Answer
Divergent
Work Step by Step
Given $$\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n-1}$$
Use the Limit Comparison Test with $a_n =\dfrac{n^{3}}{n^{4}-1}$ and $b_n=\dfrac{1}{\sqrt{n}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{n}{n-1}\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ is divergent ( $p-$series , $p<1$) , then $\displaystyle\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n-1}$ is also divergent