Computing $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}$ in an alternative way

Question

The following equality

$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{2n+1}=\frac{3}{16}\pi^3-\frac34\ln^2(2)\pi-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$

can be proved if we are allowed to use the generating function ( see Eq$(3)$)

$$\sum_{n=1}^\infty x^nH_n^2=\frac{\ln^2(1-x)+\text{Li}_2(x)}{1-x}$$

But the problem-proposer mentioned that the sum to be calculated without using this generating function.

I have no clue how to approach it with such restriction. any idea ? Thanks in advance.

Answer 1

$\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n\ H_n^2}{(2n+1)}$
Solution:
I'm used:
$\displaystyle\pmb{\sum^{+\infty}_{n=1}\mathrm{H_n^2}(-1)^{n}x^n=\frac{\mathrm{Li_2(-x)}}{1+x}+\frac{\mathrm{ln}^2(1+x)}{1+x}}$ $\displaystyle{\int_{0}^{1}}\left(\sum^{+\infty}_{n=1}\mathrm{H_n^2}(-1)^{n}x^{2n}\right){\mathrm{dx}}={\int_{0}^{1}}\left(\frac{\mathrm{Li_2(-x^2)}}{1+x^2}+\frac{\mathrm{ln}^2(1+x^2)}{1+x^2}\right){\mathrm{dx}}\xrightarrow{\textbf{then}}\sum^{+\infty}_{n=1}\frac{\mathrm{H_n^2}(-1)^{n}}{2n+1}=\underbrace{\int_{0}^{1}\frac{\mathrm{Li_2(-x^2)}}{1+x^2}\mathrm{dx}}_{\pmb{I_1}}+\underbrace{\int_{0}^{1}\frac{\mathrm{ln}^2(1+x^2)}{1+x^2}\mathrm{dx}}_{\pmb{I_2}}$

$\displaystyle I_1=\int_{0}^{1}\frac{\mathrm{Li_2(-x^2)}}{1+x^2}\mathrm{dx}=\int_{0}^{1}(Li_2(ix)+Li_2(-ix))\left(\frac{1}{1-ix}+\frac{1}{1+ix}\right)\mathrm{dx}\textbf{ ; } xi=u\longrightarrow i\mathrm{d(x)}=\mathrm{d(u)}$

$\displaystyle I_1=\frac{1}{i}(2Li_3(1-xi)-2Li_3(1+xi)-2Li_2(1-xi)\cdot ln(1-xi)-Li_2(xi)\cdot ln(1-xi)-Li_2(-xi)\cdot ln(1-xi)$

$\displaystyle +Li_2(-xi)\cdot ln(1+xi)+Li_2(xi)\cdot ln(1+xi)+2Li_2(1+xi)\cdot ln(1+xi)-ln(xi)\cdot ln^2(1-xi)+ln(-xi)\cdot ln^2(1+xi))\left.\frac{}{}\right|_0^1$ $\displaystyle I_1=\left[\frac{\pi^2}{3}-ln(1+x^2)\cdot ln(x)\right]2\arctan(x) -\pi\arctan^2(x)+\frac{\pi}{4}\cdot ln^2(1+x^2)+2\mathrm{Ti_2}(x)\cdot ln(1+x^2)-4\ \mathrm{Im}(Li_3(1+ix))\left.\frac{}{}\right|_0^1$

$\displaystyle \pmb{I_1=\frac{5\pi^3}{48}+\frac{\pi}{4}ln^2(2)+2G\cdot ln(2)-4\ \mathrm{Im}(Li_3(1+i)) }$

In $I_2$: $\displaystyle\int_{0}^{1}\frac{ln^2(1+x^2)}{1+x^2}\mathrm{dx}=\int_{0}^{\pi/4} ln^2(1+tan^2(x))\ dx=4\int_{0}^{\pi/4} ln^2(cos(x))\ dx=4\int_{\pi/4}^{\pi/2} ln^2(sin(x))\ dx=$

$\displaystyle\underbrace{4\int_{\pi/4}^{\pi/2} ln^2(2sin(x))\ dx}_{\pmb{\displaystyle I_{2,1}}}\underbrace{-8ln(2)\int_{\pi/4}^{\pi/2} ln(2sin(x))\ dx}_{\displaystyle\pmb{I_{2,2}}}+\underbrace{4\int_{\pi/4}^{\pi/2}ln^2(2)\ dx}_{\displaystyle\pmb{I_{2,3}}}$

$\displaystyle \pmb{I_{2,1}}=4\int_{\pi/4}^{\pi/2} ln^2(2sin(x))\ dx=4\int_{\pi/4}^{\pi/2} \left(2\left(\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n}-\frac{1}{n^2}\right)cos(2nx)\right)+\left(\frac{\pi}{2}-x\right)^2\right)\ dx=$

