<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] Taking the derivative of
$$
\sum_{n=0}^\infty x^n=\frac1{1-x}\tag{1}
$$
and multiplying by $x$ yields
$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}\tag{2}
$$
Therefore,
$$
\hspace{-1cm}\begin{align}
\frac83\sum_{n=1}^\infty\frac{n}{2^{2n}}\zeta(2n)
&=\frac83\sum_{n=1}^\infty\frac{n}{4^n}\sum_{k=1}^\infty\frac1{k^{2n}}\tag{3a}\\
&=\frac83\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{n}{\left(4k^2\right)^n}\tag{3b}\\
&=\frac83\sum_{k=1}^\infty\frac{4k^2}{\left(4k^2-1\right)^2}\tag{3c}\\
&=\frac23\sum_{k=1}^\infty\left[\frac1{(2k-1)^2}+\frac1{(2k+1)^2}+\frac1{2k-1}-\frac1{2k+1}\right]\tag{3d}\\
&=\frac23\sum_{k=1}^\infty\left[\frac2{(2k-1)^2}-\left(\frac1{(2k-1)^2}-\frac1{(2k+1)^2}\right)+\left(\frac1{2k-1}-\frac1{2k+1}\right)\right]\tag{3e}\\
&=\frac23\sum_{k=1}^\infty\frac2{(2k-1)^2}\tag{3f}\\[6pt]
&=\zeta(2)\tag{3g}
\end{align}
$$
Explanation:
$\text{(3a)}$: expand $\zeta(2n)$
$\text{(3b)}$: change the order of summation
$\text{(3c)}$: apply $(2)$
$\text{(3d)}$: partial fractions
$\text{(3e)}$: unify the sum of the odd squares by introducing a telescoping series
$\text{(3f)}$: the telescoping series cancel: $-1+1=0$
$\text{(3g)}$: the sum of the reciprocals of the odd squares is $\frac34$ of the sum of the reciprocals of all the squares
</s>