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Can we relate $\displaystyle\sum_{n=1}^\infty(-1)^n f(2n)$ to $\displaystyle \sum_{n=1}^\infty (-1)^n f(2n+1$?

I am trying to evaluate $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}$ using (if possible) the value of $\displaystyle \sum_{n=1}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}$ which I managed to evaluate here. I came across $S$ while I was trying to solve $\displaystyle \int_0^{\pi/2}x\ln^2(\tan x)\ dx\ $ in a different way than Song's solution.

Another related integral to $S$ is $\displaystyle \int_0^1\frac{\operatorname{Li}_3(x)}{x(1+x^2)}\ dx$.

Thanks.


Note: Solution should be done without using the generating function $\displaystyle\sum_{n=1}^\infty x^n\frac{H_n}{n^3}$.

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    $\begingroup$ Can we relate $a_{2n}$ to $a_{2n+1}$? ;) $\endgroup$ – metamorphy Sep 2 '19 at 10:47
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    $\begingroup$ @user514787 without using the generating function . I forgot to mention that . $\endgroup$ – Ali Shather Sep 2 '19 at 11:04
  • $\begingroup$ @AliShather sorry. $\endgroup$ – user514787 Sep 2 '19 at 11:08
  • $\begingroup$ @user514787 thanks for reminding me. $\endgroup$ – Ali Shather Sep 2 '19 at 11:09
  • $\begingroup$ @metamorphy alternating even series and alternating odd series. $\endgroup$ – Ali Shather Sep 2 '19 at 11:11
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This post is not intended as an answer, but perhaps provides some insight that is probably easier to read than if posted as a comment.

Mathematica indicates the following:

$$\sum_{n=1}^{\infty }\frac{(-1)^n H_{2 n}}{(2 n)^3}=\frac{1}{32} \left(2\,\text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}(\{1,1,1,1\},\{2,2,2\},-1)+\sqrt{\pi}\,\text{HypergeometricPFQRegularized}^{(\{0,0,0,0,0\},\{0,0,0,1\},0)}\left(\left\{1,1,1,1,\frac{3}{2}\right\},\left\{2,2,2,\frac{3}{2}\right\},-1\right)-3\,\zeta(3)\,(\gamma+\log(2))\right)$$

Mathematica can't seem to evaluate $\sum\limits_{n=1}^{\infty }\frac{(-1)^n H_{2 n+1}}{(2 n+1)^3}$, but since you apparently already have a result for this second infinite series perhaps you can use the definition of the HypergeometricPFQRegularized function to determine a relationship between the two infinite series.

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  • $\begingroup$ $$\sum_{n=1}^{\infty }\frac{(-1)^n H_{2 n}}{(2 n)^3}=\frac{5}{8}Li_4(\frac{1}{2})+\frac{35}{64}\zeta(3)Log2-\frac{5}{192}{\pi^2}Log^22+\frac{5}{192}Log^42-\frac{13}{1536}{\pi^4}$$ See the site Boris Gourévitch www.pi314.net L'univers de pi "series harmoniques" $\endgroup$ – user178256 Sep 3 '19 at 19:52
  • $\begingroup$ @user178256 The two representations seem equivalent as they both seem to evaluate the same numerically, but the representation you quoted is definitely less opaque to me. Proving the equivalence of the two representations might be an interesting problem. $\endgroup$ – Steven Clark Sep 3 '19 at 22:42

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