<s>[INST] Question on simplification of $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$? I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
 [/INST] Since $2 = (2n+3) - (2n+1)$ then
\begin{align}
\frac{2}{(2n+1)(2n+3)} &= \frac{(2n+3) - (2n+1)}{(2n+1)(2n+3)} = \frac{1}{2n+1} - \frac{1}{2n+3}.
\end{align} 
The series then becomes
\begin{align}
\sum_{n=1}^{\infty} \frac{2}{(2n+1)(2n+3)} &= \sum_{n=1}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+3} \right) \\
&= \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \cdots \\
&= \frac{1}{3}. 
\end{align}
</s>