MATH 141 CALC ANLY GEOM II In-Class Worksheet Week 10 Page. Solutions. Winter 2023

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In-Class Worksheet Week 10

Math 141

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Question 1 For each of the following series, determine whether they converge or diverge. You must name the test(s) used, as well as show the details of the test(s). (a)

∞ X 21/n √ n n=1

As n → ∞, 21/n → 20 = 1, and so we expect this series is similar to ∞ X 1 √ . n n=1

We attempt the Limit Comparison Test.

1/n 2√

lim

n→∞

n

√1 n

= lim 21/n = 1, which is finite and nonzero. n→∞

Since

∞ X 1 √ n n=1

is a divergent p-series (p = 1/2 ≤ 1), the original series also ∞ X 1 √ . diverges by the Limit Comparison Test with n n=1 (b)

∞ X 3 + 2 cos n

n2 The range of cosine is [−1, 1]. Therefore,

n=1

3 + 2 cos n 5 1 ≤ ≤ 2. n2 n2 n Based on the p-value of 2, the only useful inequality above is 3 + 2 cos n 5 ≤ 2, n2 n and because 5

∞ X 1 n=1

n2

converges (p = 2 > 1),

the original series also converges by the Direct Comparison Test with

∞ X 5 n=1

(c)

∞ X

n2

..

1 n(ln n)1/4 n=2 The series is not alternating, so the Alternating Series Test is not helpful. The terms go to zero so the nth-Term Divergence test doesn’t have anything to say either. So we’re down to Direct Comparison, Limit Comparison, or Integral Test (the two additional tests we’ll discuss in 10.5 don’t help here either).

This doesn’t behave close enough to anything we know about for Limit Comparison to work (we won’t get a finite nonzero limit), and there aren’t any useful inequalities for Direct Comparison either. At some point you can also start to notice a slight hint that we know how to anti-differentiate this function using a substitution.

In-Class Worksheet Week 10

Math 141 Indeed,

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1 is continuous, positive, and decreasing for all x ≥ 2. Then we calculate x(ln x)1/4 Z ∞ 2

1 dx = lim t→∞ x(ln x)1/4

Z t

1 dx x(ln x)1/4 2 Z ln t 1 = lim du t→∞ ln 2 u1/4

4 3/4 ln t = lim u t→∞ 3 ln 2

4 4 (ln t)3/4 − (ln 2)3/4 = lim t→∞ 3 3 = ∞,

and so the improper integral and hence the original series are divergent by the Integral Test .

1 (We used the substitution u = ln x, du = dx in the integral calculation.) x ∞ X 3n2 + 5n + 2 √ (d) n6 + n2 + 1 n=1 As n → ∞ the most important term in the numerator is 3n2 , and in the denominator it is n6 . So we expect the series to behave similarly to ∞ ∞ X X 1 3n2 √ =3 .

6 n n n=1 n=1

This is when we should try the Limit Comparison Test: 2

3n +5n+2

2 √ 6 2 3n + 5n + 2 n +n +1

√ lim = lim n→∞ n→∞ 3n2 n6 + n2 + 1 √

√

n6 3n2

n6

3n2 + 5n + 2 3n2

!

√

n6 √ = lim n→∞ n6 + n2 + 1 r

2 n6 3n + 5n + 2 = lim 2 6 n→∞ 3n n + n2 + 1 r 3 1 = 3 1

!

= 1, which is finite and nonzero.

Then, since

∞ ∞ X X 3n2 1 √ =3 6 n n n=1 n=1

diverges, harmonic series, p = 1 ≤ 1,

the original series also diverges by the Limit Comparison Test to

∞ X 3 n=1

n

.

Question 2 For each of the following series, determine whether they converge absolutely, converge conditionally, or diverge. You must name the test(s) used, as well as show the details of the test(s). (a)

∞ X (−1)n

n2 + 4

n=1

(b)

∞ X

(−1)n

n=2

n (ln n)1/4

In-Class Worksheet Week 10

Math 141 (c)

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∞ X (−1)n √ n+4 n=1

For these, it is helpful to remember the following. ∞ X ∞ X

∞ X

|an |

n=1

an

n=1

an converges absolutely means:

converges

=⇒

converges

an converges conditionally means:

diverges

=⇒

⇐=

converges

diverges

⇐=

diverges

n=1 ∞ X n=1

∞ X

an diverges means:

n=1

• This means that to show

∞ X

an converges absolutely, you only have to show

n=1

∞ X

|an | converges.

n=1

(The idea is forcing all the terms to be positive by taking absolute values makes it harder for the series to converge since you won’t have positive and negative terms working against each other.

If the convergence is “strong enough” so that it still converges even when the signs are all made positive — it will also converge if the positive and negative terms are allowed to work against each other. In summary, absolute convergence is a strictly stronger condition than convergence.) ∞ ∞ X X an diverges. (A common mistake an diverges, you only have to show • Similarly, to show is instead showing

n=1 ∞ X

n=1

|an | diverges and thinking this implies

∞ X

an diverges too! This is NOT

n=1

n=1

TRUE because conditional convergence exists!) This is actually just the (equivalent) contrapositive of the above idea that absolute convergence implies convergence. ∞ ∞ X X • However, to show a series an conditionally converges, you must show BOTH an converges and

∞ X

n=1

n=1

|an | diverges — neither one is implied by the other!

n=1

(a) If we are clever and first show the convergence of ∞ ∞ X X (−1)n 1 , = 2+4 2+4 n n n=1 n=1

then the original series converges absolutely and we don’t have to consider the alternating version.

