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

Question

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.

Thanks in advance.

Answer 1

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.

Answer 2

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.