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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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I'm not sure this answers the question you really want to ask, but meromorphic functions always have analytic continuations. – S. Carnahan Feb 19 '13 at 2:37
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. – Venkataramana Feb 19 '13 at 2:44
... and a chance to reject it. – Włodzimierz Holsztyński Feb 19 '13 at 3:00
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. – Marc Palm Feb 19 '13 at 9:20

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