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?
 A: Starting from the Laurent series of the cotangent function:
$$\pi z\cot \left( \pi z \right) =1-2\,\sum _{k=0}^{\infty }\zeta 
 \left( 2\,k+2 \right) {z}^{2k+2} \tag{1}$$
apply the differential operator:
$$\hat{D}=z^2\dfrac{d^2}{dz^2}-2z\dfrac{d}{dz}+2 \tag{2}$$
to get:
$${z}^{3}{\pi }^{3}\cot \left( \pi z \right)  \left( 1+ \cot
 \left( \pi z \right)   ^{2} \right) =1-\sum _{k=0}^{\infty }2k \left( 2k+1 \right)
\,\zeta  \left( 2\,k+2 \right) {z}^{2k+2}\tag{3}$$
which, by the ratio test, has a radius of convergence of $|z|<1$. Then from:
$$z=\dfrac{1}{4}, \quad \cot\left(\dfrac{\pi}{4}\right)=1 \tag{4}$$
we have:
$$\dfrac{{\pi }^{3}}{32}=1-\sum _{k=0}^{\infty }{\frac {2k \left( 2\,k+1
 \right) \zeta  \left( 2\,k+2 \right) }{{4}^{2k+2}}}\tag{5}$$
A: 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 $\dfrac{1}{\cos z}$:
$$\sum_{\nu=0}^\infty \frac{(-1)^\nu}{(2\nu+1)^3} = - \frac{\pi^3}{32} E_2 = \frac{\pi^3}{32}.$$
A: $$ 
\begin{align} 
S\, &=\,\sum_{k=1}^{\infty}\frac{2k\,(2k+1)\,\zeta(2k+2)}{4^{2k+2}} \,=\,\sum_{k=1}^{\infty}\frac{(2k+1)!\,\zeta(2k+2)}{(2k-1)!\,4^{2k+2}} \\[4mm] 
&=\,\sum_{k=1}^{\infty}\frac{(1/4)^{2k+2}}{(2k-1)!}\,\int_{0}^{\infty}\frac{x^{2k+1}}{e^x-1}\,dx \,=\,4^{-3}\int_{0}^{\infty}\frac{x^2}{e^x-1}\,\left(\sum_{k=1}^{\infty}\frac{(x/4)^{2k-1}}{(2k-1)!}\right)\,dx \\[4mm] 
&=\,4^{-3}\int_{0}^{\infty}\frac{x^2}{e^x-1}\,\sinh\left(\frac{x}{4}\right)\,dx \,=\,\frac{4^{-3}}{2}\int_{0}^{\infty}\frac{x^2}{1-e^{-x}}\,\left({\Large e}^{-\frac34x}\,-\,{\Large e}^{-\frac54x}\right)\,dx \\[4mm] 
&=\,4^{-3}\left[\zeta\left(3,\frac34\right)\,-\,\zeta\left(3,\frac54\right)\right] \,=\,1-4^{-3}\left[\zeta\left(3,\frac14\right)\,-\,\zeta\left(3,\frac34\right)\right] \\[4mm] 
&=\,1-\beta(3) \,=\,{\color\red{1-\frac{\pi^3}{32}}} \\ 
\end{align} 
$$ 

$\,\zeta(s,q) \,\,$ : Hurwitz Zeta Function
$\,\beta(s)\quad\,$ : Dirichlet Beta Function

