The given series is ∑n=1∞1n2cos1n. Let an=1n2cos1n. Note that, the bound of cosine function is…

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Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.2: Exponents And Radicals

Problem 46E

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Step 1

The given series is $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} \cos (\frac{1}{n})$ . Let $a_{n} = \frac{1}{n^{2}} \cos (\frac{1}{n})$ .

Note that, the bound of cosine function is $- 1 \leq \cos x \leq 1$ . That is,

$|\cos x| \leq 1 |\cos (\frac{1}{n})| \leq 1 |\sum_{n = 1}^{\infty} \frac{1}{n^{2}} \cos (\frac{1}{n})| \leq |\sum_{n = 1}^{\infty} \frac{1}{n^{2}}|$

Recall the fact that the comparison test states that if the infinite series $\sum_{n = 1}^{\infty} b_{n}$ converges and $0 \leq a_{n} \leq b_{n}$ for all $n > N$ , where N is some fixed number, then $\sum_{n = 1}^{\infty} a_{n}$ also converges.

Step 2

Here, $b_{n} = \frac{1}{n^{2}}$ and it satisfies the condition $0 \leq a_{n} \leq b_{n}$ .

The convergence can be tested using comparison test.

Consider the series $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ .

Recall the fact that the p test states that $\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$ converges if and only if $p > 1$ .

In $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ , the value of p is 2 which is greater than 1. Thus, by p test $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ converges.

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