Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.6 - Alternating Series and Conditional Convergence - Exercises 10.6 - Page 603: 6

Answer

Diverges

Work Step by Step

Let us consider $a_n=(-1)^{n+1}\dfrac{n^2+5}{ n^2+4}$ and $u_n$ refers to all positive values of $n$. Here, $\lim\limits_{n \to \infty} a_n= \lim\limits_{n \to \infty}\dfrac{n^2+5}{ n^2+4}=\lim\limits_{n \to \infty}\dfrac{1+5/n^2}{1+4/n^2}=1$ Also, $\lim\limits_{n \to \infty} (-1)^{n+1}\dfrac{(n^2+5)}{ (n^2+4)}=DNE$ Hence, the series diverges by the Nth Term Test for Divergence.
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