Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.2 - Series - 11.2 Exercises - Page 717: 66

Answer

$\frac{1}{96}$

Work Step by Step

$\Sigma^{\infty}_{n=3} \frac{1}{n^{5}-5n^{3}+4n} = \Sigma^{\infty}_{n=3} \frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ $\frac{1}{4}\Sigma^{\infty}_{n=3} \frac{(n+2)-(n-2)}{(n-2)(n-1)n(n+1)(n+2)}$ $\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\frac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)}]$ $\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{1}{(n-2)(n-1)n(n+1)}-\frac{1}{(n-1)n(n+1)(n+2)}]$ $\lim\limits_{N \to \infty}\frac{1}{4}\Sigma^{\infty}_{n=3} [\frac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\frac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)}]$ $=\lim\limits_{N \to \infty} \frac{1}{4} [\frac{1}{1 \times 2 \times 3\times 4}-\frac{1}{2 \times 3 \times 4 \times 5}] +\frac{1}{4} [\frac{1}{2 \times 3 \times 4\times 5}-\frac{1}{3 \times 4 \times 5\times 6}] + \frac{1}{4}[\frac{1}{3 \times 4 \times 5\times 6} -\frac{1}{4 \times 5 \times 6 \times 7}] +...+ \frac{1}{4}[\frac{1}{(N-3) \times (N-2) \times (N-1)(N)} - \frac{1}{(N-2) \times (N-1) \times (N) \times (N+1)}] + \frac{1}{4} [\frac{1}{(N-2) \times (N-1) \times (N)(N+1)} - \frac{1}{(N-1) \times (N) \times (N+1)\times (N+2)}]$ $=\lim\limits_{N \to \infty} \frac{1}{4} [\frac{1}{1 \times 2 \times 3 \times 4}-\frac{1}{(N-1) \times (N) \times (N+1) \times (N+2)}]$ $=\frac{1}{4} [\frac{1}{1 \times 2 \times 3 \times 4} -0] = \frac{1}{96}$
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