Is the Riemann Zeta Function of 0 Infinity or -1/2?

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The riemann zeta function of 0 is supposed to be -1/2. But isn't it infinity?

1/1^0 +1/2^0+1/3^0+1/4^0+...
1/1+1/1+1/1+1/1+...
1+1+1+1+...
infinity

But many sources claim that it is -1/2
 
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dimension10 said:
The riemann zeta function of 0 is supposed to be -1/2. But isn't it infinity?

1/1^0 +1/2^0+1/3^0+1/4^0+...
1/1+1/1+1/1+1/1+...
1+1+1+1+...
infinity

But many sources claim that it is -1/2

Yes, you are correct, but the definition af the Riemann-zeta function as

[tex]\zeta (s)=\sum_{n=1}^{+\infty}{\frac{1}{n^s}}[/tex]

is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].
 
micromass said:
is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].

The sum has infinitely many continuous extensions to the real line, but only one analytical extension to the complex plane.
 
disregardthat said:
The sum has infinitely many continuous extensions to the real line, but only one analytical extension to the complex plane.

Well, a meromorphic extension really, since there is a pole in 1.
 
micromass said:
Well, a meromorphic extension really, since there is a pole in 1.

You are right about that :redface:
 
micromass said:
Yes, you are correct, but the definition af the Riemann-zeta function as

[tex]\zeta (s)=\sum_{n=1}^{+\infty}{\frac{1}{n^s}}[/tex]

is only defined as s>1 (or in the complex case, as Re(s)>1). The Riemann-zeta function for [itex]s\leq 1[/itex] is defined as the "analytic continuation" of the above function. That is: the unique function that looks most like [itex]\zeta(s),s>1[/itex].

Thanks, but is there a formula for s<1?