## Calculus 8th Edition

Given $$\sum_{n=1}^{\infty} \frac{n-1}{n^{3}+1}$$ Use the Limit Comparison Test with $a_{n}=\dfrac{n-1}{n^{3}+1}$ and $b_{n}=\dfrac{1}{ n^2}$ \begin{align*} \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{ n^3-n^2 }{ n^3+1}\\ &=1 \end{align*} since $\displaystyle\sum_{n=1}^{\infty} \frac{1}{ n^2}$ is convergent $(p-\text { series } p>1),$ then $\displaystyle\sum_{n=1}^{\infty} \frac{n-1}{n^{3}+1}$ also convergent