Answer
Converges
Work Step by Step
Consider $u_n=\dfrac{n}{ n^2+1}$
Here, $u_n$ is positive for all the values of $n$.
Now, $f(n)=\dfrac{n}{ n^2+1}$
and $f'(n)=\dfrac{-n^2+1}{(n^2+1)^2} \leq 0$
The negative sign shows that, the sequence $u_n$ is not increasing.
Thus, $\lim\limits_{n \to \infty} u_n=\dfrac{n}{ n^2+1}=0$
Hence, by the Alternating Series Test, the series converges.
