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?
Using the formula for a geometric series,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{x^{2k}}
&amp;=\frac1{x^2-1}\\
&amp;=\frac12\left(\frac1{x-1}-\frac1{x+1}\right)\tag{1}
\end{align}
$$
Differentiating $(1)$ twice,
$$
\sum_{k=1}^\infty\frac{2k(2k+1)}{x^{2k+2}}
=\frac1{(x-1)^3}-\frac1{(x+1)^3}\tag{2}
$$
Changing the order of summation and applying $(2)$,
$$
\begin{align}
1-\sum_{k=1}^\infty\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}
&amp;=1-\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\
&amp;=1-\sum_{j=1}^\infty\sum_{k=1}^\infty\frac{2k(2k+1)}{(4j)^{2k+2}}\\
&amp;=1-\sum_{j=1}^\infty\left(\frac1{(4j-1)^3}-\frac1{(4j+1)^3}\right)\\
&amp;=1-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)^3}\\
&amp;=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\tag{3}
\end{align}
$$
The sum in $(3)$ can be generalized as
$$
\beta(s)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}\tag{4}
$$
and is known as the Dirichlet beta function. As shown below, $\beta(3)=\dfrac{\pi^3}{32}$.



We can develop a recurrence for $\beta(2k+1)$. First, the generating function for $\beta(2k+1)$ is
$$
\begin{align}
f(x) &amp;= \sum_{k=0}^\infty \beta(2k+1) x^{2k+1}\\
     &amp;= \sum_{n=0}^\infty \sum_{k=0}^\infty (-1)^n\left(\frac{x}{2n+1}\right)^{2k+1}\\
     &amp;= \sum_{n=0}^\infty (-1)^n\frac{\frac{x}{2n+1}}{1-\left(\frac{x}{2n+1}\right)^2}\\
     &amp;= \frac{x}{2} \sum_{n=0}^\infty (-1)^n \left(\frac{1}{2n+1+x}+\frac{1}{2n+1-x}\right)\\
     &amp;= \frac{x}{2} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{2n+1+x}\\
     &amp;= \frac{x}{4} \sum_{n=-\infty}^{+\infty}(-1)^n \frac{1}{n+\tfrac{1+x}{2}}\\
     &amp;= \frac{x}{4} \pi \csc\left(\pi\tfrac{1+x}{2}\right)\\[3pt]
     &amp;= \frac{\pi}{4} x \sec\left(\frac{\pi}{2}x\right)\tag{5}
\end{align}
$$
where we use $(7)$ from this answer to get
$$
\begin{align}
\sum_{n=-\infty}^{+\infty}\frac{(-1)^n}{n+z}
&amp;=\sum_{n=-\infty}^{+\infty}\frac2{2n+z}-\sum_{n=-\infty}^{+\infty}\frac1{n+z}\\[3pt]
&amp;=\pi\cot(\pi z/2)-\pi\cot(\pi z)\\[9pt]
&amp;=\pi\csc(\pi z)\tag{6}
\end{align}
$$
We can use equation $(5)$ to develop a recurrence relation:
$$
\begin{align}
\frac{\pi}{4} x
&amp;= \cos\left(\frac{\pi}{2} x\right) f(x)\\
&amp;= \sum_{n=0}^\infty\sum_{k=0}^n (-1)^k \frac{(\frac{\pi}{2} x)^{2k}}{(2k)!}\;\beta(2n-2k+1)x^{2n-2k+1}\\
&amp;= \sum_{n=0}^\infty\sum_{k=0}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)x^{2n+1}\tag{7}
\end{align}
$$
For $n=0$, we can use the arctan series to get
$$
\beta(1) = \frac{\pi}{4}\tag{8}
$$
and for $n\gt0$, $(7)$ gives
$$
\beta(2n+1) = -\sum_{k=1}^n \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2n-2k+1)\tag{9}
$$
Recursion $(9)$ yields
$$
\begin{align}
\beta(1)&amp;=\frac{\pi}{4}\\
\beta(3)&amp;=\frac{\pi^3}{32}\\
\beta(5)&amp;=\frac{5\pi^5}{1536}\\
\beta(7)&amp;=\frac{61\pi^7}{184320}\\
\beta(9)&amp;=\frac{277\pi^9}{8257536}\\
\beta(11)&amp;=\frac{50521\pi^{11}}{14863564800}\\
\beta(13)&amp;=\frac{540553\pi^{13}}{1569592442880}\\
\beta(15)&amp;=\frac{199360981\pi^{15}}{5713316492083200}\\
\beta(17)&amp;=\frac{3878302429\pi^{17}}{1096956766479974400}\\
\beta(19)&amp;=\frac{2404879675441\pi^{19}}{6713375410857443328000}
\end{align}
$$
First, Mathematica confirms this. For the proof, I would assume that expanding out the zeta and then changing the order of summation will do the trick (you will get sums of the sort
$\sum_{k=1}^\infty 2k(2k+1)\frac{1}{{(4 i)}^{2k+2}},$ which are easy to evaluate (by integrating the sum of the geometric series a couple of times):

$$\frac{2 \left(48 i^2+1\right)}{\left(16 i^2-1\right)^3}$$