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I have asked a question or two like this one before and I've tryed to use similar methods to prove this identity(?), but I failed. By using WA it seems that numerically the LHS=RHS

$$ \left(\frac{\pi}{3} \right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k+1)\zeta(2k+2)}{3^{2k+2}} $$ But a rigorous proof would be better!

The other similar identity is $$ \left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}} $$

Thanks in advance.

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    $\begingroup$ Why, pray tell, close this question? $\endgroup$ – Ron Gordon Feb 19 '14 at 18:07
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Exactly the same idea as in the linked similar identity by user "Grigory M" works for this.

$$\pi\cot(\pi x)=\frac{1}{x}-2\sum_{n=1}^\infty \zeta(2n)x^{2n-1}.$$

Taking derivatives:

$$\frac{\pi^2}{\sin^2(\pi x)}=x^{-2}+2\sum_{n=1}^\infty (2n-1)\zeta(2n)x^{2n-2}$$

Change the index of summation to $2n=2k+2$. Thus, write the sum

$$\sum_{n=1}^\infty (2n-1)\zeta(2n)x^{2n-2}=\zeta(2)+\sum_{k=1}^\infty (2k+1)\zeta(2k+2)x^{2k}.$$

Then $\zeta(2)=\pi^2/6$, and evaluate both sides at $x=1/3$. You have:

$$\frac{4\pi^2}{3}=9+\frac{\pi^2}{3}+2\sum_{k=1}^\infty \frac{(2k+1)\zeta(2k+2)}{3^{2k}}$$

Carrying over the $\pi^2/3$ and dividing both sides by 9 gives:

$$\left(\frac{\pi}{3}\right)^2=1+2\sum_{n=1}^\infty \frac{(2k+1)\zeta(2k+2)}{3^{2k+2}}$$

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  • $\begingroup$ In the same spirit what would you sugest for $\left(\frac{\pi}{4}\right)^{2}=1+\cdots$? $\endgroup$ – Neves Feb 19 '14 at 17:05
  • $\begingroup$ @Neves: you have two options at your disposal: 1) Taking $k$ derivatives which gives you factors like (2n-1)(2n-2)...(2n-k) infront of your zeta 2) Shifting index of summation For $\pi=4$, a good thing to try is the exact same procedure above, and plug in $x=1/4$. $\endgroup$ – Alex R. Feb 19 '14 at 17:07
  • $\begingroup$ Can that first cotangent identity be proven without the aid of complex analysis? $\endgroup$ – Lucian Feb 20 '14 at 4:27
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The sum

$$\sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = \sum_{n=1}^{\infty} \frac1{9 n^2}\sum_{k=1}^{\infty} \frac{2 k+1}{(9 n^2)^k}$$

from the definition of the zeta function. The inner sum is a geometrical sum and its derivative. In essence, it is simple to derive the following:

$$\sum_{k=1}^{\infty} (2 k+1) r^k = \frac{r (3-r)}{(1-r)^2}$$

Here, $r=1/(9 n^2)$, so we have the sum

$$\sum_{n=1}^{\infty}\frac1{9 n^2} \frac{27 n^2-1}{(9 n^2-1)^2} = -\sum_{n=1}^{\infty} \frac1{9 n^2} + \sum_{n=1}^{\infty} \frac{9 n^2+1}{(9 n^2-1)^2}$$

The first sum is easy ($-\pi^2/54$), so let's attack the second. This may be accomplished via residues. Without going into details, I state the result:

$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{k} \operatorname*{Res}_{z=z_k} \pi \, \cot{(\pi z)} \, f(z)$$

Here, $f(z) = (9 z^2+1)/(9 z^2-1)^2$. The poles of $f$ are at $z_{\pm}=\pm 1/3$. The sum of the above residues at these poles are

$$\pi \sum_{\pm} \left [\frac{d}{dz} \frac{\cot{(\pi z)} (9 z^2+1)}{(3 z \pm 1)^2} \right ]_{z=\pm 1/3} = -\frac{4 \pi^2}{27}$$

$$\implies \sum_{n=1}^{\infty} \frac{9 n^2+1}{(9 n^2-1)^2} = \frac{2 \pi^2}{27} - \frac12 $$

Thus,

$$1+2 \sum_{k=1}^{\infty} \frac{(2 k+1) \zeta(2 k+2)}{3^{2 k+2}} = -\frac{\pi^2}{27} + \frac{4 \pi^2}{27} = \frac{\pi^2}{9}$$

as was to be shown.

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