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Does the following infinite sum converges for real x satisfying $0<x<1$ and if so are there any closed form for it ?

$$ -\sum_{k=1}^{\infty} \zeta(1-2k)x^{2k} $$

Thanks for your help.

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    I think the anwer is no. –  Nov 07 '20 at 13:24
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    Possibly useful: https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function#Negative_integers – preferred_anon Nov 07 '20 at 13:27
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    Have you tried ratio test to derive the radius of convergence of this sum? – uranix Nov 07 '20 at 14:08
  • $x/(e^x-1)=\sum_k x^k B_k/k!$ whose radius of convergence is $\le 2\pi$, from which you'll find the radius of convergence of $\sum_k \zeta(-k) x^k=\sum_k x^k (-1)^kB_{k+1}/(k+1)$. – reuns Nov 07 '20 at 14:31
  • Thanks for the comments regarding the radius of convergence. I also was wondering if there is a closed form for the series, like in this title="answer">answer for example. It seems that the closed form written down in that answer would match with this series if there were no gamma function in the denominator. –  Nov 07 '20 at 15:34
  • Sorry for the link: https://math.stackexchange.com/a/2133475/ –  Nov 07 '20 at 15:46
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    why does the series converge anywhere but zero? $\log B_k$ behaves asymptotically like $k \log k$ so $B_kx^k \to \infty$ for all $x>0$ – Conrad Nov 07 '20 at 18:43

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