#### 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 $0$.