Answer
Divergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{\sqrt{n^{4}+1}}{n^{3}+n^{2}}$$
Use the Limit Comparison Test with $a_n =\dfrac{\sqrt{n^{4}+1}}{n^{3}+n^{2}}$ and $b_n=\dfrac{1}{n}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n\sqrt{n^{4}+1}}{n^{3}+n^{2}}\\
&=\lim _{n \rightarrow \infty} \frac{n/n^3\sqrt{n^{4}+1}}{n^{3}/n^3+n^{2}/n^3}\\
&=\lim _{n \rightarrow \infty} \frac{1/n^2\sqrt{n^{4}+1}}{1+1/n }\\
&=\lim _{n \rightarrow \infty} \frac{ \sqrt{1+1/n^{4}}}{1+1/n }\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ is divergent , then $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n^{4}+1}}{n^{3}+n^{2}}$ is also divergent.