3 1 8 6 1  = ·     2  k  (2  k  1)  (2  k  2)  (2  k  4) 24  2  k 2  k  1 2  k  2 2  k  4  1

=

=

1

1 1 · (3  x 2k 1  8  x 2k  6  x 2k 1  x 2k  3 )  dx = · [(3  8  x  6  x 2  x 4 )  x 2k 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  ( x  3)  x 2k 1 ]  dx .

24

 0

So, n



1 1 lim = · lim n  (2  k )  (2  k  1)  (2  k  2)  (2  k  4) 24 n   k 1 =

1 · lim 24 n  

1 = · lim 24 n  

  1 3 k 1 · lim  (1  x )  ( x  3)  x   dx = n  24   k  1   0

1



1

=

(1  x )  x  ( x  3) 1 1  dx · · lim 1 x 24 24 n   2

 0 1

1

 0

 x  x  5 x  3 x  1   dx = 1 ·   1 x 24  24  0

=

1  1 1  1  23  ·  8  5    8  ln 2  = ·   8  ln 2  , 2 4 24   24  4 



3

2

 ( x  3)  x 2k 1 ]  dx

k 1 0

 1

 0



(1  x ) 2  ( x  3)  x  (1  x 2n )  dx 1 x

(1  x )  ( x  3)  x 2n 1  dx 1 x

=

4

3

n   3 x 2k 1   dx (1  x )  ( x  3)   k 1  0



 1 xn  1 3 · lim (1  x )  ( x  3)    dx = n  1  x 24     0

  [(1  x)

1

n

1

n 1

2

 1 x2 x4 · 8  x  5    8  ln(x  1)   0 2 4  

since: 1

lim

n 

 0

(1  x ) 2  ( x  3)  x 2n 1  dx =0.

1 x

Example E2.3: The following equality holds: n

lim

n 

 (3  k)  (3  k  1)  (3  k  2)  (3  k  4) = 48 ·  2  3  ln 3  5  1

1

25

k 1

 3 .

3

(E2.3)

Proof: Equality (E2.3) is obtained from equality (E2), for: q=3, or by direct calculation. Thus, for every kN, 1  3 8 6 1  1    = ·  3  k  (3  k  1)  (3  k  2)  (3  k  4) 24  3  k 3  k  1 3  k  2 3  k  4  1

1

1 1 = · (3  x 3k 1  8  x 3k  6  x 3k 1  x 3k  3 )  dx = · [(3  8  x  6  x 2  x 4 )  x 3k 1 ]  dx 24 24 =





0 1

0

1 · [(1  x )3  ( x  3)  x 3k 1 ]  dx .

24

 0

So,

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n

lim

n 

=

=

  [(1  x)



1 1 = · lim n  ( 3  k )  ( 3  k  1 )  ( 3  k  2 )  ( 3  k  4 ) 24 k 1 1 n

k 1 0

n   3 x 3k 1   dx (1  x )  ( x  3)   k 1  0

 [(1  x)

1 · lim 24 n  

3n   1 3 2 1 x  dx = · lim (1  x )  ( x  3)  x  3  24 n   1  x   0

 ( x  3)  x 3k 1 ]  dx =

0 k 1 1

1 · lim 24 n  







1

1

 ( x  3)  x 3k 1 ]  dx

1

1 · lim 24 n  

3

3

1



(1  x ) 2  ( x  3)  x 2  (1  x 3n ) 1 x  x2

0

(1  x ) 2  ( x  3)  x 3n  2

(1  x ) 2  x 2  ( x  3) 1 1  dx · · lim 24 24 n   1 x  x2



=

 x  x  5 x  3 x 1 ·  24  1 x  x2 0



=

3 2  x  1 15 1 1     2 ·  x3  6  x  9     dx 2 2 1 x  x 2 x  x 1  24  0

=

1 24

=

1  1 1 3   1  25 3  ·  9  6     ln 3  5  3   = ·    ln 3  5  3   2 4 2 6  24  4 2 6 24 

=

 1  25 ·   3  ln 3  5  3   , 3 48  2

=

 0 1



5

4

3

2

1 x  x2

0 1

 dx

 dx

   dx = 1 ·  x 3  6  x  9  3  x  9   dx  24  1  x  x2   0

1



  2  x 1 1 x2 x4 3   · 9 x  6    ln(1  x  x 2 )  5  3  arctg   2 4 2 3   0  

since: 1

lim

n 



(1  x ) 2  ( x  3)  x 3n  2 1 x  x2

0

 dx =0.

Example E2.4: The following equality holds: n

lim

n 

 (4  k)  (4  k  1)  (4  k  2)  (4  k  4) = 24 ·  4  3  ln 2    .

1

1

21

(E2.4)

k 1

Proof: Equality (E2.4) is obtained from equality (E2), for: q=4, or by direct calculation. Thus, for every kN, 1  3 1 8 6 1     = ·  4  k  (4  k  1)  (4  k  2)  (4  k  4) 24  4  k 4  k  1 4  k  2 4  k  4  1

=

=

1

1 1 · (3  x 4k 1  8  x 4k  6  x 4k 1  x 4k  3 )  dx = · [(3  8  x  6  x 2  x 4 )  x 4k 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  ( x  3)  x 4k 1 ]  dx .

24

 0

So, n

lim

n 

=

=



1 1 = · lim n  ( 4  k )  ( 4  k  1 )  ( 4  k  2 )  ( 4  k  4 ) 24 k 1 1 n

  [(1  x)

1 · lim 24 n  

4 n   1 3 3 1 x  dx = · lim (1  x )  ( x  3)  x  4  n  24 1  x     0

0 k 1 1



 ( x  3)  x 4k 1 ]  dx

k 1 0

n   3 x 4k 1   dx (1  x )  ( x  3)   k 1  0



1 · lim 24 n  

3

1

1 · lim 24 n  

[(1  x )3  ( x  3)  x 4k 1 ]  dx =

n 1





1

 0

(1  x ) 2  ( x  3)  x 3  (1  x 4n ) 1  x  x 2  x3

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=

(1  x ) 2  x 3  ( x  3) 1 1  dx · · lim 2 3 24 24 n   1  x  x  x 0



1



(1  x ) 2  ( x  3)  x 4n  3 1  x  x 2  x3

0 1

 dx

 x6  x5  5  x 4  3  x3  1   dx = 1 ·  x 3  6  x  8  4  2  x  4  1   dx = ·  2 3  x 1 x2 1 24  24  1 x  x  x x2 1   0 0 1





 1 x2 x4 · 8  x  6    4  ln(x  1)  ln(x 2  1)  4  arctgx   0 2 4  

=

1 24

=

1  1 1   1  21  ·  8  6    4  ln 2  2  ln 2   = ·   3  ln 2    , 2 4 4  24  4 24  

since: 1

(1  x ) 2  ( x  3)  x 4n  3



lim

n 

1  x  x 2  x3

0

 dx =0.

Corollary E3: If qN, then: 1

1 1 (1  x )3  ( x  1)  x q 1 lim = ·  dx . n  (q  k )  (q  k  1)  (q  k  3)  (q  k  4) 12 1  xq k 1 n





(E3)

0

Proof: Equality (E3) is obtained from Theorem (E), for: p=1, r=3 and s=4, or by direct calculation. Thus, for every kN, 1  1 1 2 2 1   = ·     q  k  (q  k  1)  (q  k  3)  (q  k  4) 12  q  k q  k  1 q  k  3 q  k  4  1

=

=

1

1 1 · ( x q k 1  2  x q k  2  x q k  2  x q k  3 )  dx = · [(1  2  x  2  x 3  x 4 )  x qk 1]dx 12 12





0 1

0

1 · [(1  x )3  ( x  1)  x qk 1 ]  dx .

12

 0

So, n

lim

n 

=

=



1 1 = · lim n  ( q  k )  ( q  k  1 )  ( q  k  3 )  ( q  k  4 ) 12 k 1 1 n

1 · lim 12 n  



1 · lim 12 n  

n q   3 q 1 1  x ( 1  x )  ( x  1 )  x   dx  q  1  x     0

1 = · lim 12 n  

[(1  x )3  ( x  1)  x qk 1 ]  dx =

0 k 1 1

1 · lim 12 n  

n 1

  [(1  x)

3

 ( x  1)  x qk 1 ]  dx

k 1 0

n   3 x qk 1   dx (1  x )  ( x  1)   k 1  0

1





 1

 0

(1  x )3  ( x  1)  x q 1 1 xq

1  dx - · lim 12 n  

1

 0

(1  x )3  ( x  1)  x ( n 1)q 1 1 xq

 dx

1

=

1 (1  x )3  ( x  1)  x q 1  dx , · 12 1  xq

 0

because, according to (Vălcan, 2016, Proposition), 1

lim

n 



(1  x )3  ( x  1)  x ( n 1)q 1 1  xq

0

 dx =0.

Example E3.1: The following equality holds: n

lim

n 

 k  (k  1)  (k  3)  (k  4) = 144 .

1

5

(E3.1)

k 1

Proof: Equality (E3.1) is obtained from equality (E3), for:

International Journal of Arts Humanities and Social Sciences Studies

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals q=1, or by direct calculation. Thus, for every kN, 1 1 1 2 2 1  = ·     k  (k  1)  (k  3)  (k  4) 12  k k  1 k  3 k  4  1

1

1 1 = · ( x k 1  2  x k  2  x k  2  x k  3 )  dx = · [(1  2  x  2  x 3  x 4 )  x k 1 ]  dx 12 12 =





0 1

0

1 · [(1  x )3  ( x  1)  x k 1 ]  dx .

