Answer
Converges
Work Step by Step
Let us consider $a_n=\dfrac{n}{ n^2+1}$and $u_n$ refers to all positive values of $n$.
Here, $f(n)=\dfrac{n}{ n^2+1}$;$f'(n)=\dfrac{-(n^2-1)}{(n^2+1)^2} \leq 0$
The negative sign simplifies that the sequence $u_n$ is not increasing.
Then, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{n}{(n^2+1)}=\lim\limits_{n \to \infty} \dfrac{1/n}{(1+1/n^2)}=0$
Hence, the series converges by the Alternating Series Test.