Indeed, 1 1 ≤ 2, n2 + 4 n ∞ ∞ X X 1 1 and since converges (p = 2 > 1), we know converges too. Therefore, 2 2 n n +4 n=1 n=1 ∞ X (−1)n n=1

n2 + 4

converges absolutely by the Direct Comparison Test with

∞ X 1 n=1

(b) If we are clever and first show the divergence of ∞ X

(−1)n

n=2

n (ln n)1/4

n2

.

In-Class Worksheet Week 10

Math 141

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then we are done since this guarantees the series of absolute values diverges too. Indeed, 1/4  4   n n  = ∞. lim = lim   1/4 n→∞ (ln n) n→∞ ln n  | {z } →∞

To see why the above works, we can do n4 x4 H 4x3 = lim = lim 1 = lim 4x4 = ∞. n→∞ ln n x→∞ x→∞ ln x x→∞ x lim

Then, we can deduce lim (−1)n

n→∞

and so

∞ X

(−1)n

n=1

n does not exist, (ln n)1/4

n diverges by the nth Term Divergence Test . (ln n)1/4

(c) This one turns out to converge conditionally, meaning there is no clever way to investigate just ∞ ∞ X X one series — we must show BOTH an converges and |an | diverges.

For

∞ X

n=1

an , we see that since

√

n=1

n is increasing and lim

√

n→∞

n=1

n = ∞ guarantees √

1 is decreasing n+4

∞ X (−1)n √ converges by the Alternating Series Test. n+4 n=1 ∞ ∞ ∞ X X X 1 1 √ √ . . For this, we can use the limit comparison test with For |an | we consider n + 4 n n=1 n=1 n=1 We have

√ √1 n 1 n+4

= 1, which is finite and nonzero, lim = lim √ = lim n→∞ n→∞ 4 n + 4 n→∞ √1 √ 1 + n n | {z }

with limit zero. Therefore,

→0

∞ ∞ X X 1 1 1 √ diverges (p = √ ≤ 1), diverges by the Limit Comparison Test.

2 n n +4 n=1 n=1 1 1 (The Direct Comparison Test doesn’t work since the inequality √ ≤ √ is useless — havn+4 n ing terms less than those of a divergent series says nothing!) Putting it all together, ∞ ∞ X X (−1)n 1 √ √ . converges conditionally by the Alternating Series Test and the Limit Comparison Test with n+4 n n=1 n=1

and since

Question 3 Later in the course we will discuss the fact that π=

∞ X n=1

(−1)n−1

4 4 4 4 4 = 4 − + − + − ··· .

2n − 1 3 5 7 9

(a) Find the minimum number of terms, N , such that the Alternating Series Error Bound guarantees |π − SN | ≤ 10−3 , where SN =

N X

(−1)n−1

n=1

4 is the N th partial sum.

2n − 1

In-Class Worksheet Week 10

Math 141

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(b) What is the actual error |π − SN | for the N you obtained above? In Desmos you can type “sum” and it should be clear how to set this up without needing to code a for loop from scratch.

(a) The alternating Series Error Bound states that if 0 ≤ bn+1 ≤ bn for all n and lim bn = 0, n→∞

π−

N X

(−1)n−1 bn ≤ bN +1 .

n=1

Here this means we seek the smallest value of N so that 4 ≤ 10−3 .

2(N + 1) − 1 We then have 4000 ≤ 2N + 1, 3999 ≤ 2N, 3999 ≤ N, 2 and since we only consider integer indices N = 2000 is the smallest number of terms needed to attain the guaranteed error bound . (b) We have 2000 X

(−1)n−1

n−1

4 ≈ 3.141092, 2n − 1

π ≈ 3.141592,

meaning the error approximately 0.0005 so it is actually twice as accurate as guaranteed. This is ok — the error bound provides a “worst case scenario”.

Question 4 (a) Show that if an ≥ 0 and

∞ X

an converges, that

n=1

∞ X

(an )2 does too.

n=1

(b) Show that the converse of the above statement is false by providing a counterexample — that is, ∞ ∞ ∞ X X X an diverges. Hint: (an )2 converges but an such that find a series with positive terms n=1

n=1

n=1

You can do this with an appropriate p-series.

(a) Recall the nth term divergence test: lim an 6= 0 =⇒

n→∞

The contrapositive is

∞ X

∞ X

an diverges.

n=1

an converges =⇒ lim an = 0. n→∞

n=1

(Be careful! The converse of the above is NOT TRUE! There are divergent series whose terms approach zero!!! The harmonic series is our favorite example of such a series!!!! All dogs are mammals, but not all mammals are dogs!!!!!) In any case then, we know that lim an = 0,

n→∞

which means that for n large enough that an < 1.

In-Class Worksheet Week 10

Math 141

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Then, since 0.52 = 0.25 < 0.5 for example, 0 ≤ a2n ≤ an for n large enough (recall we are told that an > 0 too). Then the X Direct Comparison Test shows that a2n converges too . (b) Picking the good ol’ harmonic series an = (p = 2 > 1) while

∞ X 1 n=1

n

∞ ∞ 2 X X 1 1 1 = converges works here. We have 2 n n n n=1 n=1

diverges (p = 1 ≤ 1).