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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,

TheEconomist
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1 Answers1

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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 $-1/12$.

pshmath0
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  • Thank you! It makes sense now. – TheEconomist Feb 23 '15 at 10:57
  • See also this question which shows an analytic continuation of the Riemann zeta function to $\Re (s)>0$... http://math.stackexchange.com/questions/256992/derivative-of-the-riemann-zeta-function-for-res0... an improvement on $\Re(s)>1$, but still not the whole story ! – pshmath0 Jun 24 '15 at 15:54