12

 0

So, n

lim

n 



1 1 = · lim n  k  ( k  1 )  ( k  3 )  ( k  4 ) 12 k 1

1 = · lim 12 n   =

1 · lim 12 n  

1 n

 [(1  x)

3

 ( x  1)  x

k 1

0 k 1 1





1 1 · [(1  x ) 2  ( x  1)]  dx · lim 12 12 n  



1 1 · ( x 3  x 2  x  1)  dx = 12 12 0

=

3

 ( x  1)  x k 1 ]  dx

k 1 0

 1 xn  1 3 (1  x )  ( x  1)    dx = · lim 1  x  12 n    0

0 1

=

 [(1  x)

1 ]  dx = · lim 12 n  

1

=

n 1

n   3 x k 1   dx (1  x )  ( x  1)   k 1  0

1



 1

 [(1  x)

2

 ( x  1)  (1  x n )]  dx

0

1

[(1  x)

2

 ( x  1)  x n ]  dx

0

 x 2 x 3 x 4  1 · x     2 3 4  0 

1  1 1 1 1 5 5 · 1     = · = , 12  2 3 4  12 12 144

since: 1

lim

n 

[(1  x)

2

 ( x  1)  x n ]  dx =0.

0

Example E3.2: The following equality holds: n

lim

n 

 (2  k)  (2  k  1)  (2  k  3)  (2  k  4) = 144 .

1

1

(E3.2)

k 1

Proof: Equality (E3.2) is obtained from equality (E3), for: q=2, or by direct calculation. Thus, for every kN, 1  1 2 2 1  1    = ·  2  k  (2  k  1)  (2  k  3)  (2  k  4) 12  2  k 2  k  1 2  k  3 2  k  4  1

=

=

1

1 1 · ( x 2k 1  2  x 2k  2  x 2k  2  x 2k  3 )  dx = · [(1  2  x  2  x 3  x 4 )  x k 1 ]  dx 12 12





0 1

0

1 · [(1  x )3  ( x  1)  x k 1 ]  dx .

12

 0

So, n



1 1 lim = · lim n  (2  k )  (2  k  1)  (2  k  3)  (2  k  4) 12 n   k 1 =

1 · lim 12 n  

1 n

 0 k 1

[(1  x )3  ( x  1)  x k 1 ]  dx =

1 · lim 12 n  

n 1

  [(1  x)

3

 ( x  1)  x 2k 1 ]  dx

k 1 0

n   3 x 2k 1   dx (1  x )  ( x  1)   k 1  0

1



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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals =

1 · lim 12 n  

 1  x 2 n  1 3  dx = · lim (1  x )  ( x  1)  x  2  n  12 1  x   0

 [(1  x)

0

0

x x x ·  2   2 3 4  2

3

4

0

(1  x ) 2  ( x  1)  x  (1  x 2n )  dx 1 x 1

1



1 1 = · [(1  x ) 2  x ]  dx · lim 12 12 n   1 12





1

=

1

1

2

x

2n 1



1 ]  dx = · ( x 3  2x 2  x )  dx 12 0

1 1  1 2 1  1 1  = ·    = · = 1 ,  0 12  2 3 4  12 12 144 

since: 1

lim

n 

 [(1  x)

2

 x 2n 1 ]  dx =0.

0

Example E3.3: The following equality holds: n

lim

n 

 31

 (3  k)  (3  k  1)  (3  k  3)  (3  k  4) = 24 ·  6  3  ln 3  1

1

k 1

  3  .

3 

(E3.3)

Proof: Equality (E3.3) is obtained from equality (E3), for: q=3, or by direct calculation. Thus, for every kN, 1  1 1 2 2 1     = ·  3  k  (3  k  1)  (3  k  3)  (3  k  4) 12  3  k 3  k  1 3  k  3 3  k  4  1

1

=

=

1 1 · ( x 3k 1  2  x 3k  2  x 3k  2  x 3k  3 )  dx = · [(1  2  x  2  x 3  x 4 )  x 3k 1 ]  dx 12 12





0 1

0

1 · [(1  x )3  ( x  1)  x 3k 1 ]  dx .

12

 0

So, n



1 1 lim = · lim n  (3  k )  (3  k  1)  (3  k  3)  (3  k  4) 12 n   k 1 =

=

  [(1  x)

n   3 x 3k 1   dx (1  x )  ( x  1)   k 1  0

1 · lim 12 n  

3n   1 3 2 1 x  dx = · lim (1  x )  ( x  1)  x  3  n  12 1  x   0

0 k 1 1



(1  x )  ( x  1)  x 1 1  dx - · lim · 2 12 12 n   1 x  x



1 · lim 12 n  

2

2

0 1

1

 0

 ( x  1)  x 3k 1 ]  dx

k 1 0



[(1  x )3  ( x  1)  x 3k 1 ]  dx =

3

1

1 · lim 12 n  

1

=

1 n

n 1





1



(1  x ) 2  ( x  1)  x 2  (1  x 3n )

0

(1  x )  ( x  1)  x 3n  2 2

1 x  x2

=

3 2  x 1 3 1 1     2 ·  x 3  2x 2  3   2   dx 2 x  x 1 2 x  x 1  12  0

=

1 12

  2  x 1 1 x3 x 4 3     ln(x 2  x  1)  3  arctg  · 3 x  2    3 4 2 3   0  

=

1 12

 2 1 3   3  1  31   3   3  ln 3  ·  3     ln 3  = · ,    3 4 2 6  24  6 3  

1 x  x2

 dx

 dx



since: 1

lim

n 

 0

(1  x ) 2  ( x  1)  x 3n  2 1 x  x2

 dx =0.

Example E3.4: The following equality holds:

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lim

n 

 (4  k)  (4  k  1)  (4  k  3)  (4  k  4) = 24 ·  6    .

1

1

19

(E3.4)

k 1

Proof: Equality (E3.4) is obtained from equality (E3), for: q=4, or by direct calculation. Thus, for every kN, 1  1 1 2 2 1  = ·     4  k  (4  k  1)  (4  k  3)  (4  k  4) 12  4  k 4  k  1 4  k  3 4  k  4  1

=

=

1

1 1 · ( x 4k 1  2  x 4k  2  x 4k  2  x 4k  3 )  dx = · [(1  2  x  2  x 3  x 4 )  x 4k 1 ]  dx 12 12





0 1

0

1 · [(1  x )3  ( x  1)  x 4k 1 ]  dx .

12

 0

So, n



1 1 lim = · lim n  (4  k )  (4  k  1)  (4  k  3)  (4  k  4) 12 n   k 1 =

1 · lim 12 n  

1 = · lim 12 n   1

=

1 n



[(1  x )3  ( x  1)  x 4k 1 ]  dx =

0 k 1 1

  [(1  x)

3

1

=

1  1 1  ·  x 3  2x 2  2  2  2   dx = 12  12 x  1  0

=

1  2 1   1  19  · 2    2  = ·    , 3 4 4  24  6 12  



1

 0

 ( x  1)  x 4k 1 ]  dx

k 1 0

n   3 x 4k 1   dx (1  x )  ( x  1)   k 1  0





1



2

3

1

4 n   1 3 3 1 x  dx = · lim (1  x )  ( x  1)  x  4  n  12 1  x   0

(1  x )  ( x  1)  x 1 1  dx - · lim · 2 3 12 12 n   1 x  x  x 0



1 · lim 12 n  

n 1



(1  x ) 2  ( x  1)  x 3  (1  x 4n )

0

(1  x )  ( x  1)  x 4n  3 2

1  x  x 2  x3

1  x  x 2  x3

 dx

 dx

 1 x3 x 4 · 2 x  2   2  arctgx   0 3 4  

since: 1

lim

n 



(1  x ) 2  ( x  1)  x 4n  3

0

1  x  x 2  x3

 dx =0.

Corollary E4: If qN, then: 1

1 1 (1  x )3  (3  x  1)  x q 1 lim  dx . = · n  (q  k )  (q  k  2)  (q  k  3)  (q  k  4) 24 1 xq k 1 n





(E4)

0

Proof: Equality (E4) is obtained from Theorem (E), for: p=2, r=3 and s=4, or by direct calculation. Thus, for every kN, 1  1 1 6 8 3   = ·     q  k  (q  k  2)  (q  k  3)  (q  k  4) 24  q  k q  k  1 q  k  3 q  k  4  1

=

=

1

1 1 · ( x qk 1  6  x qk  8  x q k  2  3  x q k  3 )  dx = · [(1  6  x  8  x 3  3  x 4 )  x qk 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  (3  x  1)  x qk 1 ]  dx .

