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I would like to seek your assistance in computing the sum $$\sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2}$$ I am stumped by this sum. I have tried summing the residues of $\displaystyle f(z)=\frac{\pi\csc(\pi z)(\gamma+\psi(-z))}{(2z+1)^2}$, unfortunately the sum disappears when I add the residues up. Another idea that came to my mind would be to use $$\sum^\infty_{n=1}(-1)^nH_nx^{2n}=-\frac{\ln(1+x^2)}{1+x^2}$$ and integrate once, divide by $x$ then integrate again. However, it seems that performing these successive integrations would turn out to be disastrous. Therefore, I would like to know if any other methods can be employed to crack this sum. I am particularly interested in finding out if the contour integration method still applies for these sort of sums, and in the case that it is still viable, what then would be the appropriate kernel to use?


Thank you for your help.

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    $\begingroup$ It should be possible to get a closed form from Gery Huvent's result in Section 13.3.4 of pi314.net/eng/hypergse13.php#x15-12200013. (Note I am no specialist I found that page from following 'G. Huvent' mentioned on the Wolfram Function Site.) $\endgroup$ Sep 3, 2014 at 13:28
  • $\begingroup$ @gammatester: you just made me discover a real treasure! $\endgroup$ Sep 3, 2014 at 13:48
  • $\begingroup$ Check related techniques. $\endgroup$ Sep 3, 2014 at 19:35
  • $\begingroup$ @gammatester Thank you for the link, it was really useful. $\endgroup$ Sep 3, 2014 at 22:58
  • $\begingroup$ Have you tried summagrating by parts? $\endgroup$
    – David H
    Sep 8, 2014 at 12:53

2 Answers 2

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Since $$\int_{0}^{1}x^{2n}\log x\,dx = -\frac{1}{(2n+1)^2}$$ we have $$ S = \sum_{n=1}^{+\infty}\frac{(-1)^n H_n}{(2n+1)^2}=\int_{0}^{1}\frac{\log(1+x^2)\log x}{1+x^2}\,dx $$ for which Mathematica gives: $$\frac{1}{192} \left(-3 \pi ^3-192 K \log 2-10 i \pi ^2 \log 2-12 \pi\log^2 2+2 i \left(4 \log^3 2-192 \text{Li}_3\left(\frac{1+i}{2}\right)+105 \zeta(3)\right)\right).$$ Thanks to @gammatester, it looks that the last formula follows setting $x=i$ in the line after $(608)$ in http://www.pi314.net/eng/hypergse13.php#x15-134002r658.

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    $\begingroup$ Where $K$ is Catalan's constant, so that the numerical answer is -0.073554. Mathematica is usually better than I am at integration! $\endgroup$
    – almagest
    Sep 3, 2014 at 13:32
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    $\begingroup$ The real part of the complex polylogarithm from this expression possesses a closed form, so perhaps its imaginary part does so as well. $\endgroup$
    – Lucian
    Sep 3, 2014 at 15:28
  • $\begingroup$ @FelixMarin : sorry, I forgot a $\log x$. Thanks for pointing that out. $\endgroup$ Nov 30, 2014 at 16:55
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Using @Jack D'Aurizio's idea, there is a simplification of his closed-form. $$ \sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2} = 2\,\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]-\frac{\pi^3}{64}-\frac{\pi}{16}\ln^2 2 - G \ln 2, $$ where $G$ is Catalan's contant, and $\operatorname{Li_3}$ is the trilogarithm function.

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