Answer
Diverges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{n }{(n^2+1)^{3/5}}$$
Since $f(n)= \frac{n }{(n^2+1)^{3/5}},\ \ f'(n) = \frac{-n^2+5}{5\left(n^2+1\right)^{\frac{8}{5}}} $, is positive, decreasing and continuous for $n\geq1$, then by using the integral test
\begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{n }{(n^2+1)^{3/5}}d n \\
&=\left.\lim _{t \rightarrow \infty}\frac{ 5}{4}(n^2+1)^{2/5}\right|_{1} ^{t} \\
&= \lim _{t \rightarrow \infty} \frac{ 5}{4}(t^2+1)^{2/5}-\lim _{t \rightarrow \infty} \frac{ 5}{4}(2)^{2/5} \\ &=\infty\end{aligned}
Thus the series diverges.
