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I'm looking for a closed form for the expression above. I know that Ramanujan gave a closed form for $$ \sum_{k=1}^{\infty}\frac{1}{(2k)^3-2k}= \ln(2)-\frac{1}{2} $$

I wonder if it is possible to find such a similarly simple and nice closed form for the above case. Wolfram guives answers involving Digamma or Polygamma functions, but I'm looking for a cleaner answer. It would be nice if someone could find such a thing...

Thanks.

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  • $\begingroup$ The question is rather clear. He is interested in the sum in the title. The one in the body is an example of a sum that have a closed form. $\endgroup$ – Umberto May 23 '16 at 12:20
  • $\begingroup$ @Umberto My apologies, I misread the question. Either way, interesting question. $\endgroup$ – zz20s May 23 '16 at 12:28
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    $\begingroup$ And... which mathematical steps have you taken to solve this? $\endgroup$ – Did May 23 '16 at 12:30
  • $\begingroup$ @zz20s Indeed it is... I simplified the term to be summed to $1/(2k \cdot ((2k)^2+1))$ but this seems also difficult to evalute (at least as the OP wants). $\endgroup$ – Umberto May 23 '16 at 12:30
  • $\begingroup$ You do recall how to do partial fraction decomposition, yes? $\endgroup$ – J. M. isn't a mathematician May 23 '16 at 12:31
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Hint. One may recall that,

$$ \sum_{k=1}^N\frac{1}{k+a}=\psi(a+N+1)-\psi(a+1) \tag1 $$ where $\psi$ is the digamma function $\psi:=\Gamma'/\Gamma$.

Then observe that $$ \frac{4}{(2k)^5-2k}=\frac{1}{ 2 k-1}+\frac{1}{ 2 k+1}-\frac{2}{ k}+\frac{1}{ 2 k+i}+\frac{1}{ 2 k-i} \tag2 $$ summing from $k=1$ to $k=N$, gives $$ \begin{align} &8\sum_{k=1}^N\frac1{(2k)^5-2k} \\\\&=2\psi(1/2+N)+\frac2{2N+1}-\psi(1/2)-\psi(3/2)-4\psi(1+N)+4\psi(1) \\\\&+2\:\Re\:\psi(1+i/2+N)-2\:\Re\:\psi(1+i/2) \tag3 \end{align}$$ By letting $N \to \infty$, one gets a closed form formula in terms of special values of the digamma function.

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    $\begingroup$ Quote: "Wolfram g(i)ves answers involving Digamma or Polygamma functions, but I'm looking for a cleaner answer". $\endgroup$ – Did May 23 '16 at 12:30
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    $\begingroup$ Applying $(1)$ to $(2)$ makes $(3)$. $\endgroup$ – Olivier Oloa May 23 '16 at 12:34
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    $\begingroup$ @Did, I doubt that the OP will find a digamma-free expression for his answer; the solution I get involves $\psi\left(1+\frac{i}{2}\right)$ and its conjugate, and as GEdgar notes in the comments, only its imaginary part has a known elementary form. $\endgroup$ – J. M. isn't a mathematician May 23 '16 at 12:39
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    $\begingroup$ @Did, since the OP did not show what he got that was so unsatisfactory (sadly, a usual matter), maybe a reminder is in order, or maybe the digamma expression OP got is not the simplest it could be. $\endgroup$ – J. M. isn't a mathematician May 23 '16 at 12:43
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    $\begingroup$ One question, how can you get a "closed form" if the digamma function does not converge on infinity? $\endgroup$ – Masacroso May 23 '16 at 13:06

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