Can I get a closed-form of $$\frac{\zeta(2) }{2}-\frac{\zeta (4)}{2^3}+\frac{\zeta (6)}{2^5}-\frac{\zeta (8)}{2^7}+\cdots$$
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4Same as what was done here and put $x = \dfrac{i}{2}$, where $i = \sqrt{-1}$. – r9m Dec 29 '14 at 19:38
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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}
kobe
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It appears to be $$ \frac{\pi}{2}\coth\frac{\pi}{2}-1=\frac{\pi}{e^\pi-1}+\frac{\pi}{2}-1. $$
Charles
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PARI/GP and RIES. I haven't proved it, though -- I just verified it to 10,000 decimal places. – Charles Dec 29 '14 at 20:17