<s>[INST] Proving that $\left(\frac{\pi}{3} \right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k+1)\zeta(2k+2)}{3^{2k+2}}$. 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.
 [/INST] 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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