24

 0

So,

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n

lim

n 

1 n

1 · lim 24 n  

 [(1  x)

=

1 · lim 24 n  

n q   3 q 1 1  x (1  x )  (3  x  1)  x    dx 1  x q   0

=

1 · lim 24 n  

=

3

 (3  x  1)  x qk 1 ]  dx =

0 k 1 1

1 · lim 24 n  

3

 (3  x  1)  x qk 1 ]  dx

k 1 0

n   3 x qk 1   dx (1  x )  (3  x  1)   k 1  0

1





 1



(1  x )3  (3  x  1)  x q 1 1 xq

0

1

=

  [(1  x)



1 1 = · lim n  ( q  k )  ( q  k  2 )  ( q  k  3 )  ( q  k  4 ) 24 k 1

1 (1  x )  (3  x  1)  x · 24 1 xq 0 3



q 1

 dx -

1 · lim 24 n  

1



(1  x )3  (3  x  1)  x ( n 1)q 1 1 xq

0

 dx

 dx ,

because, according to (Vălcan, 2016, Proposition), 1

lim

n 



(1  x )3  (3  x  1)  x ( n 1)q 1 1  xq

0

 dx =0.

Example E4.1: The following equality holds: n

lim

n 

 k  (k  2)  (k  3)  (k  4) = 288 .

1

7

(E4.1)

k 1

Proof: Equality (E4.1) is obtained from equality (E4), for: q=1, or by direct calculation. Thus, for every kN, 1 1 1 6 8 3    = ·   k  (k  2)  (k  3)  (k  4) 24  k k  2 k  3 k  4  1

=

=

1

1 1 · ( x k 1  6  x k 1  8  x k  2  3  x k  3 )  dx = · [(1  6  x 2  8  x 3  3  x 4 )  x k 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  (3  x  1)  x k 1 ]  dx .

24

 0

So, n

lim

n 



1 1 = · lim n  k  ( k  2 )  ( k  3 )  ( k  4 ) 24 k 1

1 = · lim 24 n   =

1 · lim 24 n  

1 n



[(1  x )  (3  x  1)  x 3

k 1

0 k 1 1

=

k 1 0



1 1 · (3  x 3  5  x 2  x  1)  dx = 24 24 0

 (3  x  1)  x k 1 ]  dx

n   3 x k 1   dx (1  x )  (3  x  1)   k 1  0

1





1







3

 1 xn  1 3 ( 1  x )  ( 3  x  1 )  · lim [(1  x )2  (3  x  1)  (1  x n )]  dx    dx = n  1  x 24   0 0

1

=

  [(1  x)

1 ]  dx = · lim 24 n  

1 1 = · [(1  x ) 2  (3  x  1)]  dx · lim 24 24 n   0 1

n 1

1

 [(1  x)

2

 (3  x  1)  x n ]  dx

0

 x2 x3 x4 1  5  3  ·x   2 3 4  0 

1  1 5 3 1 7 7 · 1     = · = , 24  2 3 4  24 12 288

since:

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lim

n 

[(1  x)

2

 (3  x  1)  x n ]  dx =0.

0

Example E4.2: The following equality holds: n

lim

n 

 (2  k)  (2  k  2)  (2  k  3)  (2  k  4) = 24 ·  8  ln 2  12  .

1

1

65

(E4.2)

k 1

Proof: Equality (E4.2) is obtained from equality (E4), for: q=2, or by direct calculation. Thus, for every kN, 1  1 1 6 8 3  = ·     2  k  (2  k  2)  (2  k  3)  (2  k  4) 24  2  k 2  k  2 2  k  3 2  k  4  1

1

=

=

1 1 · ( x 2k 1  6  x 2k 1  8  x 2k  2  3  x 2k  3 )  dx = · [(1  6  x 2  8  x 3  3  x 4 )  x 2 k 1 ]  dx 24 24





0 1

0

1 · [(1  x )3  (3  x  1)  x 2k 1 ]  dx .

24

 0

So, n

lim

n 

=



1 1 = · lim (2  k )  (2  k  2)  (2  k  3)  (2  k  4) 24 n   k 1

1 · lim 24 n  

1 = · lim 24 n   1

=

1 n



[(1  x )3  (3  x  1)  x 2k 1 ]  dx =

0 k 1 1

  [(1  x)

0 1

1

 0





1

 0

(1  x ) 2  (3  x  1)  x  (1  x n )  dx 1 x

(1  x )  (3  x  1)  x n 1  dx 1 x 2

=

 3  x 4  5  x3  x 2  x  1   dx = 1 ·   x  1 24  24  0

=

1  9 8 3 65   1  ·   8     8  ln 2  = ·  8  ln 2   , 2 3 4 12  24   24 



 (3  x  1)  x 2k 1 ]  dx

n   3 x 2k 1   dx (1  x )  (3  x  1)   k 1  0



2

3

k 1 0 1

 1  x 2 n  1 3  dx = · lim (1  x )  (3  x  1)  x  2  24 n   1  x   0

(1  x )  x  (3  x  1) 1 1  dx · · lim 1 x 24 24 n  



1 · lim 24 n  

n 1

 1 x2 x3 x4 8  3  8  ln(x  1)  ·   8  x  9  2 3 4  0

since: 1

lim

n 

 0

(1  x ) 2  (3  x  1)  x n 1  dx =0 1 x

Example E4.3: The following equality holds: n

lim

n 

 49

 (3  k)  (3  k  2)  (3  k  3)  (3  k  4) = 48 ·  6  9  ln 3  1

1

k 1

  3  .

3 

(E4.3)

Proof: Equality (E4.3) is obtained from equality (E4), for: q=3, or by direct calculation. Thus, for every kN, 1  1 6 8 3  1    = ·  3  k  (3  k  2)  (3  k  3)  (3  k  4) 24  3  k 3  k  2 3  k  3 3  k  4  1

=

1

1 1 · ( x 3k 1  6  x 3k 1  8  x 3k  2  3  x 3k  3 )  dx = · [(1  6  x 2  8  x 3  3  x 4 )  x 3k 1 ]  dx 24 24





0

0

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=

1 · [(1  x )3  (3  x  1)  x 3k 1 ]  dx .

24

 0

So, n

lim

n 



1 1 = · lim n  ( 3  k )  ( 3  k  2 )  ( 3  k  3 )  ( 3  k  4 ) 24 k 1 1 n

1 · lim 24 n  

 [(1  x)

=

1 · lim 24 n  

3n   3 2 1 x (1  x )  (3  x  1)  x    dx 1  x 3   0

=

1 · lim 24 n  

=

3

 (3  x  1)  x 3k 1 ]  dx =

0 k 1 1

  [(1  x)

3

 (3  x  1)  x 3k 1 ]  dx

k 1 0 1

n   3 x 3k 1   dx (1  x )  (3  x  1)   k 1  0





 1



(1  x ) 2  (3  x  1)  x 2  (1  x 3n ) 1 x  x2

0

1

(1  x ) 2  x 2  (3  x  1) 1 1  dx = · · lim 2 24 24 n   1 x  x

 0 1

=

1 · lim 24 n  

n 1

 dx 1



(1  x ) 2  (3  x  1)  x 3n  2

 dx

1 x  x2

0

 3  x5  5  x 4  x3  x 2  1   dx ·  2  24  x  x  1  0

 1

9 2  x 1 3 1 1     2 = ·  3  x3  8  x 2  6  x  3   2   dx 2 x  x 1 2 x  x 1  24  0



=

1 24

  2  x 1 1 x2 x3 x4 9   · 3 x  6   8  3   ln(x 2  x  1)  3  arctg   2 3 4 2 3   0  

=

1 24

 6 8 3 9   3  1 ·  3      ln 3  =  2 3 4 2 6  48 

 49   3  ·   9  ln 3  ,  6 3  

since: 1

lim

n 



(1  x ) 2  (3  x  1)  x 3n  2

0

1 x  x2

 dx =0.

Example E4.4: The following equality holds: n

lim

n 

 (4  k)  (4  k  2)  (4  k  3)  (4  k  4) = 24 ·  12  3  ln 2    .

1

1

13

(E4.4)

k 1

Proof: Equality (E4.4) is obtained from equality (E4), for: q=4, or by direct calculation. Thus, for every kN, 1  1 1 6 8 3     = ·  4  k  (4  k  2)  (4  k  3)  (4  k  4) 24  4  k 4  k  2 4  k  3 4  k  4  1

1

1 1 = · ( x 4k 1  6  x 4k 1  8  x 4k  2  3  x 4k  3 )  dx = · [(1  6  x 2  8  x 3  3  x 4 )  x 4k 1 ]  dx 24 24 =





0 1

0

1 · [(1  x )3  (3  x  1)  x 4k 1 ]  dx .

24

 0

So, n

lim

n 



1 1 = · lim n  ( 4  k )  ( 4  k  2 )  ( 4  k  3 )  ( 4  k  4 ) 24 k 1

n 1

  [(1  x)

3

 (3  x  1)  x 4k 1 ]  dx

k 1 0

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1 · lim 24 n  



=

1 · lim 24 n  

4 n   3 3 1 x (1  x )  (3  x  1)  x    dx 1  x 4   0

=

1 · lim 24 n  

=

[(1  x )3  (3  x  1)  x 4k 1 ]  dx =

0 k 1 1

1 · lim 24 n  

n   3 x 4k 1   dx (1  x )  (3  x  1)   k 1  0

1





 1



(1  x ) 2  (3  x  1)  x 3  (1  x 4n ) 1  x  x 2  x3

0

1

 dx 1

(1  x ) 2  (3  x  1)  x 4n  3

=

(1  x ) 2  x 3  (3  x  1) 1 1  dx · · lim 2 3 24 24 n   1  x  x  x 0

=

 3 x  5 x  x  x 1 ·  24  x3  x 2  x  1 0

=

1 1 x  1   4 2  2 2 ·  3  x3  8  x 2  6  x  4    dx x 1 24  x 1 x 1 0

=

1 24

=

1 6 8 3  1  13  ·     4  ln 2    ln 2  = ·   3  ln 2    , 12 24  2 3 4 24   

 1



6

5

4

3

 0

1  x  x 2  x3

 dx

   dx  

1



 x2 1 x3 x4 1 · 6  8  3  4  ln(x  1)  4  arctgx  2   ln(x 2  1)   2 0 3 4 2  

since: 1

lim

n 



(1  x ) 2  (3  x  1)  x 4n  3

0

1  x  x 2  x3

 dx =0.

