Homework 5 Solutions PDF

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Math 1320 Homework 5 Due: 6/19

1. Consider the series

∞ X n=2

n2

n . −1

(a) Use the integral test to show that the series is divergent. (Hint: Your integral will need to start at x = 2, explain why.) Solution: We can use the integral test because the comparison function x has positive terms and is decreasing on the interval [1, ∞). x2 −1 R ∞ xThe integral test says that this series will diverge if the integral 1 x2 −1 dx is divergent. So we need to compute the improper integral: Z ∞ Z b x x dx dx = lim 2 b→∞ 2 x − 1 x2 − 1 2 Z b2 −1 1 du = lim u- sub with u = x2 − 1 b→∞ 3 u 2  b2 −1 1 = lim ln(u) b→∞ 2 3  1 = lim ln(b2 − 1) − ln(3) b→∞ 2 =∞ Because the integral diverges so does the series. Rather than change the bounds on the integral we could also solve the indefinite integral first: Z Z 1 du 1 1 x dx = = ln|u| = ln(x2 − 1) 2 u 2 2 x −1 2

Followed by the improper integral, Z b Z ∞ x x dx dx = lim b→∞ 2 x2 − 1 x2 − 1 2  b 1 2 = lim ln(x − 1) b→∞ 2 2  1 = lim ln(b2 − 1) − ln(22 − 1) b→∞ 2 =∞ (b) Use the comparison test to show that the series is divergent. Solution:

n n 1 > = n2 − 1 n2 n We know the harmonic series diverges, so the comparison test says that ∞ X

∞ X n 1 =∞ ≥ 2−1 n n n=2 n=2

and the original series diverges too. 2. Determine if the following series converge or diverge. ∞ X n+1 (a) √ 3 n=1 ( n) Solution:

n+1 1 n n √ 3> √ 3= 3 = 1 n2 n2 ( n) ( n)

∞ X 1 And we know 1 diverges because it is a p-series with p = 2 n n=1 the original series diverges as well.

(b)

∞ X n=1

1 −4

5n2

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1 2

< 1, so

Solution: We wish to show that the given series has the same behavior as ∞ X 1 , which converges because it is a p-series with p = 2 > 1. the series n2 n=1

1 n2 lim 1 n→∞ 5n2 −4

5n2 − 4 n→∞ n2

= lim

5 − n42 = lim n→∞ 1 =5

Thus, the series have the same behavior and

∞ X n=1

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1 converges. 5n2 − 4...