## Calculus 8th Edition

Published by Cengage

# Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 5

#### Answer

$0$

#### Work Step by Step

Given:$a_{n}=\frac{n sin( n)}{n^{2}+1}$ A sequence is said to be converged if and only if $\lim\limits_{n \to \infty}a_{n}$ is a finite constant. We know that $-1\leq sinx\leq1$ Note that the numerator varies from $-n$ to $n$. We can get two sequences that bound $a_{n}$ from above and below: $-\frac{n }{n^{2}+1}\leq a_{n}\leq \frac{n }{n^{2}+1}$ $-\frac{n }{n^{2}+1}\leq\frac{n sin( n)}{n^{2}+1} \leq \frac{n }{n^{2}+1}$ The limits of both of these are $0$ because the degrees of their denominators are greater than the degrees of their numerators. $\lim\limits_{n \to \infty}\frac{-n }{n^{2}+1}=\lim\limits_{n \to \infty} \frac{n}{n^{2}+1}=0$ Then by the Squeeze Theorem (also known as Sandwich Theorem) $\lim\limits_{n \to \infty}\frac{n sin( n)}{n^{2}+1}=0$ Because the limits exists , the sequence converges. Hence, the given sequence converges to $0$.

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