## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.5 - Alternating Series - 11.5 Exercises: 17

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, $b_{n}=sin\frac{\pi}{n}$ which satisfies both conditions of the Alternating Series Test as follows: 1. $b_{n}=sin\frac{\pi}{n}$, is decreasing. Here, $\frac{\pi}{n}$ is decreasing and $sinx$ is increasing on $[0,\pi/2]$ which implies $sin\frac{\pi}{n+1}\lt sin\frac{\pi}{n}$ for $n\gt 2$ 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}sin\frac{\pi}{n}$ $=sin(0)$ $=0$ Hence, the given series is convergent by the Alternating Series Test.

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