# Riemann Zeta Function and Including Complex Numbers [duplicate]

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,

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

• A little bit of fun, related. "the infinite summation is only valid for $s\gt 1$" - Where here we have $s=-1$ – user142198 Feb 23 '15 at 8:30
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 $-1/12$.
• 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 ! – Antinous Jun 24 '15 at 15:54