Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$ After numerical analysis it seems that 
$$
\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
$$
Could someone prove the validity of such identity?

Yes, we can prove it. We can change the order of summation in
$$\begin{align}
\sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
&= \sum_{k=1}^\infty \frac{2k(2k+1)}{4^{2k+2}}\sum_{n=1}^\infty \frac{1}{n^{2k+2}}\\
&= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{2k(2k+1)}{(4n)^{2k+2}}\\
&= \sum_{n=1}^\infty r''(4n),
\end{align}$$
where, for $\lvert z\rvert > 1$, we define
$$r(z) = \sum_{k=1}^\infty \frac{1}{z^{2k}} = \frac{1}{z^2-1} = \frac12\left(\frac{1}{z-1} - \frac{1}{z+1}\right).$$
Differentiating yields $r''(z) = \frac{1}{(z-1)^3} - \frac{1}{(z+1)^3}$, so
$$1 - \sum_{k=1}^\infty \frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} = \sum_{\nu = 0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3},$$
and the latter sum is by an earlier answer using the partial fraction decomposition of $\dfrac{1}{\cos z}$:
$$\sum_{\nu=0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3} = - \frac{\pi^3}{32} E_2 = \frac{\pi^3}{32}.$$
