Is there a closed-form of $\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$ The value of $$\frac{\zeta (2)}{\pi ^2}-\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}-.....=\frac{1}{e^2-1}$$
Is there a closed-form of $$\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$$ 
any help, thanks
 A: Recalling that $$\zeta\left(2n\right)=\left(-1\right)^{n+1}\frac{B_{2n}2^{2n-1}\pi^{2n}}{\left(2n\right)!},\ n\geq1$$the Laurent expansion of $\cot$ function at the origin in terms of the Riemann zeta function is $$\cot\left(z\right)=\frac{1}{z}-2\sum_{k\geq1}\zeta\left(2k\right)\frac{z^{2k-1}}{\pi^{2k}},\ 0<\left|z\right|<\pi$$ so for your series take $z=1$.
A: $$\begin{align}
\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}
&=\sum_{n=1}^{\infty}\frac{1}{\pi^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}\tag{1}\\
&=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{2}\\
&=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{3}\\
&=\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)-1}\tag{4}\\
&=\frac{1}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2-\frac1{\pi^2}}\tag{5}\\
&=\frac{1}{2\pi}\left[\pi-\pi\cot\left(\frac\pi\pi\right)\right]\tag{6}\\
&=\frac{1-\cot1}{2}\tag{7}
\end{align}$$

$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}=\frac{1-\cot1}{2}$$


$\text{Explanation :}$
$\text{6}$ using 
$$\sum_{k=1}^\infty\frac{1}{k^2-z^2}
=\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z}
=\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]$$
