## Calculus (3rd Edition)

Given $$\sum_{n=2}^{\infty} \frac{1}{n^2(\ln n)^{3}}$$ Compare with convergent series $\sum_{n=2}^{\infty} \frac{1}{n^2 }$ ($p-$series $p>1$) , then 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^2(\ln n)^{3}}\\ &= \lim_{n\to \infty} \frac{1}{ (\ln n)^{3}}\\ &=0 \end{align*} Thus the given series also converges.