We now move on to another category of sequences.

F) Theorem F: If p, q, r, s, tN and prst, then: 1

n

lim

n 

1  F( x )   dx , =· lim    q n  (q  k  p)  (q  k  r)  (q  k  s)  (q  k  t) 1  x  k 0 0

(F)

where:

F(x)=(1-2xr-p+3xs-p-4xt-p)xp-1, 1 = , (r  p)  (s  p)  ( t  p)  (s  r )  ( t  r )  ( t  s) 1=(s-r)(t-r)(t-s), 2=(s-p)(t-p)(t-s), 3=(r-p)(t-p)(t-r), 4=(r-p)(s-p)(s-r).

Proof: For every kN,  1 2 3 4  1  =·     (q  k  p)  (q  k  r )  (q  k  s)  (q  k  t ) q  k  p q  k  r q  k  s q  k  s   1



=· (1  x q k  p 1   2  x q k  r 1  3  x qk  s 1   4  x qk  t 1 )  dx 0 1



=· (1   2  x r  p  3  x s  p   4  x t  p )  x q k  p 1  dx , 0

where: =

1 , (r  p)  (s  p)  ( t  p)  (s  r )  ( t  r )  ( t  s)

1=(s-r)(t-r)(t-s),

2=(s-p)(t-p)(t-s),

3=(r-p)(t-p)(t-r),

4=(r-p)(s-p)(s-r).

So,

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 (q  k  p)  (q  k  r)  (q  k  s)  (q  k  t) =

lim

n 

1

k 0

n 1

=· lim

n 

 [(  

2 x

r p

 3  x s  p   4  x t  p )  x qk  p 1 ]  dx

 [(  

2 x

r p

 3  x s  p   4  x t  p )  x qk  p 1 ]  dx

1

k 0 0 1 n

=· lim

1

n 

0 k 0 1

n   r p sp t p p 1 x qk   dx (1   2  x  3  x   4  x )  x   k 0  0





=· lim

n 

( n 1)q   r p sp t p p 1 1  x (     x    x    x )  x   1   dx 2 3 4 q 1  x    0

1



=· lim

n 

 (1   2  x r  p  3  x s  p   4  x t  p )  x p 1     dx q 1  x    0

1



=· lim

n 

 (1   2  x r  p  3  x s  p   4  x t  p )  x ( n 1)q  p 1     dx 1  xq   0

1

-· lim

n 



 (   2  x r  p  3  x s  p   4  x t  p )  x p 1  =·  1   dx , 1  xq   0 because, according to (Vălcan, 2016, Proposition), 1



 (1   2  x r  p  3  x s  p   4  x t  p )  x ( n 1)q  p 1     dx =0.

1  xq   0

1

lim

n 



Here, too, the following remark is necessary: Remark F0: A simple calculation shows us that: 1-2+3-4=0, which tells us that none of the above integrals is not improper.

Corollary F1: If qN, then: 1

n

lim

n 

1 1 (1  x )3 =   dx .  (q  k  1)  (q  k  2)  (q  k  3)  (q  k  4) 6 0 1  x q k 0

Proof: Equality (F1) is obtained from Theorem (F), for: p=1, r=2, or by direct calculation. Thus, for every kN,

1 1 = (q  k  1)  (q  k  2)  (q  k  3)  (q  k  4) 6

s=3

(F1)

and

t=4

 1 3 3 1   ·      q  k 1 q  k  2 q  k  3 q  k  4 

1

=



1 · ( x q k  3  x q k 1  3  x q k  2  x q k  3 )  dx 6 0 1

=



1 · [(1  3  x  3  x 2  x 3 )  x qk ]  dx .

6 0

So, n

lim

n 

 (q  k 1)  (q  k  2)  (q  k  3)  (q  k  4) = 1

k 0

1 = · lim 6 n 

n 1

 [(1  3  x  3  x

2

 x 3 )  x qk ]  dx

k 0 0

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1 · lim 6 n 

=

1 · lim 6 n 

=

1 · lim 6 n 

=

1 · lim 6 n 

1

n

  [(1  3  x  3  x

2

 x 3 )  x qk ]  dx

0 k 0

n   2 3 x qk   dx (1  3  x  3  x  x )   k 0  0

1





( n 1)q   2 3 1 x ( 1  3  x  3  x  x )     dx q 1  x    0

1

 1



1  3  x  3  x 2  x3 1 x

0

q

 dx -

1 · lim 6 n 

 (1  3  x  3  x 2  x 3 )  x ( n 1)q     dx q 1  x    0

1



1

1 (1  x )3 = ·  dx , 6 1 xq 0



because, according to (Vălcan, 2016, Proposition),

 (1  3  x  3  x 2  x 3 )  x ( n 1)q     dx =0. n  1 xq   0 Example F1.1: The following equality holds: n 1 1 lim = . n  18 ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 ) k 0 1

lim





(F1.1)

Proof: Equality (F1.1) is obtained from equality (F1), for: q=1, or by direct calculation. Thus, for every kN, 1 3 3 1  1  1    = ·  (k  1)  (k  2)  (k  3)  (k  4) 6  k  1 k  2 k  3 k  4  1

=

1





0

0

1 1 · ( x k  3  x k 1  3  x k  2  x k  3 )  dx = · [(1  3  x  3  x 2  x 3 )  x k ]  dx .

6 6

So, n

lim

n 

1 1 = · lim  ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 ) 6 n  k 0

1 = · lim 6 n  =

1 · lim 6 n 

=

1 · lim 6 n 

1 n



[(1  3  x  3  x 2  x 3 )  x k ]  dx =

0 k 0 1

 [(1  3  x  3  x

2

 x 3 )  x k ]  dx

k 0 0 1

n   2 3 ( 1  3  x  3  x  x )  x k   dx   k 0  0





n 1   2 3 1 x (1  3  x  3  x  x )    dx 1  x   0

 1

 0

1  3  x  3  x 2  x3 1  dx - · lim 1 x 6 n 

1

=

1 · lim 6 n 

n 1

1

 (1  3  x  3  x 2  x 3 )  x n 1     dx 1 x   0

1



1 1 1 (1  x )3 ·  dx = · (1  2  x  x 2 )  dx = 1 x 6 6 6





0

0

 x 2 x 3  1  · x  2  2 3  0 

1 1 1 1  = · 1  1   = , 3  0 18 6  since,

 (1  3  x  3  x 2  x 3 )  x n 1     dx =0.

1 x    0

1

lim

n 



Example F1.2: The following equality holds:

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lim

n 

 (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) = 6 ·  4  ln 2  2  .

1

5

1

(F1.2)

k 0

Proof: Equality (F1.2) is obtained from equality (F1), for: q=2, or by direct calculation. Thus, for every kN, 1 3 3 1  1  1 = ·     (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) 6  2  k  1 2  k  2 2  k  3 2  k  4  1

1





0 1

0

1 1 = · ( x 2k  3  x 2k 1  3  x 2k  2  x 2k  3 )  dx = · [(1  3  x  3  x 2  x 3 )  x 2k ]  dx 6 6 =



1 · [(1  x )3  x 2k ]  dx .

6 0

So, n

lim

n 

=



1 1 = · lim (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) 6 n   k 0

1 · lim 6 n 

1 = · lim 6 n  1

1 n

 [(1  x)

3

 x 2k ]  dx =

0 k 0 1

1 · lim 6 n 

2( n 1)

1

 0

=

4  1 1  · x 3   dx = x  1 6 6   0

 x2 1 ·  3  x  4  ln(x  1)   2 0  

=

5 1 1  1  ·   3  4  ln 2  = ·  4  ln 2   , 2 6 2  6 

0 1



0

(1  x )  x 1 x

k 0 0

(1  x ) 2  (1  x 2( n 1) )  dx 1 x





 x 2k ]  dx



(1  x ) 1 1  dx - · lim · 1 x 6 6 n 

=

2

3

n





1

  [(1  x)

  3 x 2k   dx (1  x )   k 0  0

1

2( n 1)   1 3 1 x (1  x )    dx = · lim 2 6 n  1 x   0 2

n 1

 dx

since:

1 · lim 6 n 

1

 0

(1  x ) 2  x 2( n 1)  dx =0.

1 x

Example F1.3: The following equality holds: n

lim

n 

1 

 (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4) = 6 · 1  2  ln 3  1

k 0

3

3    .