$\displaystyle 4\left(\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n^2}-\frac{1}{n^3}\right)\left((sin(n\pi)-sin\left(\frac{n\pi}{2}\right)\right)-\left.\frac{1}{3}\left(\frac{\pi}{2}-x\right)^3\right|_{\pi/4}^{\pi/2}\right)=-4\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n^2}-\frac{1}{n^3}\right)sin\left(\frac{n\pi}{2}\right)+\frac{\pi^3}{48}$ $\displaystyle -4\ \mathrm{Im}\left(\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n^2}-\frac{1}{n^3}\right)e^{in\pi/2}\right)+\frac{\pi^3}{48}=$

$\displaystyle -4\ \mathrm{Im}\left(-\mathrm{Li_{3}}(1-\mathrm{e^{\pi i/2}})+\mathrm{Li_{2}}(1-\mathrm{e^{\pi i/2}})\cdot\mathrm{\ln}(1-\mathrm{e^{\pi i/2}})+ \frac{1}{2}\mathrm{\ln}(\mathrm{e^{\pi i/2}})\cdot\mathrm{\ln}^2(1-\mathrm{e^{\pi i/2}})+\zeta(3)\right)+\frac{\pi^3}{48}$ $\displaystyle -4\ \mathrm{Im}\left(-\mathrm{Li_{3}}(1-i)+\mathrm{Li_{2}}(1-i)\cdot\mathrm{\ln}(1-i)+ \frac{1}{2}\mathrm{\ln}(i)\cdot\mathrm{\ln}^2(1-i)+\zeta(3)\right)+\frac{\pi^3}{48}$
$\displaystyle -4\ \mathrm{Im}\left(\mathrm{Li_{3}}(1+i)+\mathrm{Li_{2}}(1-i)\cdot\mathrm{\ln}(1-i)+ \frac{1}{2}\mathrm{\ln}(i)\cdot\mathrm{\ln}^2(1-i)+\zeta(3)\right)+\frac{\pi^3}{48}$ $\displaystyle-4\left(\ \mathrm{Im}\left(\mathrm{Li_{3}}(1+i)\right)-\frac{\pi^3}{64}-\frac{ln(2)\mathrm{G}}{2}-\frac{ln^2(2)\pi}{8}+\frac{\pi ln^2(2)}{16}-\frac{\pi^3}{64}\right)+\frac{\pi^3}{48}$ $\displaystyle -4\ \mathrm{Im}(\mathrm{Li_3}(1+i))+\frac{7\pi^3}{48}+2ln(2)\mathrm{G}+\frac{\pi ln^2(2)}{4}$

$\displaystyle \pmb{I_{2,2}}=-8ln(2)\int_{\pi/4}^{\pi/2} ln(2sin(x))\ dx=-8ln(2)\int_{\pi/4}^{\pi/2}\left(-\sum_{n=1}^{+\infty}\frac{cos(2nx)}{n}\right)dx=$

$\displaystyle 8\ln(2)\left(\sum_{n=1}^{+\infty}\frac{\displaystyle\left((sin(n\pi)-sin\left(\frac{n\pi}{2}\right)\right)}{2n^2}\right)=-4ln(2)\mathrm{G}$

$\displaystyle \pmb{I_{2,3}}=4\int_{\pi/4}^{\pi/2}ln^2(2)\ dx= \pi\ ln^2(2)$
$\displaystyle\pmb{I_2}=-4\ \mathrm{Im}(\mathrm{Li_3}(1+i))+\frac{7\pi^3}{48}+2ln(2)\mathrm{G}+\frac{\pi ln^2(2)}{4}-4ln(2)\mathrm{G}+\pi\ ln^2(2)$ $\displaystyle\pmb{I_2=-4\ \mathrm{Im}(\mathrm{Li_3}(1+i))+\frac{7\pi^3}{48}-2ln(2)\mathrm{G}+\frac{5\pi ln^2(2)}{4}}$

Finally:
$\displaystyle S=I_1+I_2=\frac{5\pi^3}{48}+\frac{\pi}{4}ln^2(2)+2G\cdot ln(2)-4\ \mathrm{Im}(Li_3(1+i))-4\ \mathrm{Im}(\mathrm{Li_3}(1+i))+\frac{7\pi^3}{48}-2ln(2)\mathrm{G}+\frac{5\pi ln^2(2)}{4}$ $\pmb{\displaystyle S=\frac{\pi^3}{4}+\frac{3\pi\ln^2(2)}{2}-8\mathrm{Im(Li_3(}1+i\mathrm{))}}$

