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