Summation using telescopic method

Question

How can I find the telescopic sum of this summation using the telescopic method?
$${\sum_{n = 0}^{\infty}{1 \over (2n+1)^{2}}}$$

Given that the answer is $\pi^2/8$ that will require some very ingenious telescoping... — Myself, Sep 16 '16 at 14:50
I think you're better off just noticing that $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2} \leq \sum_{n=1}^\infty \frac{1}{n^2}$$ if all you're doing is trying to show that the sum is convergent. — Edward Evans, Sep 16 '16 at 14:52
Perhaps not 'true telescoping' but $;\displaystyle \sum_{k = 1}^\infty {1 \over k^{2}}=\sum_{n = 0}^\infty {1 \over (2n+1)^{2}}+{1 \over (2n+2)^{2}};$ should help if you know the series at the left (and thus at the right...) — Raymond Manzoni, Sep 16 '16 at 15:02
I have solved a lot of summation using Telescopic method but this seems a bit complicated, this is why I asked whether it could be done easily or not. How did you guess the answer $\pi^/ 8$ easily? :O — Numerical Person, Sep 16 '16 at 15:15
From the Basel problem we have $;\dfrac{\pi^2}6=\zeta(2)=S+\dfrac{\zeta(2)}4$. A neat direct proof was exposed by Zagier (see $(3)$). — Raymond Manzoni, Sep 16 '16 at 15:50

score 4 · Accepted Answer · edited Apr 13 '17 at 12:21

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.

Summation using telescopic method

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