## Calculus (3rd Edition)

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