Answer
Convergent.
Work Step by Step
$$
\int_{3}^{\infty} \frac{1}{(x-2)^{3 / 2}} d x
$$
Observe that the given integral is improper
$$
\begin{aligned} \int_{3}^{\infty} \frac{1}{(x-2)^{3 / 2}} d x &=\lim _{t \rightarrow \infty} \int_{3}^{t}(x-2)^{-3 / 2} d x \\
&=\lim _{t \rightarrow \infty}\left[-2(x-2)^{-1 / 2}\right]_{3}^{t} \\
& \quad\quad\quad [u=x-2, d u=d x] \\
&=\lim _{t \rightarrow \infty}\left(\frac{-2}{\sqrt{t-2}}+\frac{2}{\sqrt{1}}\right) \\
&=0+2 \\
&=2 \end{aligned}
$$
It follows that the integral
$$
\int_{3}^{\infty} \frac{1}{(x-2)^{3 / 2}} d x
$$
is convergent.
