# 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.

• This is really about partial fractions decomposition. The hint of O.L is of course right, but if you do not see that, PFD is the way. – imranfat Oct 17 '14 at 17:29
• Search Partial fraction decomposition It is mostly used for simplifying integrals! – Zaid Ajaj Oct 17 '14 at 18:08
• @ZaidAjaj @ imranfat Thanks! I see how to do it now. – Kenshin Oct 17 '14 at 21:29

## 2 Answers

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}

Hint: $$\qquad(2n+3)-(2n+1)=2.\qquad$$