<s>[INST] Is $\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta (2)$? 
$$\sum_{n=1}^{\infty }\frac{8n\cdot\zeta (2n)}{3\cdot 2^{2n}}=\zeta(2)$$

By using numerical calculation, I found this relationship between the values of zeta function at even integers and $\zeta(2)$, but this needs proving, any help?
 [/INST] Ok, let give me a slightly different answer the Jack,
Our sum is
$$
S=\frac{8}{3}\sum_{n=1}^{\infty}\frac{n}{2^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}=\frac{8}{3}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{n}{(2^2 k^2)^n}\underbrace{=}_{(1)}\frac{8}{3}\sum_{k=1}^{\infty}\frac{4 k^2}{\left(4 k^2-1\right)^2}\underbrace{=}_{(2)}\\\frac{2}{3}\left(\partial_x\cot(1/x)-1+\underbrace{2\sum_{k=1}^{\infty}\frac{1}{k^2\pi^2 x^2-1}}_{-Q(x)}\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(3)}\frac{2}{3}\left(\partial_x\cot(1/x)\right)\big|_{x=\frac{2}{\pi}}\underbrace{=}_{(4)}\frac{\pi^2}{6}
$$
in (1) we just differentiated a geometric series. In (2) we used the Mittag-Leffler expansion of $\cot(1/z)$ and differentiated it
$$
\partial_z\cot(1/z)=\partial_z\left(z+2z\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}\right)=1+\underbrace{2\sum_{k=1}\frac{1}{1-\pi^2 k^2 z^2}}_{Q(z)}+\underbrace{4\sum_{k=1}\frac{\pi^2 k^2 z^2}{(1-\pi^2 k^2 z^2)^2}}_{\frac{3}{2}S \quad\text{if}\quad z=2/\pi}
$$ 
In (3) we again used the Mittag Leffler expansion to show that the sum on the left is $Q(2/\pi)=-1+\cot(1/x)/x\big|_{\frac{2}{\pi}}=-1$ 
In (4) we used $ \partial_z \cot(1/z)=\frac{z^2}{\sin^2(1/z)}$ and $\sin(\pi/2)=1$
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