# What's the value of $\sum_{k=1}^{\infty}(\zeta(2k+1)-\zeta(2k+2))$?

I'm confused about this. I have this expression $$\frac{1}{2}=\sum_{k=1}^{\infty}(\zeta(2k)-\zeta(2k+1))$$ Now if I want claculate $\zeta(2)$ I'll do the apropriate manipulations to get $$\zeta(2)=\frac{1}{2}+\sum_{k=1}^{\infty}(\zeta(2k+1)-\zeta(2k+2))$$ but this seems to be wrong, the correct expression seems to be $$\zeta(2)=\frac{3}{2}+\sum_{k=1}^{\infty}(\zeta(2k+1)-\zeta(2k+2))$$ Whats hapening here, what's wrong and what's right? Why can't I go from one expression to the other?

Note, that I've been doing this numerically with WolframAlpha.

• The general case is is explained in this question: $$\sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}(-1)^{n-1}\,\zeta(s+n) = \sum_{n=1}^{\infty}\frac{n^{-s}}{n+1} - \sum_{n=1}^{\infty}\left[\color{red}{\zeta(s+2n-1)-\zeta(s+2n)}\right] \\ \quad = \lim_{N\rightarrow\infty}\sum_{n=1}^{\infty}\frac{n^{-(s+N)}}{n+1} = \color{red}{\frac{1}{2}} \quad\colon\space Re\{s\}\ge0 \quad\{\small\text{holds for s=0 too}\normalsize\}$$ – Hazem Orabi Dec 13 '16 at 22:30

Starting from the first expression, we have

\begin{align*} \frac{1}{2} &= \sum_{k=1}^{\infty} (\zeta(2k) - \zeta(2k+1)) \\ &= \sum_{n=2}^{\infty} (-1)^n (\zeta(n) - 1) \\ &= \zeta(2) - 1 - \sum_{n=3}^{\infty} (-1)^{n-1} (\zeta(n) - 1) \\ &= \zeta(2) - 1 - \sum_{k=1}^{\infty} (\zeta(2k+1) - \zeta(2k+2)) \end{align*}

Rearranging this equality, we get the correct result

$$\zeta(2) = \frac{3}{2} + \sum_{k=1}^{\infty} (\zeta(2k+1) - \zeta(2k+2)).$$

Notice that the series

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

does not converge in ordinary sense, so we need to be careful when manipulating the original series.

Instead of trying to do all of the switching of summations at once, if we take a closer look at the limit, we have for any $\varepsilon>0$ $$\left|\frac{1}{2}-\left[\sum_{k=1}^N\zeta(2k)-\zeta(2k+1)\right]\right|<\varepsilon$$ which gives $$\left|\frac{1}{2}+\left[\sum_{k=1}^{N-1}[\zeta(2k+1)-\zeta(2k+2)]-(\zeta(2)-\zeta(2N+1))\right]\right|<\varepsilon.$$In particular, we see $$\frac12+\sum_{k=1}^{N-1}\zeta(2k+1)-\zeta(2k+2)=\zeta(2)-\zeta(2N+1).$$Taking limits on both sides yields the result.