## Calculus (3rd Edition)

Given $$\sum_{n=2}^{\infty} \frac{n^2}{n^4-1}$$ Compare with the convergent series $\displaystyle \sum_{n=2}^{\infty}\frac{1}{n^2}$ and by using the Limit Comparison Test, we get: \begin{align*} \lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty} \frac{n^2}{n^4-1}n^2\\ &=1 \end{align*} Then $\displaystyle \sum_{n=2}^{\infty}\frac{n^2}{n^4-1}$ also converges