# 11 by the integral test this series converges if and

• Notes
• 18

This preview shows page 11 - 15 out of 18 pages.

11
By the Integral Test, this series converges if and only if the following improper integral converges: Z 2 1 x (ln( x )) 2 dx = lim b →∞ Z b 2 1 x (ln( x )) 2 dx. To evaluate this integral, let u = ln( x ). Then du = 1 x dx and the above integral becomes lim b →∞ Z ln( b ) ln(2) 1 u 2 du = lim b →∞ - 1 u ln( b ) ln(2) = lim b →∞ - 1 ln( b ) + 1 ln(2) = 0 + 1 ln(2) = 1 ln(2) . Therefore the integral converges and so the Integral Test says that the series of absolute values con- verges, meaning that the given series converges absolutely. 24. Find the interval of convergence for each of the following power series. (a) n =1 ( x +2) n n Answer: Using the Ratio Test on the series of absolute values, lim n →∞ ( x +2) n +1 n +1 ( x +2) n n = lim n →∞ | x + 2 | n +1 n + 1 n | x + 2 | n = lim n →∞ | x + 2 | n n + 1 = | x + 2 | . Therefore, the series definitely converges absolutely when | x + 2 | < 1. Now check the endpoints. When x + 2 = 1, the series becomes X n =1 1 n n = X n =1 1 n , which diverges. When x + 2 = - 1, the series becomes X n =1 ( - 1) n n , which converges. Therefore, the power series converges for - 1 x + 2 < 1 , or - 3 x < - 1 , which means the interval of convergence is [ - 3 , - 1). 12
(b) n =1 ( x - 3) n 1+ n 2 Answer: Using the Ratio Test on the series of absolute values, lim n →∞ ( x - 3) n +1 1+( n +1) 2 ( x - 3) n 1+ n 2 = lim n →∞ | x - 3 | n +1 1 + ( n + 1) 2 1 + n 2 | x - 3 | n = lim n →∞ | x - 3 | 1 + n 2 1 + ( n + 1) 2 = lim n →∞ | x - 3 | n 2 + 1 n 2 + 2 n + 2 = | x - 3 | . Therefore, the series definitely converges absolutely when | x - 3 | < 1. Now check the endpoints. When x - 3 = 1, the series becomes X n =1 1 n 1 + n 2 = X n =1 1 1 + n 2 < X n =1 1 n 2 , which converges, so the series converges when x - 3 = 1. When x - 3 = - 1, the series becomes X n =1 ( - 1) n 1 + n 2 , which converges by the Alternating Series Test. Therefore, the given power series converges for - 1 x - 3 1 , or 2 x 4 , so the interval of convergence is [2 , 4]. (c) n =1 ( - 1) n ( x - 1) n ne n Answer: Using the Ratio Test on the series of absolute values, lim n →∞ ( - 1) n +1 ( x - 1) n +1 ( n +1) e n +1 ( - 1) n ( x - 1) n ne n = lim n →∞ | x - 1 | n +1 ( n + 1) e n +1 ne n | x - 1 | n = lim n →∞ | x - 1 | e n n + 1 = | x - 1 | e . Therefore, the series definitely converges absolutely for | x - 1 | e < 1, meaning for | x - 1 | < e . Now, check the endpoints. When x - 1 = e , the series becomes X n =1 ( - 1) n e n ne n = X n =1 ( - 1) n n , which converges. 13
When x - 1 = - e , the series becomes X n =1 ( - 1) n ( - e ) n ne n = X n =1 ( - 1) n ( - 1) n e n ne n = X n =1 1 n , which diverges. Therefore, the given power series converges for - e < x - 1 e, or 1 - e < x 1 + e, so the interval of convergence is (1 - e, 1 + e ]. (d) n =1 n 2 x n Answer: Using the Ratio Test on the series of absolute values, lim n →∞ ( n + 1) 2 x n +1 | n 2 x n | = lim n →∞ ( n + 1) 2 n 2 | x | n +1 | x | n = lim n →∞ ( n + 1) 2 n 2 | x | = | x | , so the series definitely converges absolutely for | x | < 1. Now check the endpoints. When x = 1, the series becomes X n =1 n 2 1 n = X n =1 n 2 , which diverges since the terms are going to infinity. Likewise, when x = - 1, the series becomes X n =1 n 2 ( - 1) n , which diverges since the terms are getting large in absolute value and hence not going to zero.