Answer
Converges
Work Step by Step
Let $u_n=\dfrac{\tanh (n)}{n^2}$ and $v_n=(\dfrac{1}{ n^2})$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{\tan h (n)}{n^2}}{1/n^2}$
$ \implies \lim\limits_{n \to \infty} \tanh (n)=1 \ne 0 \ne \infty $
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{(2)}}$ shows a convergent $p$-series test with $p \gt 1$.
Hence, the series converges by the limit comparison test.