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Probably this is known, but mathworld and wolfram alpha don't recognize this potential identities.

Numerical evidence suggests:

$$ \sum_{n=2}^\infty \frac{\zeta(n)}{a^n} =? \sum_{n=1}^\infty \frac{1}{a (a n^2 -n) } = \frac{-\gamma - \psi((a-1)/a)}{a} \qquad (1) $$


$$ \sum_{n=2}^\infty \frac{(-1)^n \zeta(n)}{a^n} =? \sum_{n=1}^\infty \frac{1}{a (a n^2 + n) } = \frac{\gamma + \psi(1/a)+a}{a} $$

for $a \in \mathbb{C}$ whenever the first sum converges.

Web search found the special case $a=2$ is proved in a paper.

Are these identities true?

mathworld 52 gives related identity for even zeta:

$$ \sum_{n=0}^\infty\zeta(2n+2) x^{2n}=\frac{1-\pi x \cot(\pi x)}{2 x^2} \qquad (52) $$

Setting $a=\pi$ in (1) and $x=1/\pi$ in (52) gives: $$\sum_{n=1}^\infty\zeta(2n+1)/\pi^{2n+1} = (-\gamma -\psi((\pi-1)/\pi)/\pi - (1-\cot(1))/2 $$

Verified with mpmath.nsum and $1000$ digits of precision.

share|cite|improve this question
The first two formulas (question marks) are easy to prove by expanding the Riemann zeta-function into its definition, changing the summation order and using the formula for a geometric series. – Johan Andersson Jun 12 '13 at 9:16
Thank you Johan. – joro Jun 12 '13 at 10:49
But of course absolute convergence must be checked to reverse the order of summation. Not hard. – Gerald Edgar Jun 12 '13 at 13:30

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