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I want to compute $$ \sum ^{\infty }_{n=1}\dfrac {\left( 2n+k-2\right) !\zeta \left( 2n\right) \left( -1\right) ^{n-1}}{\left( 2\pi \right) ^{2n}}. $$

This series is surely not convergent for any value of $k$,but moreover it is a divergent series. We shall assign a value to this series by adapted Borel sums. It also grows faster than $n^{2n}$.I need a function depended on "k" This series might be considered from a different angle.I think regularized Borel-sum works ,perhaps you guys give me a little help to simplify this series

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  • $\begingroup$ $\zeta(2n) = (-1)^{n+1} B_{2n} (2\pi)^{2n} / 2(2n)! $ so that the summands simplify to $\frac{(2n+k-2)!}{2(2n)!} B_{2n}$ which is always positive and goes to infinity. It seems to me that the only reasonable value for the series is $+\infty$, no matter what kind of regularisation you're thinking about. $\endgroup$ Jan 20, 2018 at 18:27
  • $\begingroup$ İf we let k be zero the series can be computed by Norlound sum the result is something like 0.0405307... İn deed ,I didnt get the series adapted for any value of k $\endgroup$
    – emre iris
    Jan 20, 2018 at 18:56
  • $\begingroup$ ı realized that this series always alterna series not always positive $\endgroup$
    – emre iris
    Jan 20, 2018 at 22:24
  • $\begingroup$ My personal belief always was that the method of regularization should be determined by the purpose of computing the series, and that the "problem of regularization" has absolutely no meaning by itself . Can you say something about the motivation? $\endgroup$ Jan 21, 2018 at 7:53
  • $\begingroup$ I am dealing with RH,in the progress former series of this series give the result for any value when $ k>0 $ but also I am not able to prove it coz all series diverge,that is why in my humble opinion, if professionals show the way ,at least I understand that it is eihter the key of RH or not $\endgroup$
    – emre iris
    Jan 21, 2018 at 8:09

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