Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{\tan h n}{n^2}$ and $b_n=\dfrac{1}{ n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{\tan h n}{n^2}}{\dfrac{1}{ n^2}}$
or, $ =\lim\limits_{n \to \infty} \tanh n$
and $\lim\limits_{n \to \infty} \tanh n=1 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ is a convergent series due to the p-series test with $p \gt 1$
Thus, the series converges by the limit comparison test.