Answer 2

Replace $x$ by $-x^2$ in the generating function: $$\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^{n}=\frac{\ln^2(1-x)}{1-x}$$

we get

$$\sum_{n=1}^\infty (-1)^n(H_n^2-H_n^{(2)})x^{2n}=\frac{\ln^2(1+x^2)}{1+x^2}.$$ Next, integrate both sides from $x=0$ to $1$, we have

$$\sum_{n=1}^\infty \frac{(-1)^nH_n^2}{2n+1}-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{2n+1}=\int_0^1 \frac{\ln^2(1+x^2)}{1+x^2}\mathrm{d}x=4\int_0^{\pi/4}\ln^2(\cos x)\mathrm{d}x.$$

$$=\frac{7\pi^3}{48}+\frac{5\pi}{4}\ln^2(2)-2\ln(2)G-4\Im\operatorname{Li_3}(1+i),$$ where the latter integral is calculated here .

To get the remaining sum on the left side, we start with the following integral:

\begin{gather} \int_0^1\frac{\ln(x)\arctan x}{x(1+x)}\mathrm{d}x=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{x^{2n}\ln(x)}{1+x}\mathrm{d}x\nonumber\\ =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1\frac{\partial}{\partial{n}}\frac12\frac{x^{2n}}{1+x}\mathrm{d}x\nonumber\\ =\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\mathrm{d}}{\mathrm{d}{n}}\int_0^1\frac{x^{2n}}{1+x}\mathrm{d}x\nonumber\\ =\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{\mathrm{d}}{\mathrm{d}{n}}\left(H_n-H_{2n}+\ln(2)\right)\nonumber\\ =\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(2H_{2n}^{(2)}-H_n^{(2)}-\zeta(2)\right)\nonumber\\ =\sum_{n=0}^{\infty}\frac{(-1)^nH_{2n}^{(2)}}{2n+1}-\frac12\sum_{n=0}^{\infty}\frac{(-1)^nH_{n}^{(2)}}{2n+1}-\frac12\zeta(2)\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\nonumber\\ =\sum_{n=0}^{\infty}\frac{(-1)^nH_{2n+1}^{(2)}}{2n+1}-\frac12\sum_{n=0}^{\infty}\frac{(-1)^nH_{n}^{(2)}}{2n+1}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}-\frac12\zeta(2)\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}. \end{gather} On the other hand, we have \begin{gather} \int_0^1\frac{\ln(x)\arctan x}{x(1+x)}\mathrm{d}x=\int_0^1\frac{\ln(x)\arctan x}{x}\mathrm{d}x-\int_0^1\frac{\ln(x)\arctan x}{1+x}\mathrm{d}x\nonumber\\ =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(x)\mathrm{d}x-\int_0^1\frac{\ln(x)\arctan x}{1+x}\mathrm{d}x\nonumber\\ =-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}-\int_0^1\frac{\ln(x)\arctan x}{1+x}\mathrm{d}x . \end{gather} Therefore, $$\sum_{n=0}^{\infty}\frac{(-1)^nH_{n}^{(2)}}{2n+1}=2\sum_{n=0}^{\infty}\frac{(-1)^nH_{2n+1}^{(2)}}{2n+1}-\zeta(2)\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}+2\int_0^1\frac{\ln(x)\arctan x}{1+x}\mathrm{d}x,$$

where $$\sum_{n=0}^{\infty}\frac{(-1)^nH_{2n+1}^{(2)}}{2n+1}=\Im \sum_{n=1}^{\infty}\frac{i^nH_{n}^{(2)}}{n}$$

$$=\Im\{\operatorname{Li}_3(i)+2\operatorname{Li}_3(1-i)-\ln(1-i)\operatorname{Li}_2(1-i)-\zeta(2)\ln(1-i)-2\zeta(3)\}$$

$$=-2\,\mathfrak{J}\operatorname{Li}_3(1+i)+\frac{17\pi^3}{192}+\frac{\pi}{8}\ln^2(2)+\frac12\ln(2)G,$$

$$\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}=\frac{\pi}{4},$$

and the latter integral is calculated here:

$$\int_0^1\frac{\ln(x)\arctan x}{1+x}\mathrm{d}x=\frac12G\ln(2)-\frac{\pi^3}{64}.$$

Collecting all the pieces, we have

$$\sum_{n=0}^\infty\frac{(-1)^nH_{n}^{(2)}}{2n+1}=-4\,\mathfrak{J}\operatorname{Li}_3(1+i)+\frac{5\pi^3}{48}+\frac{\pi}{4}\ln^2(2)+2\ln(2)G.$$

Thus,

$$\sum_{n=0}^\infty \frac{(-1)^nH_n^2}{2n+1}=-8\,\mathfrak{J}\operatorname{Li}_3(1+i)+\frac{\pi^3}{4}+\frac{3\pi}{2}\ln^2(2). $$