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I recently watched a video on youtube by a group called Numberphile. Theyre were discussing the Riemann Hypothesis and I was confused when they brought up 'Analytical Continuance' and 'Holomorphic Functions'.

Now where I am confused is when they bring up the analytical continuance of this function.

How does

$$ \zeta(-1)=\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{3^{-1}}+...=1+2+3+...=-\frac{1}{12}$$ Doesnt this series diverge since the sum of the natural numbers diverges??

Jyrki Lahtonen
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silversurfer
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  • That video is extremely misleading. It is true that $\zeta(-1) = -\frac1{12}$. However, it's emphatically false that $\zeta(-1) = 1+2+3+\cdots$. The normal definition of $\zeta(s)$ is valid only for $\Re(s) >1$, since, as you've correctly noticed, it diverges if $\Re(s)\le 1$. The function can be extended (analytically continued) to the whole complex plane (except for $s=1$). This is analogous to the way that $f(z) = 1+z+z^2+\cdots$ is a geometric series that converges only for $|z|<1$, however the function $g(x) = \frac{1}{1-z}$ agrees with $f$ where it is defined and is valid for $z\ne 1$. – Mathmo123 Feb 02 '16 at 09:07

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