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

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 $$\displaystyle\sum_{n=1}^\infty x^n\frac{H_n}{n^3}$$.

• Can we relate $a_{2n}$ to $a_{2n+1}$? ;) – metamorphy Sep 2 '19 at 10:47
• @user514787 without using the generating function . I forgot to mention that . – Ali Shather Sep 2 '19 at 11:04
• @AliShather sorry. – user514787 Sep 2 '19 at 11:08
• @user514787 thanks for reminding me. – Ali Shather Sep 2 '19 at 11:09
• @metamorphy alternating even series and alternating odd series. – Ali Shather Sep 2 '19 at 11:11

$$\sum_{n=1}^{\infty }\frac{(-1)^n H_{2 n}}{(2 n)^3}=\frac{1}{32} \left(2\,\text{HypergeometricPFQRegularized}^{(\{0,0,0,0\},\{0,0,1\},0)}(\{1,1,1,1\},\{2,2,2\},-1)+\sqrt{\pi}\,\text{HypergeometricPFQRegularized}^{(\{0,0,0,0,0\},\{0,0,0,1\},0)}\left(\left\{1,1,1,1,\frac{3}{2}\right\},\left\{2,2,2,\frac{3}{2}\right\},-1\right)-3\,\zeta(3)\,(\gamma+\log(2))\right)$$
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.
• $$\sum_{n=1}^{\infty }\frac{(-1)^n H_{2 n}}{(2 n)^3}=\frac{5}{8}Li_4(\frac{1}{2})+\frac{35}{64}\zeta(3)Log2-\frac{5}{192}{\pi^2}Log^22+\frac{5}{192}Log^42-\frac{13}{1536}{\pi^4}$$ See the site Boris Gourévitch www.pi314.net L'univers de pi "series harmoniques" – user178256 Sep 3 '19 at 19:52