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Dec 20, 2014 at 18:04 vote accept GH from MO
Dec 6, 2014 at 20:49 history edited GH from MO CC BY-SA 3.0
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Dec 6, 2014 at 20:22 history edited GH from MO CC BY-SA 3.0
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Dec 6, 2014 at 20:09 history edited GH from MO CC BY-SA 3.0
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Dec 6, 2014 at 19:54 comment added juan @GH from MO I have added a little explanation about how (4) implies (5) for $\sigma<0$
Dec 6, 2014 at 19:54 history edited juan CC BY-SA 3.0
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Dec 6, 2014 at 12:40 comment added GH from MO Thanks! I also like your proof, I find (7) natural and revealing. Can you add more detail, for the sake of the readers, how (4) implies (5)? I agree that a simpler proof might exist. Let us wait two weeks, and if there is no better proof, I will accept this one officially.
Dec 6, 2014 at 12:32 comment added godelian Given the fact that what you end up explaining why the series, considered as a Laurent series, predicts also the values of $\zeta$ at the negative even numbers, I think this proof is as simple as it reasonably can be expected to be. This because you are forced to consider the analytic continuation of $\zeta$ to the left half plane, not just to the critical strip, so it's reasonable that the functional equation is invoked.
Dec 6, 2014 at 12:08 history edited Emil Jeřábek CC BY-SA 3.0
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Dec 6, 2014 at 11:23 history answered juan CC BY-SA 3.0