11.8_ Power Series-solutions

# Thus the 2 1 1 interval of convergence 1 1

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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 ....
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## This document was uploaded on 09/17/2013.

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