Is the reciprocal of the Zeta function analytically continuable?
As $1/\zeta(n) = \sum_{n=1}^\infty \mu(n)/{n^s}$, this does not look obvious.
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1$\begingroup$ I'm not sure this answers the question you really want to ask, but meromorphic functions always have analytic continuations. $\endgroup$– S. Carnahan ♦Commented Feb 19, 2013 at 2:37
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$\begingroup$ If you can show that $\frac{1}{\zeta (s)}$, which is defined for $re(s)>1$, extends to an $analytic$ function for $re(s)>\frac{1}{2}$ , you will get at least a million dollars. $\endgroup$– VenkataramanaCommented Feb 19, 2013 at 2:44
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1$\begingroup$ ... and a chance to reject it. $\endgroup$– Włodzimierz HolsztyńskiCommented Feb 19, 2013 at 3:00
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2$\begingroup$ I voted to close. Yes, with little knowledge of the Riemann zeta function it is obviously not analytically, but meromorphically continuable to the complex plane. This question is too elementary for this forum. $\endgroup$– Marc PalmCommented Feb 19, 2013 at 9:20
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1 Answer
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That $1/\zeta(s)$ is absolutely and uniformly convergent and therefore analytic for $\Re(s)>1$, and that it has meromorphic continuation to the whole complex plane, is elementary. But the issue of convergence on $1/2\leq\Re(s)\leq 1$ is more interesting.
- at $\Re(s)=1/2$ it diverges
Lucia gave a very nice proof of this fact here on MO.
- at $\Re(s)=1$ it converges
This follows with a bit of work from the estimate
$$\sum_{n<x}\frac{\mu(n)}{n^s}=\frac{1}{2\pi i}\int_{c-it}^{c+it}\frac{1}{\zeta(s+w)}\frac{x^w}{w}dw+O\left(\frac{x^c}{Tc}\right)+O\left(\frac{\log x}{T}\right)$$
which is proved in more generality in section 3.12 of Titchmarsh's book on Riemann zeta function theory. The application to $a_n=\mu(n)$ is worked out in section 3.13.
- at $1/2<\Re(s)< 1$ it converges if and only if the Riemann hypothesis holds
The only if part is obvious. The converse follows from the same kind of estimate as above and the residue theorem. See section 14.25 of Titchmarsh.
