Answer
convergent
$S=\dfrac{1}{3}$
Work Step by Step
Given $$ \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}$$
Since by using partial fractions
$$\frac{1}{(k+2)(k+3)}=\frac{1}{(k+2) }-\frac{1}{ (k+3)} $$
Then
\begin{align*}
\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k+2}-\frac{1}{k+3}&
=\lim _{n \rightarrow \infty}\bigg\{\left[\frac{1}{1+2}-\frac{1}{Y+3}\right]
+\left[\frac{1}{2+2}-\frac{1}{2+3}\right]\\
&\ \ \ \ +\left[\frac{1}{3+2}-\frac{1}{3+3}\right]+\cdots +\left[\frac{1}{n+2}-\frac{1}{n+3}\right]\bigg\}\\
&=\lim _{n \rightarrow \infty}\left(\frac{1}{3}-\frac{1}{n+3}\right)\\
&=\frac{1}{3}
\end{align*}
Hence the series is convergent and has sum $\frac{1}{3}$
