# Why $\zeta(-2)$ is not $\sum_{n=1}^{\infty}\frac{1}{n^{-2}}$? [duplicate]

Let $\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{n^{s}}$ a standard formula.

I'm confused if you tell me: does this series: $\sum_{n=1}^{\infty}\frac{1} {n^{s}}$ converge?

I will answer you: this series is divergent. But if you say: $\zeta(-2)$ it will be: $\zeta(-2)= \sum_{n=1}^{\infty}\frac{1}{n^{-2}}=0$. Will be convergent. So why ?

## marked as duplicate by Cameron Williams, user147263, Grigory M, Claude Leibovici, dawJun 9 '15 at 6:25

• Similar questions has been asked before, see for example this. Note that this link talks about $\zeta(-1)$, but for what is the essence of your questions this is the same thing. See also this – Winther Jun 7 '15 at 21:32
• The fact that the analytic continuation has a finite value (namely, $0$) does not vouch for convergence of the series. A simpler example: $1+2+2^2+2^3+\ldots$ arguably blows up, yet this is the analytic continuation to $x=2$ of $1/(1-x)$, so the analytic continuation takes value $-1$. Hard to rationalize how the sum of positive reals could be negative. (Hilariously, in the $2$-adics, that conclusion is literally correct, but that is tangential to the present issue.) – paul garrett Jun 7 '15 at 21:37
See this, which gives a functional equation for $\zeta{(s)}$: $$\zeta{(s)} = 2^s \pi^{s-1} \sin{\left(\frac{\pi s}{2}\right)} \Gamma{(1-s)} \zeta{(1-s)}$$ All negative even integers are "trivial zeros" of the zeta function, because $\sin{\left(\frac{\pi s}{2}\right)}$ would equal $0$.
• I'm guessing that, when you say "in general", you mean that $s$ is a positive integer. In that case, $\Gamma{(1-s)}$ is the $\Gamma$ of a negative integer, which is undefined. – Ant Jun 7 '15 at 21:45
• @zera: Mathematicians do advise us to use analytic continuation when we talk about $\zeta(s)$ with $\mathrm{Re}\;s \le 1$. – GEdgar Jun 7 '15 at 22:37