# Closed form for $\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$

From WolframAlpha it seems that $$\frac{1}{2}=\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$$

Could someone provide a proof for this?

Thanks.

• Where in WA is written that? – DonAntonio Jan 13 '14 at 11:09
• ...WA provided $\approx 0.4999999999999...$ not exactly $=1/2$. – Neves Jan 13 '14 at 11:13
• Not that I asked, @Neves ... – DonAntonio Jan 13 '14 at 11:16
• Are you sure that the summation starts at k=0 ? Not k=1 instead ? – Claude Leibovici Jan 13 '14 at 11:16
• @claude, corrected that. – Neves Jan 13 '14 at 11:17