11.8_ Power Series-solutions

Thus the 2 1 1 interval of convergence 1 1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n! (4x − 1)n . 7. (−1, 1] 1. (−4, 4) 8. [−4, , 4] 2. converges only at x = 4 Explanation: When an 3. (−∞, ∞) 4 xn =√, n 11 4. − , 44 then √ n xn+1 ·n =√ n+1 x √ | x| n n = | x| =√ n+1 n+1 an+1 an Thus lim n→∞ an+1 an 5. converges only at x = . n=1 6. [0, 8) Explanation: The given series has the form = | x| . By the Ratio Test, therefore, the given series converges when |x| < 1, and diverges when |x| > 1. We have still to check for convergence at x = ±1. But when x = 1, the series reduces to ∞ 4 √ n 1 correct 4 ∞ n=1 an (x − a)n with a n = n! 4n , a= 1 . 4 But if x = a, lim n!4n (x − a)n = ∞ , n→∞ stiurca (mas7745) – 11.8: Power Series – meth – (91825) By the Ratio Test, therefore, the given series and so by the Divergence Test, the series (i) converges when |x| < 1, ∞ n n=1 n!4 (x − a) n (ii) diverges when |x| > 1. diverges whenever x = a. Consequently, the given series converges only at x = 010 1 . 4 which diverges by the Inegral Test. On the other hand, at x = 1, the series becomes Find the interval of convergence of the series ∞ xn . (−1)n 2n + 5 n=1 1. interval of cgce = [−5, 2] 2. interval of cgce = [−1, 1) 4. converges only at x = 0 5. interval of cgce = (−1, 1] correct 6. interval of cgce = (−2, 5] 7. interval of cgce = [−1, 1] 8. interval of cgce = (−∞, ∞) Explanation: When xn , 2n + 5 then But lim n→∞ 2n + 5 = 1, 2n + 7 in which case lim n→∞ an+1 an = | x| . ∞ n=1 (−1)n 2n + 5 which converges by the Alternating Series Test. Consequently, the interval of convergence = (−1, 1] . 3. interval of cgce = (−1, 1) xn+1 2n + 5 2n + 5 an+1 =− = | x| n an 2n + 7 x 2n + 7 We have still to check what happens at the endpoints x = ±1. At x = −1 the series becomes ∞ 1 2n + 5 n=1 10.0 points an = (−1)n 6 ....
View Full Document

This document was uploaded on 09/17/2013.

Ask a homework question - tutors are online