Answer
convergent
using Limit Comparison Test
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{n }{(n^2+2)^2}$$
Use the Limit Comparison Test with $a_n =\dfrac{n }{(n^2+2)^2}$ and $b_n=\dfrac{ 1}{ n^{3}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n^4 }{(n^2+2)^2}\\
&=\lim _{n \rightarrow \infty} \frac{n^4 }{n^4+4n^2+4}\\
&=\lim _{n \rightarrow \infty} \frac{1}{1+4/n^2+4/n^4}\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{ n^3}$ is convergent ($p-$series series $p>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{n }{(n^2+2)^2}$ is also convergent