Proving $\sum_{k=1}^\infty(-1)^{k-1}(\zeta(2k)-1)=\dfrac{\pi\coth(\pi)-2}{2}$

Question

For quite some time now I've been stuck on proving the following:

$$\sum_{k=1}^\infty(-1)^{k-1}(\zeta(2k)-1)=\dfrac{\pi\coth(\pi)-2}{2} $$

I have been working on proving this for quite some time but didn't get any satisfactory result. I'm wondering if one can prove this using contour integration and residue theorem or maybe Fourier series or integrals? I'm most interested in a proof making use of contour integration and integrals.

However any approach is most welcomed. I'm looking for a complete detailed answer.

Thanks in advance.

Answer 1

Using the Taylor series for $\coth(x)$ and the relationship between even Riemann Zeta values and Bernoulli Numbers one gets the following series $$ \sum_{k=0}^{\infty} \frac{ 2^{2k} B_{2k} z^{2k-1}}{(2k)!} = \coth(z) $$ $$ \sum_{k=0}^{\infty} (-1)^{k+1} \zeta(2k) z^{2k} = \coth(\pi z) \frac{\pi z}{2} $$

Some basic rearranging will achieve your series.