## Calculus 8th Edition

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 $(p-\text { series } p>1),$ then $\displaystyle\sum_{n=1}^{\infty} \frac{n^2+n+1}{n^{4}+n^{2}}$ also convergent