Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 11 - Infinite Sequences and Series - 11.4 Exercises - Page 750: 22

Answer

Convergent

Work Step by Step

The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$. Given: $\Sigma _{n=3}^{\infty}\frac{{n+2}}{(n+1)^{3}}$ It can be re-written as: $\Sigma _{n=3}^{\infty}\frac{{n+2}}{(n+1)^{3}}=\Sigma _{n=3}^{\infty}\frac{{n+1+1}}{(n+1)^{3}}$ $=\Sigma _{n=3}^{\infty}\frac{{n+1}}{(n+1)^{3}}+\Sigma _{n=3}^{\infty}\frac{{1}}{(n+1)^{3}}$ Adding and subtracting a finite number of terms from a series does not affect the convergence or divergence of the series. The first series is convergent because a $p-$ series with $p=2$ is convergent. The second series is convergent because a $p-$ series with $p=3$ is convergent. The sum of two convergent series is convergent.
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