## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.6 - Absolute Convergence and the Ratio and Root Tests - 11.6 Exercises: 6

#### Answer

The series is (conditionally) convergent.

#### Work Step by Step

Alternating series test: Suppose that we have a series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two conditions are satisfied the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. In the given problem, $b_{n}=\frac{n}{n^{2}+4}$ 1.$b_{n}=\frac{n}{n^{2}+4}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{n}{n^{2}+4}$ $=\lim\limits_{n \to \infty}\frac{1/n}{1+4/n^{2}}$ $=0$ Thus, the limit is zero. Hence, the given series is convergent by AST.

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