MATH 141-Written Homework 10. Solutions. Winter 2023.

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Written Homework 10

Math 141

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Question 1 Determine whether each series converges or diverges. Be sure to name the test used and the key details. (a)

∞ X 4n+1

(if this series converges, find its sum) 5n This is a geometric series — we have

2 ∞ 16 X 4n+1 16 16 4 16 4 5 = + ··· = + + = 16, 5n 5 5 5 5 5 1 − 54 n=1

n=1

since |r| = |4/5| < 1. The geometric series converges with sum 16. (b)

∞ X (n!)2

(2n)!

The factorials are a hint to try the Ratio Test:

[(n + 1)!]2 |an+1 | [2(n + 1)]!

lim = lim n→∞ |an | n→∞ (n!)2 (2n)!

[(n + 1)!]2 (2n)! = lim . n→∞ (n!)2 (2n + 2)! n=1

For the factorials, we have (n + 1)! = (n + 1) (n)(n − 1)(n − 2) · · · 3 · 2 · 1 = (n + 1)(n!), {z } | n!

(2n + 2)! = (2n + 2)(2n + 1) (2n)(2n − 1)(2n − 2) · · · 3 · 2 · 1 = (2n + 2)(2n + 1) (2n)! . {z } | (2n)!

Therefore,

lim

n→∞

[(n + 1)!] (n!)2

2

(2n)!

(2n + 2)!

2

2





(n + 1) [(n!)]  (2n)!

 (n!)2 (2n + 2)(2n + 1) (2n)!

1 = lim (n + 1)2 n→∞ (2n + 2)(2n + 1) 2 n + lower order terms = lim n→∞ 4n2 + lower order terms 1 = < 1, 4 = lim

n→∞

and so the series converges by the Ratio Test . (c)

∞ X 3n + 2

5n + 3 n=1 We have lim

3n + 2

n→∞ 5n + 3

=

3 6= 0, 5

and so the series diverges by the nth Term Divergence Test . n ∞ X 3n + 2 (d) 5n + 3 n=1 The expression of the form f (n)n is a hint to use the Root Test because the algebra generally works out well. We have

n/n 3n + 2 3n + 2 3 lim |an |1/n = lim = lim = < 1, n→∞ n→∞ 5n + 3 n→∞ 5n + 3 5 and so the series converges by the Root Test .

Written Homework 10

Math 141 (e)

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∞ X 102n+5 n!

(2n)! n=1 The factorials are a hint to try the Ratio Test. We have 2(n+1)+5

10 (n+1)!

[2(n+1)]! |an+1 |

= lim lim n→∞ n→∞ |an | 102n+5 n!

(2n)!

2n+7

10 (n + 1)!

(2n)! = lim n→∞ 102n+5 n!

(2n + 2)!

(n + 1)n!

(2n)!

2n+7−(2n+5) = lim 10 n→∞ n!

(2n + 2)(2n + 1)(2n)!

1 = lim 102 (n + 1) n→∞ (2n + 2)(2n + 1) 102 (n + 1) = lim n→∞ 4n2 + lower order terms = 0 < 1, and so the series converges by the Ratio Test .

(f)

∞ X n!

nn The factorial is a hint to try the Ratio Test. We have

(n+1)! n+1 (n+1) |an+1 |

lim = lim n! n→∞ |an | n→∞ nn

nn (n + 1)! = lim n→∞ n! (n + 1)n+1

nn (n + 1)n! = lim n→∞ n! (n + 1)(n + 1)n

nn = lim (n + 1) n→∞ (n + 1)(n + 1)n n n = lim n→∞ (n + 1)n

n n = lim n→∞ n + 1

n 1 = lim n→∞ 1 + 1 n

−n 1 = lim 1 + . n→∞ n n=1

This is an indeterminate (→ 1)→−∞ form. We have

−n

−x 1 1 1+ = lim 1 + n→∞ x→∞ n x lim

−x

= lim eln[(1+ x ) 1

]

x→∞

= lim e−x ln(1+ x ) 1

x→∞

−

= lim e x→∞

!