6 

(F1.3)

Proof: Equality (F1.3) is obtained from equality (F1), for: q=3, or by direct calculation. Thus, for every kN, 3 3 1  1  1 1    = ·  (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4) 6  3  k  1 3  k  2 3  k  3 3  k  4  1

=

=

1





0 1

0

1 1 · ( x 3k  3  x 3k 1  3  x 3k  2  x 3k  3 )  dx = · [(1  3  x  3  x 2  x 3 )  x 3k ]  dx 6 6



1 · [(1  x )3  x 3k ]  dx .

6 0

So,

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lim

n 

=

=



1 1 = · lim ( 3  k  1 )  ( 3  k  2 )  ( 3  k  3 )  ( 3  k  4 ) 6 n  k 0 1 n

1 · lim 6 n 

3( n 1)   1 3 1 x (1  x )    dx = · lim 3 6 n  1  x   0

0 k 0 1



1



1

=

1 · lim 6 n 

(1  x ) 2 1 1  dx - · lim · 2 6 6 n  1  x  x 0



1





 x 3k ]  dx

k 0 0

1 x  x2

(1  x ) 2  (1  x 3( n 1) ) 1 x  x2

0

(1  x ) 2  x 3( n 1)

0

3



 [(1  x)

 x 3k ]  dx =

  [(1  x)

n   3 3k (1  x )  x   dx  k 0  0

1

1 · lim 6 n 

3

n 1

 dx

 dx

1

=

1 1  3 2  x 1 3    2 · 1   2   dx 6  2 x  x 1 2 x  x 1  0

=

1 6



 3 3    · 1   ln 3  ,  2 6  

  2  x 1 1 1 3   = ·  x   ln(x 2  x  1)  3  arctg   2 3   0 6 

since: 1

lim

n 



(1  x ) 2  x 3( n 1)

0

1 x  x2

 dx =0.

Example F1.4: The following equality holds: n

lim

n 



 (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) = 12 ·  2  ln 2  2  .

1

1

3

(F1.4)

k 0

Proof: Equality (F1.4) is obtained from equality (F1), for: q=4, or by direct calculation. Thus, for every kN, 1 3 3 1  1  1    = ·  (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) 6  4  k  1 4  k  2 4  k  3 4  k  4  1

=

=

1





0 1

0

1 1 · ( x 4k  3  x 4k 1  3  x 4k  2  x 4k  3 )  dx = · [(1  3  x  3  x 2  x 3 )  x 4k ]  dx 6 6



1 · [(1  x )3  x 4k ]  dx .

6 0

So, n



1 1 lim = · lim n  (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) 6 n   k 0 =

=

1



1 · lim 6 n 

 [(1  x)

1 · lim 6 n 

4( n 1)   1 3 1 x (1  x )    dx = · lim 4 6 n  1 x   0

1

=

1 n

3

 x 4k ]  dx =

0 k 0 1

(1  x ) 1 1  dx - · lim · 2 3 6 6 n  1 x  x  x 0



 (1  x)   x

1 · lim 6 n 

1

k 0

(1  x )  x 2

 1 x  x 0

3

0



2

n

2

1



n 1

  [(1  x)

0

 x3

 x 4k ]  dx

k 0 0

4 k

   dx 

(1  x ) 2  (1  x 4( n 1) )

4( n 1)

3

1  x  x 2  x3

 dx

 dx

1

=

1 1 2 x 1  1    2  2   dx · 2 x  1 2 6 x  1 x 1   0



International Journal of Arts Humanities and Social Sciences Studies

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals =

1 1  1 ·  2  ln(x  1)   ln(x 2  1)  arctgx  2 6  0

=

1  1 3  1  ·  2  ln 2   ln 2   = ·   ln 2   , 2 4  12  2 2 6 

since: 1

lim

n 

(1  x ) 2  x 4( n 1)

 1 x  x 0

2

 x3

 dx =0.

Another category of sequences follows.

G) Theorem G: If p, q, r, s, tN and prst, then: n



1 lim =· lim n  n  (q  k )  (q  k  p)  (q  k  r )  (q  k  s)  (q  k  t ) k 1

1

 G(x )   dx , q  0

 1  x

(G)

where:

G(x)=(1-2xp+3xr-4xs+5xt)xq-1, 1 = , p  r  s  t  (r  p)  (s  p)  ( t  p)  (s  r )  ( t  r )  ( t  s) 1=(r-p)(s-p)(t-p)(s-r)(t-r)(t-s), 2=rst(s-r)(t-r)(t-s), 3=pst(s-p)(t-p)(t-s), 4=prt(r-p)(t-p)(t-s), 5=prs(r-p)(s-p)(s-r).

Proof: For every kN,   1 2 3 4 5   =·  1     q  k  (q  k  p)  (q  k  r )  (q  k  s)  (q  k  t )  qk qk  p qk  r qk s qk  t  1



=· (1  x qk 1   2  x q k  p 1  3  x qk  r 1   4  x qk  s 1  5  x q k  t 1 )  dx 0 1



=· [( 1   2  x p  3  x r   4  x s  5  x t )  x qk 1 ]  dx , 0

where:

1 , p  r  s  t  (r  p)  (s  p)  ( t  p)  (s  r )  ( t  r )  ( t  s) 1=(r-p)(s-p)(t-p)(s-r)(t-r)(t-s), 2=rst(s-r)(t-r)(t-s), 3=pst(s-p)(t-p)(t-s), 4=prt(r-p)(t-p)(t-r), 5=prs(r-p)(s-p)(s-r). =

So, n

lim

n 

 (q  k)  (q  k  p)  (q  k  r)  (q  k  s)  (q  k  t) = 1

k 1

n 1

=· lim

n 

 [(   1

2 x

p

 3  x r   4  x s  5  x t )  x qk 1 ]  dx

2 x

p

 3  x r   4  x s  5  x t )  x qk 1 ]  dx

k 1 0

1 n

=· lim

n 

=· lim

n 

 [(   1

0 k 1 1

n   p r s t (     x    x    x    x )  x qk 1   dx  1 2 3 4 5  k 1  0





n q   p r s t q 1 1  x (1   2  x  3  x   4  x  5  x )  x    dx 1  x q   0

1

=· lim

n 



International Journal of Arts Humanities and Social Sciences Studies

V6●

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92

Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals  (1   2  x p  3  x r   4  x s  5  x t )  x q 1     dx 1 xq   0

1



=· lim

n 

 (1   2  x p  3  x r   4  x s  5  x t )  x ( n 1)q 1     dx q 1  x    0

1

-· lim

n 



 (   2  x p  3  x r   4  x s  5  x t )  x q 1  =·  1   dx , 1 xq   0 1



because, according to (Vălcan, 2016, Proposition),

 (1   2  x p  3  x r   4  x s  5  x t )  x ( n 1)q 1     dx =0. n  1  xq   0 The following remark is required here: Remark G0: A simple calculation shows us that: 1-2+3-4+5=0, which tells us that none of the above integrals is not improper.

Corollary G1: If qN, then: 1

lim



1  x q 1  (1  x ) 4  1 1 =     dx . (G1) n  (q  k )  (q  k  1)  (q  k  2)  (q  k  3)  (q  k  4) 24  1  x q  k 1 0 Proof: Equality (G1) is obtained from Theorem (G), for: p=1, r=2, s=3 and t=5 or by direct calculation. Thus, for every kN, 1 4 6 4 1  1  1  =       q  k  (q  k  1)  (q  k  2)  (q  k  3) 24  q  k q  k  1 q  k  2 q  k  3 q  k  4 

n

lim





1

=



1 · ( x qk 1  4  x qk  6  x qk 1  4  x qk  2  x qk  3 )  dx 24 0 1

=

1





0

0

1 1 · [(1  4  x  6  x 2  4  x 3  x 4 )  x qk 1 ]  dx = · [(1  x ) 4  x q k 1 ]  dx .

24 24

So, n

lim

n 



1 1 = · lim n  24 ( q  k )  ( q  k  1 )  ( q  k  2 )  ( q  k  3 )  ( q  k  4 ) k 1 1 n

1 · lim 24 n  



=

1 · lim 24 n  

n q   4 q 1 1  x (1  x )  x    dx 1  x q   0

=

1 · lim 24 n  

=

 x q 1  (1  x ) 4  1 ·    dx , 24  1  x q  0

=

[(1  x ) 4  x qk 1 ]  dx =

0 k 1 1

1 · lim 24 n  

n 1

  [(1  x)

4

 x qk 1 ]  dx

k 1 0

n   4 x qk 1   dx (1  x )   k 1  0

1







 x q 1  (1  x ) 4  1    dx - · lim q n  24 1  x   0

1



 x ( n 1)q 1  (1  x ) 4    dx q 1  x  0

1

 

1



because, according to (Vălcan, 2016, Proposition),  x ( n 1)q 1  (1  x ) 4     dx =0. q 1  x    0

1

lim

n 



Example G1.1: The following equality holds: n

lim

n 

 k  (k  1)  (k  2)  (k  3)  (k  4) = 96 .

1

1

(G1.1)

k 1

International Journal of Arts Humanities and Social Sciences Studies

V6●

I 12 ●

93

Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals Proof: Equality (G1.1) is obtained from equality (G1), for: q=1, or by direct calculation. Thus, for every kN, 1 4 6 4 1  1 1 =       k  (k  1)  (k  2)  (k  3)  (k  4) 24  k k  1 k  2 k  3 k  4  1



1 =  ( x k 1  4  x k  6  x k 1  4  x k  2  x k  3 )  dx 24 0 1

=

1





0

0

1 1  [(1  4  x  6  x 2  4  x 3  x 4 )  x k 1 ]  dx =  [(1  x ) 4  x k 1 ]  dx .

