<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, you are correct that the sum telescopes quite nicely. The partial sum formula is$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left[\left(1+\frac 12+\cdots+\frac 1m\right)-\left(\frac 14+\frac 15+\cdots+\frac 1{m+3}\right)\right]$$Notice how anything past $\frac 14$ in the first sum is automatically canceled from the second sum. Hence, the partial sum formula is given as$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left(1+\frac 12+\frac 13-\frac 1{m+1}-\frac 1{m+2}-\frac 1{m-3}\right)$$As $m\to\infty$, the fractions containing $m$ vanish, leaving$$\sum\limits_{n\geq1}\frac 1{n(n+3)}\color{blue}{=\frac {11}{18}}$$
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