Practice exercises - Exam 4 (Sections 11.1 through 11.8 and 11.10)

1. Write a formula for the nth term in the sequence 1 2 3 4 5 2 1 , 3 1 , 4 1 , 5 1 , 6 1 ,... 2 3 4 5 6 2. Compute the following limits, if they exist.

(a) lim ( 1)n n x n+2 !1 · n2 (b) lim n x e !1

3. Given series n1=1 an and n1=1 bn with positive terms, are the following statements true or false? P P

(a) If lim an =0, the series 1 an must converge. n n=1 !1 (b) If lim an =1, the limit comparisonP test gives no information. n bn !1 1 1 1 (c) If an = pn , the series n=1 pn converges as a p-series. 1 1 1 (d) If an = 3n 2 , the seriesP n=1 3n 2 converges by comparison with the convergent 1 n geometric series 1 . n=1 3P P n 1 4. Use the integral test to determine whether the series n=1 n2+1 converges or diverges.

n 1 P 5. Use a comparison test to demonstrate that n=2 n2 3 diverges. 6. Determine whether the following series convergeP or diverge. Explain your answer, indicating the test you used to draw your conclusion.

n+1 1 (a) n=1 n+2 n 1 (b) Pn=1 2+3n3 ln n 1 (c) Pn=1 n! 2n 1 (d) Pn=1 n

P n 1 7. Use the alternate series test to conclude that the series 1 ( 1) converges. n=3 ln n n 2n 8. Determine whether the infinite series 1 ( 1) convergesP absolutely, converges n=2 · n! conditionally, or diverges. P n n 9. Find the radius of convergence for the power series 1 (n ⇡ ) x n=1 · 1 n 1 P 10. Find the interval of convergence for the series n=1 n 2n (x 1) · 11. Find the first five terms of the Maclaurin seriesP for f(x)=p1 x 12. Find the Taylor series of f(x)=lnx based at the point a = e Answers

n 1. Starting at n =1, an = (n+1) 1 (n+1) 2. (a) limit does not exist and (b) limit is 0 using L’Hopital’s rule.

3. (a) False. The divergence test states that if an 9 0, then the series must diverge. If a 0, the series could converge or could diverge. n ! (b) False. lim an =1and 1 =0, 1 = . The limit comparison test tells us that the series n bn !1 6 6 1 n1=1 an and n1=1 bn behave the same; they either both converge or both diverge. 1 1 1 (c) False. 1 = 1 1 so this is a p-series with p = . Since p 1, this p-series P n=1P pn n=1 n /2 2  diverges. P P 1 1 1 1 (d) False. Comparing an = 3n 2 and bn = 3n , we get the inequality 3n 2 > 3n . 1 The smaller series is geometric with r = 3 so it converges. The comparison test does 1 1 not give any information about the bigger series n=1 3n 2 . 1 1 1 1 (note: the series n=1 3n 2 does converge via a limit comparison test with n=1 3n ) P P x 2 P 4. We can integrate x2+1 dx as a u-substitution with u = x +1. The improper integral x 1 2 dx diverges so the original series diverges 1 x +1 R x 1 x2 (notice that 2 0 = 2 < 0 for x>1 so that the integrand is decreasing). R x +1 (x2+1)

n n n 1 1 1 5. Notice that n2 1 > n2 so that n2 1 > n . Since the harmonic series n=1 n diverges, the n 1 bigger series n=1 n2 1 must also diverge by the comparison test. P 6. (a) Series divergesP by the nth term (divergence) test since the terms go to 1 instead of 0. 1 1 (b) Series converges by limit comparison with the convergent p-series n=1 n2 (c) Series converges by the ratio test as the ratio an+1 0 which is less than 1. an ! P n (d) Series diverges by the root test as the root an 2 which is bigger than 1. | |! 1 1 p 2 1 7. Notice that lim =0and, for f(x)= , f 0(x)= 1 (ln x) < 0 for x 2. k ln n ln x · · x The terms go!1 to 0 and are decreasing, so it convergences by the alternating series test.

8. The series converges absolutely. Applying the ratio test to the series with absolute values gives a lim 2 =0< 1. n n+1 !1 9. The ratio test (with absolute values) gives ⇡ x < 1 so the radius of convergence is 1 | | ⇡ x 1 10. The ratio test (with absolute values) gives | | < 1 so that 1