24 24

So, n 1

n





1 1 [(1  x ) 4  x k 1 ]  dx =  24 k  ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 ) k 1 k 1 0

1 n



1 =  24 =

[(1  x ) 4  x k 1 ]  dx =

0 k 1 1

1  24

1 n    1 xn  1 4 k 1 ( 1  x )  x  dx =  (1  x ) 4      dx 24 1  x      k  1  0 0

1









1  [(1  x )3  (1  x n )]  dx .

24 0

It follows that: n

lim

n 



1 1 =  lim n  24 k  ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 ) k 1 1

=



1 1  (1  x )3  dx -  lim 24 24 n   0 1

=

 [(1  x)

3

 (1  x n )]  dx

0

1

 [(1  x)

3

 x n ]  dx

0



1 1  (1  3  x  3  x 2  x 3 )  dx = 24 24 0

=

1

 x2 x3 x 4  1   x  3  3    2 3 4  0 

1  3 3 1 1 1 1  1     =  = , 24  2 3 4  24 4 96

since: 1

lim

n 

 [(1  x)

3

 x n ]  dx =0.

0

Example G1.2: The following equality holds: n

lim

n 

 (2  k)  (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) = 24   12  8  ln 2  .

1

1

67

(G1.2)

k 1

Proof: Equality (G1.2) is obtained from equality (G1), for: q=2, or by direct calculation. Thus, for every kN, 1 (2  k )  (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) =

4 6 4 1  1  1       24  2  k 2  k  1 2  k  2 2  k  3 2  k  4  1



1 =  ( x 2k 1  4  x 2k  6  x 2k 1  4  x 2k  2  x 2k  3 )  dx 24 0 1

=



1



1 1  [(1  4  x  6  x 2  4  x 3  x 4 )  x 2k 1 ]  dx =  [(1  x ) 4  x 2k 1 ]  dx .

24 24 0

0

International Journal of Arts Humanities and Social Sciences Studies

V6●

I 12 ●

94

Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals So, n 1

n





1 1 [(1  x ) 4  x 2k 1 ]  dx =  24 ( 2  k )  ( 2  k  1 )  ( 2  k  2 )  ( 2  k  3 )  ( 2  k  4 ) k 1 k 1 0

=

1 n

 [(1  x)

1  24

  1 x 2k 1   dx  (1  x ) 4  24  k 1  0 1

4

 x 2k 1 ]  dx =

0 k 1 1

n





  1  x 2 n  1  x 2 n  1 4 3  dx ( 1  x )  x  =  (1  x )  x      dx .

24  1  x  1  x 2   0 0 1



1 =  24



It follows that: n

lim

n 

 (2  k)  (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4) 1

k 1

1 =  lim 24 n  

 1  x 2 n  3 (1  x )  x    dx 1  x   0

1



=

 (1  x )3  x  1 1     dx -  lim 24  1  x  24 n   0

=

x  3  x 2  3  x3  x 4 1  1 1   dx =   8  7  x  4  x 2  x 3  8     dx 24 24  1 x 1 x  0 0

=

1 x2 x3 x 4 1   8 x  7   4   8  ln(x  1)  0 24  2 3 4 

=

1  7 4 1  1  67    8     8  ln 2  =    8  ln 2  , 24  2 3 4  24  12 

1



2n 1   3 x (1  x )    dx 1  x   0

1



1

1





since: 2n 1   3 x (1  x )    dx =0.

1  x   0

1

lim

n 



Example G1.3: The following equality holds: n

lim

n 

 (3  k)  (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4) = 48   6  3  ln 3    1

13

1

k 1

 3 . 

(G1.3)

Proof: Equality (G1.3) is obtained from equality (G1), for: q=3, or by direct calculation. Thus, for every kN, 1 4 6 4 1  1  1     =   (3  k )  (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4) 24  3  k 3  k  1 3  k  2 3  k  3 3  k  4  1

=



1  ( x 3k 1  4  x 3k  6  x 3k 1  4  x 3k  2  x 3k  3 )  dx 24 0 1

1





0

0

1 1 =  [(1  4  x  6  x 2  4  x 3  x 4 )  x 3k 1 ]  dx =  [(1  x ) 4  x 3k 1 ]  dx .

24 24 So, n 1

n





1 1 [(1  x ) 4  x 3k 1 ]  dx =  24 ( 3  k )  ( 3  k  1 )  ( 3  k  2 )  ( 3  k  3 )  ( 3  k  4 ) k 1 k 1 0

=

1  24

 [(1  x) 0 k 1

  1 x 3k 1   dx  (1  x ) 4  24  k 1  0 1

1 n

4

 x 3k 1 ]  dx =



n



International Journal of Arts Humanities and Social Sciences Studies

V6●

I 12 ●

95

Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals 3n    1  x 3n  1 1 3 2 1 x  dx ( 1  x )  x   dx .  (1  x ) 4  x 2  =    3 2 24  24  1  x 1  x  x     0 0 1

1

=





It follows that: n

lim

n 



1 1 =  lim n  24 ( 3  k )  ( 3  k  1 )  ( 3  k  2 )  ( 3  k  3 )  ( 3  k  4 ) k 1  (1  x )3  x 2  1  dx -  lim  2  24 n    1  x  x  0

1



1 =  24

3n   3 2 1 x  dx (1  x )  x  2 1  x  x     0

1



 x 3n  2  3 (1  x )    dx 1  x  x 2   0

1



1

=

x 2  3  x3  3  x 4  x5 1  dx  24 1 x  x2

 0 1

=

3 2  x 1 9 1 1     2   3  6  x  4  x 2  x3   2   dx 24  2 x  x 1 2 x  x 1  0

=

 2  x 11 x2 x3 x 4 3 1     3 x  6   4    ln(x 2  x  1)  3  3  arctg  24  2 3 4 2 3   0 

=

 1 1  6 4 1 3   3      ln 3  3  3   24  2 3 4 2 60

=

  3  1  13 1  13 3   =    3  ln 3    3  ,   ln 3    24  12 2 2  48  6 



since:

 x 3n  2  3 (1  x )    dx =0. n  1  x  x 2   0 Example G1.4: The following equality holds: 1

lim



n

lim

n 

 (4  k)  (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) = 24   12  3  ln 2  .

1

25

1

(G1.4)

k 1

Proof: Equality (G1.4) is obtained from equality (G1), for: q=4, or by direct calculation. Thus, for every kN, 1 4 6 4 1  1  1     =   (4  k )  (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) 24  4  k 4  k  1 4  k  2 4  k  3 4  k  4  1

=



1  ( x 4k 1  4  x 4k  6  x 4k 1  4  x 4k  2  x 4k  3 )  dx 24 0 1

=

1





1 1  [(1  4  x  6  x 2  4  x 3  x 4 )  x 4k 1 ]  dx =  [(1  x ) 4  x 4k 1 ]  dx .

24 24 0

0

So, n 1

n





1 1 [(1  x ) 4  x 4k 1 ]  dx =  24 ( 4  k )  ( 4  k  1 )  ( 4  k  2 )  ( 4  k  3 )  ( 4  k  4 ) k 1 k 1 0

=

1  24

1 n

 [(1  x) 0 k 1 1

  1 x 4k 1   dx  (1  x ) 4  24  k 1  0 1

4

 x 4k 1 ]  dx =

n





   1  x 4 n  1  x 4 n 1 1  dx =  (1  x )3  x 3   dx . =  (1  x ) 4  x 3  4  2 3 24  24  1  x 1  x  x  x     0 0



1



It follows that:

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lim

n 

 (4  k)  (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4) k 1

1 =  lim 24 n  

  1  x 4 n 3 3  dx (1  x )  x  2 3 1  x  x  x   0

1



 (1  x )3  x 3  1 1     dx -  lim 24 1  x  x 2  x 3  24 n   0 1

=

1

  x 4 n  3 3  dx (1  x )  2 3 1  x  x  x     0

1





1

1

x3  3  x 4  3  x5  x6 1 2 x  1 1   dx =   4  6  x  4  x 2  x 3  4   2   dx =  2 3 24 24  x 1 x 1  1 x  x  x 0 0





=

1 x2 x3 x 4 1   4 x  6  4   4  ln(x  1)  ln(x 2  1)  0 24  2 3 4 

=

6 4 1 1   1 1  25    4     4  ln 2  ln 2  =    3  ln 2  , 24  2 3 4  0 24  12 

since:

  x 4 n  3 3 ( 1  x )   dx =0.  2 3 1  x  x  x     0

1

lim

n 



By reasoning inductively, we immediately arrive at the next result, which we can obtain and by reasoning deductively, from (Vălcan, 2016, Equalities (10)).

Example G1.p: For every p, qN, the following equality holds:  x q 1  (1  x ) p  1 1 =     dx . (q  k )  (q  k  1)  (q  k  2)    (q  k  p) p!  1  x q  k 1 0 1

n

lim

n 





(G1.p)

particularly, n

lim

n 



1

1 1 =  (1  x ) p 1  dx p ! k  ( k  1 )  ( k  2 )    ( k  p ) k 1

 0

 1 1 1 1 1  =   C0p 1   C1p 1   C2p 1    (1) p  2   Cpp 12  (1) p 1   Cpp 11  . p!  2 3 p 1 p 

(G1.p)

Finally, we have reached the last category of sequences whose limit we want to calculate.