( ) ( ) .

ln 1+ 1 x 1 x

Written Homework 10

Math 141 We see we have a

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→ ln 1 →0 = form, and we can use L’Hospital’s Rule: →0 →0

1 1 − x12 1 ln 1 + x H 1+ x

= −1, = lim − lim − 1 x→∞ x→∞ − x12 x

and therefore

−n 1 1 lim 1 + = e−1 = < 1. n→∞ n e

Therefore, the series converges by the Ratio Test . (Remember e ≈ 2.718.)

Question 2 Use the Alternating Series Estimation Theorem to find the minimum number of terms of the infinite ∞ X (−1)n series we need to add to approximate the sum of the series with |error| ≤ .008. n3 n=1 The argument that proves the Alternating Series Test (was an application of Bounded Monotone Sequence Convergence) also shows any series satisfying its hypotheses also obey the Alternating Series Estimation Theorem for the N th partial sum SN : |S − SN | = S −

N X

(−1)n bn = S − (−b1 + b2 − b3 + b4 − b5 + · · · + bN ) ≤ bN +1 ,

n=1

where S is the sum of the series S = lim SN = N →∞

∞ X

(−1)n bn = −b1 + b2 − b3 + b4 − b5 + · · · .

n=1

(Here we’re using our standard notation for alternating series — that bn ≥ 0 for all n.) This is saying the size of the error of a partial sum is controlled by the size of the first term not included in the partial sum.

For this particular series, because

1 is decreasing with limit 0, we have the error bound n3

|S − SN | = S −

N X (−1)n n=1

n3

≤

1 .

(N + 1)3

Therefore, we seek the smallest N so that 1 ≤ 0.008.

(N + 1)3 Then, 1 ≤ (N + 1)3 , 0.008 1 ≤ N + 1, 0.2 5 ≤ N + 1,

√ f (x) = 3 x is increasing so preserves inequalities

and so N ≥ 4 and so we need at least 4 terms .

Question 3 Captain Picard and the rest of the crew of the Rocinante are defending their position on Tatooine against the evil robotic Cylon invaders. Each day of the battle, 1500 Cylons arrive at dawn, and throughout each day the crew is able to eliminate 60% of the Cylons on the battlefield. (a) Let Cn be the number of Cylons on the planet at the end of the day on the nth day of the battle.

Find C1 , C2 , and C3 .

Math 141

Written Homework 10

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At the beginning of the first day of battle, 1500 Cylons arrived. But only 40% remain at the end of the first day, and so C1 = 1500(0.4) .

At the beginning of the second day, 1500 more arrive and so there are 1500 + 1500(0.4). But then by the end of the second day only 40% of these remain, and so C2 = 1500(0.4) + 1500(0.42 ) .

Similarly, 1500 more arrive at the beginning of the third day, making the total 1500 + 1500(0.4) + 1500(0.42 ). But then only 40% remain by the end of the third day and C3 = 1500(0.4) + 1500(0.42 ) + 1500(0.43 ) .

(b) Find an explicit, closed-form formula for the number of Cylons at the end of the day on the nth day of the battle. (Explicit means we write Cn as a function of n alone as opposed to a recursive formula where Cn is written in terms of Cn−1 . By closed-form, we mean a formula without shorthand such as sigma notation or + · · · +.)

The previous part suggests the pattern Cn = 1500(0.4) + 1500(0.42 ) + 1500(0.43 ) + · · · + 1500(0.4n ) =

1500(0.4) − 1500(0.4n+1 ) .

1 − 0.4

Here we used the mnemonic starting term − first missing term 1 − ratio for the geometric partial sum formula. By “first missing term” we mean the term in the pattern that would’ve appeared next. (c) In the long run, how many Cylons will be on the planet after a day of battle?

The 0.4n+1 is exponentially decaying and so lim Cn =

n→∞

1500(0.4) − 0 = 1000 Cylons .

1 − 0.4

(d) If instead the Cylons want to maintain a minimum force of 5000 soldiers, how many should they send each day?

If we think about how our steps would change, if they send N per day then lim Cn =

n→∞

N (0.4) 2N = , 0.6 3

and so they would need to send 7500 per day . The fact this is the long term minimum force is because the reduction happens throughout the day — this is analogous to the medicine amount immediately before a dose from last week’s problem. At dawn right after Cylons arrive is analogous to the medicine amount immediately after a dose.