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Riemann Zeta Function and Analytic Continuation
Calculating the Zeroes of the Riemann-Zeta function

It is stated that Riemann's Zeta function has zeros at negative even numbers.For example if -2 is directly substituted, since the index is in the denominator each term will become a positive number.Then how does the value become zero

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marked as duplicate by J. M., Did, rschwieb, Martin Argerami, froggie Nov 25 '12 at 16:39

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The series you have in mind is valid only for arguments whose real parts are greater than $1$. One then uses analytic continuation to have the zeta function be defined for the rest of the complex plane; this is one way to obtain values for the negative even numbers. –  J. M. Nov 25 '12 at 12:36
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The definition $\zeta(s) = \sum{n^{-s}}$ is only valid for $\rm{Re}(s) \gt 1$. Determination of its values for other arguments requires analytic continuation. See: Zeta regularization.

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