Answer
Convergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{n-1}{n^2 \sqrt{n}}$$
Use the Limit $c$ Comparison Test with $a_n =\dfrac{n-1}{n^2 \sqrt{n}}$ and $b_n=\dfrac{1}{ n^{3/2} }$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{ n^{3/2}(n-1)}{n^2 \sqrt{n}}\\
&=\lim _{n \rightarrow \infty} \frac{ n^{5/2}/n^{5/2}-n^{3/2}/n^{5/2}}{n^{5/2} /n^{5/2}}\\
&=\lim _{n \rightarrow \infty} \frac{1-1/n }{1}\\
&=1
\end{align*}
Since the $p-$ series $p>1$ $\displaystyle \sum_{n=1}^{\infty} \frac{1}{ n^{3/2}}$ is convergent, then $\displaystyle\sum_{n=1}^{\infty}\dfrac{n-1}{n^2 \sqrt{n}}$ is also convergent.