# Riemann Zeta function - Zeta(0) != Infinity ??

• Mar 7th 2009, 08:43 PM
Mentia
Riemann Zeta function - Zeta(0) != Infinity ??
I am obviously missing something crucial. So this is how wikipedia defines the Riemann Zeta:

And then a few lines down says:

How is this possible? If I put in a zero for each s in the series, I get an infinite series of 1 + 1 + 1 + 1 + ... which must be infinity.

I realize that s can be a complex number... but i'm just not seeing this. Please help.
• Mar 8th 2009, 12:15 AM
Opalg
The formula $\displaystyle \zeta(s) = \sum_{n=1}^\infty\frac1{n^s}$ is only valid when the real part of s is greater than 1. However, the function ζ(s) defined in that way is analytic, and it can be extended by analytic continuation to the rest of the complex plane except for the point s=1 where there is a simple pole. You can find more details about the construction here.
• Mar 8th 2009, 12:25 AM
Amanda1990
If we use analytic continuation, what is should zeta(1) be? Is there a sensible way of defining 1 + 1/2 + 1/3 + ...?
• Mar 8th 2009, 12:46 AM
Opalg
Quote:

Originally Posted by Amanda1990
If we use analytic continuation, what is should zeta(1) be? Is there a sensible way of defining 1 + 1/2 + 1/3 + ...?

As I said above, there is a pole at s=1, so the function becomes infinite at that point, as you would expect from the fact that harmonic series 1 + 1/2 + 1/3 + ... diverges.
• Mar 8th 2009, 12:50 AM
Amanda1990
Fair enough. But is there *any* sensible way of assigning a finite value to the sum of the harmonic series? (Not using the zeta function). Just out of interest!