Summation using telescopic method How can I find the telescopic sum of this summation using the telescopic method?
$${\sum_{n = 0}^{\infty}{1 \over (2n+1)^{2}}}$$
 A: You cannot directly compute such series (and get $\frac{\pi^2}{8}$) by creative telescoping, but you may use creative telescoping to provide tight approximations of its value. For instance,
$$ \frac{1}{(2n+1)^2} - \color{green}{\frac{1}{(2n+1)(2n+3)}} = \frac{2}{(2n+1)^2(2n+3)} \tag{1}$$
and the green term is a telescopic term, hence
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{1}{(2n+1)^2} &=& \color{green}{\sum_{n\geq 0}\frac{1}{(2n+1)(2n+3)}}+2\sum_{n\geq 0}\frac{1}{(2n+1)^2(2n+3)}\\&=&\color{green}{\frac{1}{2}}+2\sum_{n\geq 0}\frac{1}{(2n+1)^2(2n+3)}\end{eqnarray*}\tag{2}$$
Now we may approximate $\frac{1}{(2n+1)^2(2n+3)}$ with a telescopic term, like $\color{green}{\frac{1}{4}\left(\frac{1}{(2n+1)^2}-\frac{1}{(2n+3)^2}\right)}$, and continue the above procedure, producing an accurate rational approximation of $\frac{\pi^2}{8}$:
$$ \frac{\pi^2}{8}=\color{green}{\frac{4}{3}}-\frac{8}{3}\sum_{n\geq 0}\frac{1}{(4n^2+8n+3)^3}\tag{3}$$
and an accelerated series:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2}=\color{blue}{\frac{9}{4}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}}.\tag{4}$$
The exact value of the series can be computed, for instance, by applying Parseval's theorem to
$$ \sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} \tag{5}$$
that is the Fourier series of a rectangle wave. Another proof lies in this historical thread. It is interesting to point out that the above technique (creative telescoping) also provides a rather short proof of Stirling's inequality, for instance.
