Mind-blowing Sums: Compute $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}$ and $\sum_{n=1}^\infty\frac{H_n^3}{n^22^n}$

How to prove the following two sums

\begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}&=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\&\quad-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52 \end{align}

\begin{align} \sum_{n=1}^\infty\frac{H_n^3}{n^22^n}&=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\&\quad-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52 \end{align} where $$H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$$ is the $$n$$th generalized harmonic number of order $$p$$ and $$\operatorname{Li}_s(x)=\sum_{n=1}^\infty\frac{x^n}{n^s}$$ is the polylogarithmic function.

Edit: These two sums were proposed by Cornel Ioan Valean here but no solution was submitted. I am presenting my solution in the answer section and would like to see different approaches.

Thanks.

I am going to use the following results :

$$V_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=\frac78\ln2\zeta(4)+\frac38\zeta(2)\zeta(3)-2\zeta(5)$$

$$V_2=\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac14\ln^42$$

$$V_3=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)} dx=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{285}{16}\zeta(5)-3\zeta(2)\zeta(3)$$

$$-\frac{21}{4}\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52$$

$$V_4=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^22^n}=4\operatorname{Li}_5\left(\frac12\right)+3\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{64}\zeta(5)+\frac{5}{16}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)$$ $$+\frac78\ln^22\zeta(3)-\frac{5}{12}\ln^32\zeta(2)+\frac{11}{120}\ln^52$$

We are going to establish two relations and solve them by eliminations

The first relation:

From here we have

$$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

Multiply both sides by $$\frac{1}{n2^n}$$ then sum both sides from $$n=1$$ to $$\infty$$ we get

$$\sum_{n=1}^\infty\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{(x/2)^n}{n}=\int_0^1\frac{\ln^3(1-x)\ln(1-x/2)}{x}\ dx\\\overset{x\mapsto 1-x}{=}\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}=-6V_1$$

Note that for the last step, we used this rule $$\int_0^1\frac{\ln^ax\ln(1+x)}{1-x}\ dx=(-1)^aa!\left(\ln2\zeta(a+1)+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(a+1)}}{n}\right)$$

Then

$$R_1=\sum_{n=1}^\infty\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=-6V_1$$

The second relation:

From here we have

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\tag{1}$$

Divide both sides of (1) by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2}x^{n-1}\ dx=\frac1{n2^n}$$ to get

$$\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n2^n}=-\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx=-V_2$$

Now multiply both sides of (1) by $$-\frac{\ln x}{x}$$ then integrate from $$x=0$$ to $$1/2$$ and use that fact that $$-\int_0^{1/2}x^{n-1}\ln x\ dx=\frac{\ln2}{n2^n}+\frac1{n^22^n}$$ to get

$$\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\left(\frac{\ln2}{n2^n}+\frac1{n^22^n}\right)=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx=V_3$$

$$-\ln2V_2+\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=V_3$$

Then

$$R_2=\sum_{n=1}^\infty\frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n^22^n}=V_3+\ln2V_2$$

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}=\frac{R_1-R_2}{6}=-\frac{6V_1+\ln2V_2+V_3}{6}$$

and by plugging the results of $$V_1$$, $$V_2$$ and $$V_3$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^22^n}&=2\operatorname{Li}_5\left(\frac12\right)+\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{31}{32}\zeta(5)+\frac{1}{8}\ln2\zeta(4)+\frac18\zeta(2)\zeta(3)\\&\quad-\frac{1}{12}\ln^32\zeta(2)+\frac{1}{40}\ln^52 \end{align}

As for the second sum :

$$\sum_{n^1}^\infty\frac{H_n^3}{n^22^n}=\frac{R_1+R_2}{2}=\frac{-6V_1+\ln2V_2+V_3-4V_4}{2}$$

and by plugging the results of $$V_1$$, $$V_2$$, $$V_3$$ and $$V_4$$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_n^3}{n^22^n}&=-14\operatorname{Li}_5\left(\frac12\right)-9\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{279}{16}\zeta(5)-\frac{25}{4}\ln2\zeta(4)-\frac78\zeta(2)\zeta(3)\\&\quad-\frac74\ln^22\zeta(3)+\frac{13}{12}\ln^32\zeta(2)-\frac{31}{120}\ln^52 \end{align}

.

