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: 65

Answer

$\Sigma^{\infty}_{n=1}\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}} = 1$

Work Step by Step

$\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}}=\frac{1}{n^{3}}-\frac{1}{(n+1)^{3}}$ $\Sigma^{\infty}_{n=1} \frac{1}{n^{3}}- \frac{1}{(n+1)^{3}} = \lim\limits_{N \to \infty}\Sigma^{N}_{n=1} \frac{1}{n^{3}}- \frac{1}{(n+1)^{3}}$ $= \lim\limits_{N \to \infty} \frac{1}{1^{3}}-\frac{1}{2^{3}}+\frac{1}{2^{3}}-\frac{1}{3^{3}}+\frac{1}{3^{3}}-\frac{1}{4^{3}}+...+\frac{1}{(N-1)^{3}}-\frac{1}{N^{3}}+\frac{1}{N^{3}}-\frac{1}{(N+1)^{3}}$ $\lim\limits_{N \to \infty}\frac{1}{1^{3}} - \frac{1}{(N+1)^{3}}=1-0=1$ $\Sigma^{\infty}_{n=1}\frac{3n^{2}+3n+1}{(n^{2}+n)^{3}} = 1$
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