Talk:Analytic Continuation of Riemann Zeta Function

Anybody have any ideas for the rest of part 2? It's clear that both of these series converge, since they're basically alternating series which pass the Alternating Series Test, and the trig functions of the logarithm will eventually hold the same sign for arbitrarily long periods. Nevertheless, we can't actually use the Alternating Series Test, since they will still OCCASIONALLY switch signs. Thoughts? Zelmerszoetrop 19:37, 24 April 2009 (UTC)

Slightly over my head at the moment - I've got some studying to do. I'll probably get to this in a few months or years depending on where my studies take me. --Matt Westwood 19:42, 24 April 2009 (UTC)

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x} \geq \left| \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x}\cos(y\log(n))\right|$, and $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x}$ converges as an alternating series, so the full series converge by the Comparison Test. Does that do it? --Cynic (talk) 21:25, 24 April 2009 (UTC)


 * Sadly, no. The comparison test only works when all the terms are positive; to see where it fails here, observe that your equation shows that the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x}\cos(y\log(n))$ is bounded; however, your suggestion does not rule out the possibility that the partial sums $\displaystyle s_k = \sum_{n=1}^k \frac{(-1)^{n-1}}{n^x}$ will oscillate wildly in the interval $\displaystyle \left({- \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x} }\right) \leq s_k \leq \left({ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^x} }\right)$.
 * Honestly, I'm not even sure if it's legal to split the sum; remember, we can be quite sure the series does NOT converge absolutely. I think it's acceptable, though; and while I think the computational route to proving this convergence in the critical strip lies with seperating the complex exponent into sines and cosines as done here, it does miss the geometric nature of what's going on: since $log(n) \ $ has derivative $1/n \ $, as n goes to infinity, the angle change between two consecutive terms of the series becomes very small and the alternating term will put them on opposite ends of a circle; the effect is that the partial sums will form two spirals which curl tightly into the same point.
 * I'm thinking of trying to demonstrate the series is Cauchy instead of the route I'm taking now, but that's not nearly as developed, I've got to sit down and work that idea out. It might be fruitful, it might not.

NEWS: I think I've got it! See the page! Zelmerszoetrop 16:48, 27 April 2009 (UTC)

Ok, so I've completed parts 1-3, but part 4 is the hardest part of all: demonstrating the validity of the symmetric equation. This could take a while. Zelmerszoetrop 23:15, 27 April 2009 (UTC)


 * Filled in something a little more succinct -- feel free to object/ change it back --Linus44 17:21, 30 April 2011 (CDT)


 * As I say, completely over my head at the moment, so I'll leave it up to someone else to validate. --prime mover 17:51, 30 April 2011 (CDT)