Can I get a closed-form of $\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$? Can I get a closed-form of 
$$\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$$
 A: Let $A$ be the series. Then 
\begin{align}A &= \sum_{k = 1}^\infty (-1)^k \frac{\zeta(2k)}{2^{2k-1}}\\
&= \sum_{k = 1}^\infty \frac{(-1)^k}{2^{2k-1}} \sum_{n = 1}^\infty \frac{1}{n^{2k}}\\
&= 2\sum_{n = 1}^\infty \sum_{k = 1}^\infty \left(-\frac{1}{4n^2}\right)^k\\
&= -2\sum_{n = 1}^\infty \frac{1}{4n^2+1}\\
& = -\frac{1}{4}\sum_{{n = -\infty \atop n \neq 0}}^\infty \frac{1}{n^2 + \frac{1}{4}}\\
&= \frac{1}{4}\left(\sum_{n = -\infty}^\infty \frac{1}{n^2+\frac{1}{4}} - 4\right)\\
&= \frac{1}{4}\left(\operatorname{Res}_{z = \frac{i}{2}} \frac{\pi \cot(\pi z)}{z^2 + \frac{1}{4}} + \operatorname{Res}_{z = -\frac{i}{2}} \frac{\pi \cot(\pi z)}{z^2 + \frac{1}{4}} - 4\right)\\
&= \frac{1}{4}\left(\frac{\pi\cot\left(\frac{\pi i}{2}\right)}{2\left(\frac{i}{2}\right)} + \frac{\pi \cot\left(-\frac{\pi i}{2}\right)}{2\left(-\frac{i}{2}\right)} - 4\right)\\
&= \frac{1}{4}\left(\pi \coth\left(\frac{\pi}{2}\right) + \pi \coth\left(\frac{\pi} {2}\right) - 4 \right)\\
&=\frac{\pi}{2}\coth\left(\frac{\pi}{2}\right) - 1.
\end{align}
A: It appears to be
$$
\frac{\pi}{2}\coth\frac{\pi}{2}-1=\frac{\pi}{e^\pi-1}+\frac{\pi}{2}-1.
$$
