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This question already has an answer here:

I'm a high-school senior attempting to make sense of the zeta function. I know Riemann regularized it to include complex numbers. Apparently, from this we could obtain that the sum of natural numbers may be assigned the value '-1/12' (zeta of -1)

What I don't get is how did extending the domain to complex numbers help in getting this result.

Thanks,

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marked as duplicate by mrf, Willie Wong, Shobhit, Davide Giraudo, user 170039 Feb 23 '15 at 10:05

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  • $\begingroup$ A little bit of fun, related. "the infinite summation is only valid for $s\gt 1$" - Where here we have $s=-1$ $\endgroup$ – user142198 Feb 23 '15 at 8:30
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The definition $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is valid only for $\Re(s)>1$. It simply does not converge otherwise. By analytically continuing the Riemann zeta function we obtain a function which agrees with $\zeta(s)$ for $\Re(s)>1$, and which is also defined for all $s\in\mathbb{C}\setminus\{1\}$. To answer your question, $-1$ is the complex number $-1+0i$. The analytic continuation of the Riemann zeta function is defined at $-1+0i$ and it evaluates to $-1/12$.

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  • $\begingroup$ Thank you! It makes sense now. $\endgroup$ – TheEconomist Feb 23 '15 at 10:57
  • $\begingroup$ See also this question which shows an analytic continuation of the Riemann zeta function to $\Re (s)>0$... math.stackexchange.com/questions/256992/…... an improvement on $\Re(s)>1$, but still not the whole story ! $\endgroup$ – Antinous Jun 24 '15 at 15:54

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