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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 ?

Daniel Fischer
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zeraoulia rafik
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  • 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 07 '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 07 '15 at 21:37

2 Answers2

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When people say "zeta function has zeros on negative even integers", they are talking about the analytic continuation of the Riemann zeta function, and the naive formula only works for Re(s)>1

Saberization
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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 $0$.

Ant
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  • ok, so why mathematicians didn't advice us to use the above formula in general , i understand somethings according to the received answers that :zeta(-2) has no general formula and diverge . – zeraoulia rafik Jun 07 '15 at 21:41
  • 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 07 '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 07 '15 at 22:37