Answer
Divergent
Work Step by Step
Given $$ \sum_{n=1}^{\infty} \frac{n^{2}}{n^{3}+1}$$
Since $f(x)=\dfrac{x^{2}}{x^{3}+1}$ is continuous and positive on $[2, \infty)$ and
$$ f^{\prime}(x)=\frac{x\left(2-x^{3}\right)}{\left(x^{3}+1\right)^{2}}<0$$
for all $x\geq 2 $, then by using the Integral Test, we get
\begin{align*}
\int_{2}^{\infty} \frac{x^{2}}{x^{3}+1} d x&=\frac{1}{3}\lim _{t \rightarrow \infty}\left( \ln \left(x^{3}+1\right)\right)\bigg|_{2}^{t}\\
&=\frac{1}{3} \lim _{t \rightarrow \infty}\left(\ln \left(t^{3}+1\right)-\ln 9\right)\\
&=\infty
\end{align*}
Hence $\displaystyle\sum_{n=2}^{\infty} \frac{n^{2}}{n^{3}+1}$ diverges and so does the given series $\displaystyle\sum_{n=1}^{\infty} \frac{n^{2}}{n^{3}+1}$
