<s>[INST] Find $ \sum_{n=1}^{\infty} \frac{1}{n(n+2)} $ I am doing the telescoping technique with partial sum of this infinite series because I want to find an upper bound of its partial sum.
 \begin{equation} 
\sum_{n=1}^{\infty} \frac{1}{n(n+2)}
\end{equation} 
First I set $ \frac{1}{n(n+2)}= \frac{A}{n}+\frac{B}{n+2}$ and solve to get $A=\frac{1}{2}$  and  $B=-\frac{1}{2}$.
So I'm finding $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{2n+4})$
and proceed with the partial sum expansion.
At the end I am left with $S_N=\frac{1}{2}+\frac{1}{4}-\frac{1}{2N+4}$ by cancelling terms in between and then $\lim_{N \to \infty}S_N=\frac{3}{4}$.
Am I doing it correctly? Thank you for your time.
 [/INST] $$\sum^{\infty}_{n=1}\frac{1}{n(n+2)} = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\frac{1}{n}-\frac{1}{n+2}\bigg]$$
$$ = \frac{1}{2}\sum^{\infty}_{n=1}\bigg[\bigg(\frac{1}{n}-\frac{1}{n+1}\bigg)+\bigg(\frac{1}{n+1}-\frac{1}{n+2}\bigg)\bigg]$$
$$ = \frac{1}{2}\bigg[\frac{1}{1}+\frac{1}{2}\bigg] = \frac{3}{4}$$
</s>