Answer
Diverges
Work Step by Step
Let $u_n=\dfrac{n( n+1)}{(n^2+1)(n-1)}$ and $v_n=(\dfrac{ 1}{n})$
Now, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\dfrac{n( n+1)}{(n^2+1)(n-1)}}{1/n}$
$\implies \lim\limits_{n \to \infty} \dfrac{n^2(n+1)}{(n^2+1)(n-1)}=1$
But $v_n=\dfrac{ 1}{n}$ is a divergent series.
Thus, the series diverges due to the limit comparison test.