Proofs of the results:

$$V_1$$ can be found here and $$V_2$$ can be found using wolfram but here is the steps

\begin{align} V_2&=\int_0^{1/2}\frac{\ln^3(1-x)}{x(1-x)}\ dx\overset{x\mapsto 1-x}{=}\int_{1/2}^1\frac{\ln^3x}{x(1-x)}\ dx\\ &=\int_{1/2}^1\frac{\ln^3x}{x}\ dx+\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\ &=-\frac14\ln^42+\sum_{n=1}^\infty \int_{1/2}^1x^{n-1}\ln^3x\ dx, \quad \text{apply integration by parts}\\ &=-\frac14\ln^42+\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=-\frac14\ln^42+\ln^42+3\ln^22\operatorname{Li}_2\left(\frac12\right)+6\ln2\operatorname{Li}_3\left(\frac12\right)+6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)\\ &\boxed{=6\operatorname{Li}_4\left(\frac12\right)-6\zeta(4)+\frac{21}{4}\ln2\zeta(3)-\frac32\ln^22\zeta(2)+\frac14\ln^42} \end{align}

where we used $$\operatorname{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ and $$\operatorname{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$

\begin{align} V_3&=\int_0^{1/2}\frac{\ln^3(1-x)\ln x}{x(1-x)}\ dx\overset{x\mapsto 1-x}{=}\int_{1/2}^{1}\frac{\ln^3x\ln(1-x)}{x(1-x)}\ dx\\ &=-\sum_{n=1}^\infty H_n\int_{1/2}^1 x^{n-1}\ln^3x\ dx\\ &=-\sum_{n=1}^\infty H_n\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\ &=-\ln^32\sum_{n=1}^\infty\frac{H_n}{n2^n}-3\ln^22\sum_{n=1}^\infty\frac{H_n}{n^22^n}-6\left(\color{blue}{\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}+\sum_{n=1}^\infty\frac{H_n}{n^42^n}}\right)+6\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align}

I managed here to prove

$$\color{blue}{\ln2\sum_{n=1}^\infty\frac{H_n}{n^32^n}+\sum_{n=1}^\infty\frac{H_n}{n^42^n}}=-\frac12\ln^22\sum_{n=1}^\infty\frac{H_n}{n^22^n}-\frac16\ln^32\sum_{n=1}^\infty\frac{H_n}{n2^n}+\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac{47}{32}\zeta(5)+\frac1{15}\ln^52+\frac13\ln^32\operatorname{Li}_2\left(\frac12\right)+\ln^22\operatorname{Li}_3\left(\frac12\right)+2\ln2\operatorname{Li}_4\left(\frac12\right)+2\operatorname{Li}_5\left(\frac12\right)$$

and $$V_3$$ simplifies into

$$\small{V_3=3\sum_{n=1}^\infty\frac{H_n}{n^4}+\frac{141}{16}\zeta(5)-\frac25\ln^52-2\ln^32\operatorname{Li}_2\left(\frac12\right)-6\ln^22\operatorname{Li}_3\left(\frac12\right)-12\operatorname{Li}_4\left(\frac12\right)-12\operatorname{Li}_5\left(\frac12\right)} \\ \small{\boxed{=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{285}{16}\zeta(5)-3\zeta(2)\zeta(3)-\frac{21}{4}\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52\quad}}$$

where we substituted $$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$ along with the values $$\operatorname{Li}_2\left(\frac12\right)$$ and $$\operatorname{Li}_3\left(\frac12\right)$$.

To prove the last result of $$V_4$$, we use this generating function

$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)$$

Divide both sides by $$x$$ then integrate from $$x=0$$ to $$1/2$$ and use the fact that $$\int_0^{1/2} x^{n-1}\ dx=\frac1{n2^n}$$ we get

\begin{align} V_4&=\int_0^{1/2}\frac{\operatorname{Li}_4(x)}{x}\ dx-\underbrace{\int_0^{1/2}\frac{\ln(1-x)\operatorname{Li}_3(x)}{x}\ dx}_{IBP}-\frac12\int_0^{1/2}\frac{\operatorname{Li}^2_2(x)}{x}\ dx\\ &=\operatorname{Li}_5\left(\frac12\right)+\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-\frac32\int_0^{1/2}\frac{\operatorname{Li}^2_2(x)}{x}\ dx \end{align}

@Song proved in this solution

$$\int_0^{1/2}\frac{\left(\operatorname{Li}_2(x)\right)^2}{x}\ dx=\frac12\ln^32\zeta(2)-\frac78\ln^22\zeta(3)-\frac58\ln2\zeta(4)+\frac{27}{32}\zeta(5)+\frac78\zeta(2)\zeta(3)\\-\frac{7}{60}\ln^52-2\ln2\operatorname{Li}_4\left(\frac12\right)-2\operatorname{Li}_5\left(\frac12\right)$$

by plugging this result along with the values of $$\operatorname{Li}_2(1/2)$$ and $$\operatorname{Li}_3(1/2)$$ the closed form of $$V_4$$ follows.