H) Theorem H: If p, q, r, s, t, uN and prstu, then: n



1 lim =· lim n  n  (q  k  p)  (q  k  r )  (q  k  s)  (q  k  t )  (q  k  u ) k 0

1

 H( x )   dx , q  0

 1  x

(H)

where:

H(x)=(1-2xr-p+3xs-p-4xt-p+5xu-p)xp-1, 1 = , (r  p)  (s  p)  ( t  p)  (u  p)  (s  r )  ( t  r )  (u  r )  ( t  s)  (u  s)  ( t  s) 1=(s-r)(t-r)(u-r)(t-s)(u-s)(u-t), 2=(s-p)(t-p)(u-p)(t-s)(u-s)(u-t), 3=(r-p)(t-p)(u-p)(t-r)(u-r)(u-t), 4=(r-p)(s-p)(u-p)(s-s)(u-s)(u-s), 5=(r-p)(s-p)(t-p)(s-r)(t-r)(t-s).

Proof: For every kN, 1 = (q  k  p)  (q  k  r )  (q  k  s)  (q  k  t )  (q  k  u )

 1 2 3 4 5   =·       qk  p qk  r qk s qk  t qk  u  1



=· (1  x q k  p 1   2  x qk  r 1  3  x qk  s 1   4  x q k  t 1  5  x q k  u 1 )  dx 0

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

=· [(1   2  x r  p  3  x s  p   4  x t  p  5  x u  p )  x qk  p 1 ]  dx , 0

where: =

1 , (r  p)  (s  p)  ( t  p)  (u  p)  (s  r )  ( t  r )  (u  r )  ( t  s)  (u  s)  ( t  s)

1=(s-r)(t-r)(u-r)(t-s)(u-s)(u-t), 3=(r-p)(t-p)(u-p)(t-r)(u-r)(u-t), 5=(r-p)(s-p)(t-p)(s-r)(t-r)(t-s).

2=(s-p)(t-p)(u-p)(t-s)(u-s)(u-t), 4=(r-p)(s-p)(u-p)(s-s)(u-s)(u-s),

So, n

lim

n 

 (q  k  p)  (q  k  r)  (q  k  s)  (q  k  t)  (q  k  u) = 1

k 0

n 1

 [(  

=· lim

1

n 

2

 x r p  3  x sp   4  x t p  5  x u p )  x qk p1 ]  dx

k 1 0

1 n

 [(  

=· lim

1

n 

2 x

r p

 3  x s  p   4  x t  p  5  x u  p )  x qk  p 1 ]  dx

0 k 1 1

n   r p sp t p u p p 1 x qk   dx (1   2  x  3  x   4  x  5  x )  x   k 1  0



=· lim

n 



( n 1)q   r p sp t p u p p 1 1  x (1   2  x  3  x   4  x  5  x )  x    dx 1  x q   0

1



=· lim

n 

 (1   2  x r  p  3  x s  p   4  x t  p  5  x u  p )  x p 1     dx 1 xq   0

1



=· lim

n 

 (1   2  x r  p  3  x s  r   4  x t  p  5  x u  p )  x p 1  x ( n 1)q     dx q 1  x    0

1

-· lim

n 



 (   2  x r  p  3  x s  p   4  x t  p  5  x u  p )  x p 1  =·  1   dx , q 1  x    0 1



because, according to (Vălcan, 2016, Proposition),

 (1   2  x r  p  3  x s  r   4  x t  p  5  x u  p )  x p 1  x ( n 1)q     dx =0.

1 xq   0

1

lim

n 



The following remark is required here: Remark H0: A simple calculation shows us that: 1-2+3-4+5=0, which tells us that none of the above integrals is not improper.

Corollary H1: If qN, then: n

lim

n 



Proof: Equality (H1) is obtained from Theorem (H), for: p=1, r=2, s=3, t=4 or by direct calculation. Thus, for every kN, 1 = (q  k  1)  (q  k  2)  (q  k  3)  (q  k  4)  (q  k  5) =

1

1 1 (1  x ) 4 =   dx . q (q  k  1)  (q  k  2)  (q  k  3)  (q  k  4)  (q  k  5) 24 1  x k 1 0



and

(H1)

u=5

4 6 4 1  1  1       24  q  k  1 q  k  2 q  k  3 q  k  4 q  k  5 

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=



1 · ( x qk  4  x q k 1  6  x qk  2  4  x q k  3  x q k  4 )  dx 24 0 1

=

1





1 1 · [(1  4  x  6  x 2  4  x 3  x 4 )  x qk ]  dx = · [(1  x ) 4  x qk ]  dx .

24 24 0

0

So, n

lim

n 

=

=

=



1 1 = · lim n  24 (q  k  1)  (q  k )  (q  k  3)  (q  k  4)  (q  k  5) k 1

1 · lim 24 n  

1 n

 [(1  x) 0 k 1 1



1 · lim 24 n  

 (1  x)

1 · lim 24 n  

(1  x ) 4

0 1

 1 x 0

q

4



4

1 · lim 24 n  

 x qk ]  dx =

1 x

( n 1)q

1 x

 dx -

q

1



n

 (1  x)   x 4

q k

k 1

0

n 1

  [(1  x)

4

 x qk ]  dx

k 1 0

   dx 

   dx 

1 · lim 24 n  

 x ( n 1)q  (1  x ) 4  1 (1  x ) 4 = ·  dx  dx ,   q q 24 1  x 1  x     0 0

1

1





because, according to (Vălcan, 2016, Proposition),  x ( n 1)q  (1  x ) 4     dx =0. q 1  x    0

1



lim

n 

Example H1.1: The following equality holds: n

lim

n 

 (k  1)  (k  2)  (k  3)  (k  4)  (k  5) = 96 .

1

1

(H1.1)

k 0

Proof: Equality (H1.1) is obtained from equality (H1), for: q=1, or by direct calculation. Thus, for every kN, 1 4 6 4 1  1  1     =   (k  1)  (k  2)  (k  3)  (k  4)  (k  5) 24  k  1 k  2 k  3 k  4 k  5  1

=



1  ( x k  4  x k 1  6  x k  2  4  x k  3  x k  4 )  dx 24 0 1

=

1





0

0

1 1  [(1  4  x  6  x 2  4  x 3  x 4 )  x k ]  dx =  [(1  x ) 4  x k ]  dx .

24 24

So, n 1

n





1 1 [(1  x ) 4  x k ]  dx =  24 ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 )  ( k  5 ) k 1 k 0 0

1 n    1  x n 1  1 1 1 4 k 4 k [(1  x )  x ]  dx =  (1  x )  x   dx =  (1  x ) 4  =    dx 24 k 1 24  24  1  x   k 1 0 0 0 1 n

1









1

=



1  [(1  x )3  (1  x n 1 )]  dx .

24 0

It follows that: n

lim

n 



1 1 =  lim n  24 ( k  1 )  ( k  2 )  ( k  3 )  ( k  4 )  ( k  5 ) k 0 1

=

 [(1  x)

3

 (1  x n 1 )]  dx

0

1



 [(1  x)

0

0

1 1  (1  x )3  dx -  lim 24 24 n  

1

3

 x n 1 ]  dx

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=

x2 x3 x 4  1 1 1   (1  3  x  3  x 2  x 3)  dx =   x  3   3   24 24  2 3 4  0 0

=

1  3 3 1 1 1 1  1     =  = , 24  2 3 4  24 4 96



since: 1

lim

n 

 [(1  x)

3

 x n 1 ]  dx =0.

0

Example H1.2: The following equality holds: n

lim

n 

1 

16 

 (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4)  (2  k  5) = 6   8  ln 2  3  .

1

(H1.2)

k 0

Proof: Equality (H1.2) is obtained from equality (H1), for: q=2, or by direct calculation. Thus, for every kN, 1 = (2  k  1)  (2  k  2)  (2  k  3)  (2  k  4)  (2  k  5) =

4 6 4 1  1  1       6  2  k 1 2  k  2 2  k  3 2  k  4 2  k  5 

=

1 1  ( x 2k  4  x 2k 1  6  x 2k  2  4  x 2k  3  x 2k  4 )  dx =  [(1  x )3  x 2k ]  dx .

6 6

1

1





0

0

So, 1

n

lim

n 

1 1 n = lim  [(1  x ) 4  x 2k ]  dx n  6 ( 2  k  1 )  ( 2  k  2 )  ( 2  k  3 )  ( 2  k  4 )  ( 2  k  5 ) k 0 k 0



 0

=

=

1 n

1  lim 6 n 

 [(1  x)

1  lim 6 n 

2 n  2   1 4 1 x  dx =  lim (1  x )  2  6 n  1  x   0

1

4

 x 2k ]  dx =

0 k 0 1

3

 0

1  lim 6 n 

1

 0

(1  x )  x 1 x 3



n

 (1  x)   x

2 n  2

4

k 0

0



1 1 (1  x ) =   dx -  lim 6 6 n  1 x =

1

1

 0

2 k

   dx 

(1  x )3  (1  x 2n  2 )  dx 1 x 1

8  1  2   dx  dx =    x  4  x  7  x 1  6  0



16  1  1  1      2  7  8  ln 2  =   8  ln 2   , 6  3 6 3  

since:

1  lim 6 n 

1

 0

(1  x )3  x 2n  2  dx =0.

