Answer
The series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges.
Work Step by Step
Using the limit comparison test by $b_n=\frac{1}{n^2}$, now we have
$$L=\lim_{n\to \infty} \frac{a_n}{b_n}= \lim_{n\to \infty} \frac{1/(n^2+\sin n)}{1/n^2}\\
=\lim_{n\to \infty} \frac{n^2}{n^2+\sin n}=\lim_{n\to \infty} \frac{1}{1+(\sin n/n^2)}=1$$
Since $\Sigma_{n=1}^{\infty}\frac{1}{n^2}$ is a convergent p-series and $L=1\gt 0$ then the series $\Sigma_{n=1}^{\infty} \frac{1}{n^2+\sin n}$ converges.
