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[tex]\sum_{n = 0}^{ + {\infty}} \exp {in{\theta}} = \frac {1}{2} + \frac {1}{2} i\cot {\frac {{\theta}}{2}}[/tex]

I was able to obtain (by differentiation) valid expressions for the analytic continuation of the [tex]{\zeta}[/tex] series for negative integers(and 0)

Putting [tex]{\theta} = {\pi} + {\theta}[/tex]

in the real part of the above equation I have,

[tex]\sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} \cos {n{\theta}} = \frac {1}{2}[/tex]

I proceeded to integrate this equation 0 to [tex]{\theta}[/tex] 2k times in the hopes of finding the general formula for [tex]{\zeta}(2k)[/tex]

However I am stuck at,

[tex]\frac {( - 1)^{k}{\theta}^{2k}}{2{(2k)}!} = \sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} n^{ - 2k} \cos {n{\theta}} + \sum_{n = 1}^{ + {\infty}} ( - 1)^{n - 1} n^{ - 2k} \sum_{r = 0}^{k - 1} ( - 1)^{r - 1} \frac {n^{2r}}{{(2r)}!} {\theta}^{2r}[/tex]

No idea what to make of that troublesome finite series or how to introduce the Bernoulli numbers to the mix!

Any assistance would be greatly appreciated! Thanks!

PS- people who'll feed me the convergence issues shouldn't even bother replying! I need this derivation only, I know many other proofs but none so interesting. I think the Bernoulli numbers might come into the mix somewhere between the calculation!