<s>[INST] Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _{n=1}^{\infty }\:\frac{1}{n}-\frac{1}{n+3} = \frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots\right)
$$
Wich means some values are zeroed, does that mean it's a telescoping sum?
I also know that  $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} \le \sum _{n=1}^{\infty }\:\frac{1}{n^2}\:$ so the series converges by the harmonic p-series.
 [/INST] Yes, it is telescoping,
\begin{align}
\sum_{n=1}^N \frac{1}{n(n+3)}&= \frac13 \sum_{n=1}^N \left[ \frac1{n}-\frac{1}{n+3}\right]\\
&= \frac13\left[1-\frac{1}{4} +\frac12-\frac1{5}+\frac13-\frac16+\ldots +\frac1{N-2}-\frac1{N+1}+\frac1{N-1}-\frac1{N+2}+\frac1N-\frac1{N+3}\right]\\
&=\frac13\left[1+\frac12+\frac13-\frac1{N+1}-\frac{1}{N+2}-\frac1{N+3} \right]
\end{align}
Now take limit $N \to \infty$,
$$\sum_{n=1}^N \frac{1}{n(n+3)}=\frac13\left[1+\frac12+\frac13 \right]$$
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