Answer
Converges
Work Step by Step
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.