1 x

Example H1.3: The following equality holds: n

lim

n 

1 7



 (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4)  (3  k  5) = 6   2  3  ln 3 .

1

(H1.3)

k 0

Proof: Equality (H1.3) is obtained from equality (H1), for: q=3, or by direct calculation. Thus, for every kN, 1 = (3  k  1)  (3  k  2)  (3  k  3)  (3  k  4)  (3  k  5) =

4 6 4 1  1  1       6  3 k 1 3 k  2 3 k  3 3 k  4 3 k  5 

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=

1

1 1  ( x 3k  4  x 3k 1  6  x 3k  2  4  x 3k  3  x 3k  4 )  dx =  [(1  x ) 4  x 3k ]  dx .

6 6





0

0

So, 1

n

lim

n 

1 1 n = lim  [(1  x ) 4  x 3k ]  dx n  6 ( 3  k  1 )  ( 3  k  2 )  ( 3  k  3 )  ( 3  k  4 )  ( 3  k  5 ) k 0 k 0



 0

=

=

1  lim 6 n 

1 n

 [(1  x) 0 k 0 1



1  lim 6 n 

 (1  x)

1

3

0

4



4

 x 3k ]  dx =

=



1



(1  x )  x 3

4

k 0

0

1



3k

   dx 

(1  x )3  (1  x 3n  3 ) 1 x  x2

0

3 n  3

1 x  x2

0

n

 (1  x)   x

1 1  x 3n  3    dx =  lim 6 n  1  x 3 

1 1 (1  x ) =   dx -  lim 2 6 1 x  x 6 n  0



1  lim 6 n 

1

 dx

1

2  x 1  1    dx  dx =    x  4  3  2 6  x  x 1  0



1  1  1 7      4  3  ln 3  =    3  ln 3  , 6  2 6 2   

since:

1  lim 6 n 

1



(1  x )3  x 3n  3 1 x  x2

0

 dx =0.

Example H1.4: The following equality holds: n

lim

n 



1 



 (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4)  (4  k  5) = 6   4  ln 2  2 1 .

1

(H1.4)

k 0

Proof: Equality (H1.4) is obtained from equality (H1), for: q=4, or by direct calculation. Thus, for every kN, 1 = (4  k  1)  (4  k  2)  (4  k  3)  (4  k  4)  (4  k  5) =

4 6 4 1  1  1      .

6  4  k 1 4  k  2 4  k  3 4  k  4 4  k  5 

=

1 1  ( x 4k  4  x 4k 1  6  x 4k  2  4  x 4k  3  x 4k  4 )  dx =  [(1  x ) 4  x 4k ]  dx 6 6

1

1





0

0

So, 1

n

lim

n 

1 1 n = lim  [(1  x ) 4  x 4k ]  dx n  6 ( 4  k  1 )  ( 4  k  2 )  ( 4  k  3 )  ( 4  k  4 )  ( 4  k  5 ) k 0 k 0



 0

n   1 1 =  lim [(1  x ) 4  x 4k ]  dx  lim (1  x ) 4  x 4k   dx 6 n  k 0 6 n     k 0 0 0 1

1 n



1 =  lim 6 n  1

=

4 n 1   1 4 1 x  dx =  lim (1  x )  4  6 n  1  x   0

1



1 1 (1  x )3   dx -  lim 2 3 6 1 x  x  x 6 n  0







1

 0

(1  x )3  (1  x 4n 1 ) 1  x  x 2  x3

(1  x )3  x 4n 1

 1 x  x 0

1

2

 x3

 dx

 dx

1

=

1 1    1  1   2  2   dx =   4  ln 2   1 ,   1  4  x 1 6  6 2  x  1   0



since:

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals 1  lim 6 n 

1

(1  x )3  x 4n 1

 1 x  x 0

2

 x3

 dx =0.

By reasoning inductively, we immediately arrive at the next result, which we can obtain and by reasoning deductively, from (Vălcan, 2016, Equalities (10)).

Example H1.p: For every p, qN, the following equality holds: 1

n

lim

n 

1 1 (1  x ) p =   dx . (q  k  1)  (q  k  2)    (q  k  p  1) p!

1  xq k 0 0





(H1.p)

particularly, 1

n

lim

n 



1 1 =  (1  x ) p1  dx (k  1)  (k  2)   (k  p  1) p! k 0

 0

 1 1 1 1 1  0   Cp1   C1p1   C2p1    (1) p2  (H1.p)  Cp2  (1) p1   Cpp11  .

2 3 p  1 p1 p p!   Equality (H1.p) results from equality (H1.p) or from (Vălcan, 2016, Equalities (10)) and taking into account the fact that, for every kN, =

1



1 1 =  [(1  x ) p  x k ]  dx (k  1)  (k  2)  ... (k  p  1) p!

0

=

1  p!

 x 1

k

 C1p  x k 1  C 2p  x k  2  ...  (1) p 1  C pp 1  x k  p 1  (1) p  C pp  x k  p



0

1 1 1 1 1 1 =   +...+(-1)p  .

0!p! k  1 1!(p  1)! k  2 p!0! k  p  1 So, n

lim

n 

=

=



1 1 =  lim ( k  1 )  ( k  2 )    ( k  p  1 ) p ! n  k 0 1 n

1  lim p! n 

n  1 p 1 x  (1  x )    dx =  lim 1  x  p! n   0

0 k 1 1

1

=

=



1 1  (1  x ) p 1  dx -  lim p! p! n   1  p!

0 1

 C

0 p 1

 (1  x) 1

p 1

 x k ]  dx

k 0 0





1  lim p! n 



p

n   p x k   dx (1  x )   k 1  0

 [(1  x)

 x k ]  dx =

  [(1  x)

1

1  lim p! n 

p

n 1

 (1  x) 1

p 1



 (1  x n )  dx

0



 x n  dx

0



 x  C1p 1  x 2  C 2p 1    (1) p  2  x p  2  C pp 12  (1) p 1  x p 1  C pp 11  dx

0

 1 1 1 1 1   Cpp 12  (1) p 1   Cpp 11  , =   C0p 1   C1p 1   C2p 1    (1) p  2  2 3 p 1 p p!   since: 1

lim

n 

 [(1  x

p 1

)  x n ]  dx =0.

0

III.

Conclusions

Now we can say that we have solved the problem of calculating the limits of the sequences of type (H1.p). I worked inductively, starting from small values of p, q and k. The attentive and interested reader of these issues will notice that one can also work deductively. On the other hand, teachers, in class, or training circles / camps, with students capable of performance, can take other values for these numbers, only then the calculations will be more complicated and the results more beautiful, more spectacular.

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Again About Calculating Of Limits Of Some Sequences With The Help Of Integrals References: [1].

[2].

[3].

[4].

[5].

[6].

[7].

Mocica, Gh., (1998), Probleme de funcţii speciale, (Problems of special functions), București, Editura Didactică şi Pedagogică.

Sireţchi, Gh., (1985), Calcul diferenţial şi integral, Vol. I, Noţiuni fundamentale, (Differential and Integral Calculus, Vol. I, Fundamental notions), București, Editura ştiinţifică şi Enciclopedică.

Sireţchi, Gh., (1985), Calcul diferenţial şi integral, Vol. II, Exerciții, (Differential and Integral Calculus, Vol. II, Exercises), București, Editura ştiinţifică şi Enciclopedică.

Vălcan, D., (2016), An Example of the Training a Methodological Competences in Mathematical Analysis, in The European Proceedings of Social & Behavioural Sciences EpSBS, Published by the Future Academy, Volume XVIII, 2016, p. 631-639.

Vălcan, D., (2017), From Integrable Functions To Limits Of Sequences, in The European Proceedings of Social & Behavioural Sciences EpSBS, Published by the Future Academy, Volume XLI, International Conference of the Departament of Educational Sciences, „ERD 2017 - Education, Reflection, Development” – Five Edition, 07 – 08 July 2017, editori: Chiş, V., Albulescu, I., 2018, p.

422-431.

Vălcan, D., (2017) From limits of sequences to definite integrals, în „Educația din perspectiva valorilor”, Tom XII: Summa Paedagogica, p. 139-145, volumul Conferinței Internaționale „Educația din perspectiva valorilor”, Ediția a IX – a, organizată de Universitatea de Stat din Moldova și Universitatea „1 Decembrie 1918” din Alba-Iulia, Chişinău, 28-30, Septembrie, 2017, Editori: Opriș, D., Scheau, I., Moșin, O., București, Editura EIKON. Vălcan, D., (2018), The spectacole of Mathematics ... continues, în „Educația din perspectiva valorilor”, Tom XIV: Summa Paedagogica, p. 85-91, volumul Conferinței Internaționale „Educația din perspectiva valorilor”, Ediția a X – a, organizată de Universitatea „1 Decembrie 1918” din Alba-Iulia și Universitatea de Stat din Moldova, Alba-Iulia, 10-11, Octombrie, 2018, Editori: Opriș, D., Scheau, I., Moșin, O., București, Editura EIKON. http://mathworld.wolfram.com/zetafunction

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