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How to prove:

$$S_1=\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)-\frac{581}{128} \zeta (5)-\frac{2}{3}\ln ^32 \zeta (2)+\frac{7}{4} \ln^22\zeta (3)\\ +\frac{2}{15} \ln ^52 +4\ln2 \operatorname{Li}_4\left(\frac{1}{2}\right) +4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$

$$S_2=\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)+\frac{917 }{128}\zeta (5)+\frac{2}{3} \ln ^32\zeta (2)-\frac{7}{4} \ln ^22\zeta (3)\\-\frac{2}{15} \ln ^52 -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$

where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}, \ m\ge1,$ represents the $n$th generalized harmonic number of order $m$, $\zeta$ is the Riemann zeta function and $\operatorname{Li}_n$ is the polylogarithm function.


These two sums were proposed by Cornel here where he asked if these results and others ( in the link) exist in the literature.

I managed to find a relation between $S_1$ and $S_2$ so we need to find another relation or evaluate of them separately.


Here is how I got the relation:

From here we have

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

multiply both sides by $\frac{H_n}{2n}$ then sum them from $n=1$ to $\infty$ to get

$$\sum_{n=1}^\infty\frac{H_n}{(2n)^2}(H_{2n}^2+H_{2n}^{(2)})=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty(x^2)^n\frac{H_n}{n}\ dx\\=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(\operatorname{Li}_2(x^2)+\frac12\ln^2(1-x^2)\right)\ dx\\=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(2\operatorname{Li}_2(x)+2\operatorname{Li}_2(-x)+\frac12\ln^2(1-x^2)\right)\ dx\\=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}+\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(-x)}{x}+\frac14\int_0^1\frac{\ln^2(1-x)\ln^2(1-x^2)}{x}\ dx\\=A+B+\frac14C$$


\begin{align} A&=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}\ dx=\int_0^1\frac{\ln^2x\operatorname{Li}_2(1-x)}{1-x}\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^2x}{1-x}-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}-\int_0^1\frac{\ln^2x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=2\zeta(2)\zeta(3)+\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3x-\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\zeta(2)\zeta(3)-6\sum_{n=1}^\infty\frac{H_n}{n^4}+6\zeta(5)-2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+2\zeta(5)\\ &=\boxed{2\zeta(2)\zeta(3)-\zeta(5)} \end{align}

where we used $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$


\begin{align} B&=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(-x)}{x}\ dx\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1x^{n-1} \ln^2(1-x)\ dx\\ &=\sum_{n=1}^\infty(-1)^n\frac{H_n^2+H_n^{(2)}}{n^3}\\ &=\boxed{\small{\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac{3}4\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}} \end{align}

where the results $$\small{\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}=\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac{3}4\zeta(2)\zeta(3)+\frac{15}{16}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}$$

and

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)$$

were used. Both series can be found here.


To find $C$, we are going to use the algebraic identity:

$$a^2(a+b)^2=\frac43a^4-\frac23b^4+\frac5{24}(a+b)^4+\frac{13}{24}(a-b)^4-(a-b)^3b$$

with $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write

$$C=\frac43\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac5{24}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}-\frac{13}{24}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}\\C=\small{\frac{23}{16}\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{24\zeta(5)}-\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac{13}{12}\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}4\zeta(5)}}-2\int_0^1\frac{\ln^2x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx\\C=\frac{149}{16}\zeta(5)-\frac23\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$

The integrals $K$ and $J$ are calculated here

$$K=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)$$

$$J=\frac{279}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)$$

Combining $K$ and $J$ gives

$$\boxed{\small{C=\frac{43}{4}\zeta(5)-\frac{21}4\zeta(2)\zeta(3)-\frac83\ln^32\zeta(2)+7\ln^22\zeta(3)+\frac8{15}\ln^52+16\ln2\operatorname{Li}_4\left(\frac12\right)+16\operatorname{Li}_5\left(\frac12\right)}}$$

now combine the boxed results of the integrals $A$, $B$ and $C$ we get

$$\sum_{n=1}^\infty\frac{H_n}{(2n)^2}(H_{2n}^2+H_{2n}^{(2)})=\frac{23}{16}\zeta (2) \zeta (3)+\frac{21}{8} \zeta (5)$$

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Using the fact that

$$\int_0^1 x^{2n-1}\ln x\ln(1-x)\ dx=\frac{H_{2n}^{(2)}}{2n}+\frac{H_{2n}}{(2n)^2}-\frac{\zeta(2)}{2n}$$

which comes from differentiating both sides of $\int_0^1x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$ with respect to $n$.

multiply both sides by $\frac{H_{n}}{2n}$ then sum them from $n=1$ to $\infty$ we get

$$\sum_{n=1}^\infty\frac{H_nH_{2n}^{(2)}}{(2n)^2}+\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n)^3}-\frac{\zeta(2)}{4}\underbrace{\sum_{n=1}^\infty\frac{H_n}{n^2}}_{2\zeta(3)}=\frac12\int_0^1\frac{\ln x\ln(1-x)}{x}\sum_{n=1}^\infty (x^2)^n\frac{H_n}{n}\ dx\\ =\frac12\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac12\ln^2(1-x^2)+\operatorname{Li}_2(x^2)\right)\ dx\\=\small{\frac14\int_0^1\frac{\ln x\ln(1-x)\ln^2(1-x^2)}{x}+\int_0^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(x)}{x}+\int_0^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(-x)}{x}}\ dx\\=\frac14A+B+C$$

Rearranging the terms we have

$$\sum_{n=1}^\infty\frac{H_nH_{2n}^{(2)}}{(2n)^2}=\frac12\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n)^3}+\frac14A+B+C\tag{1}$$


From here we have

$$\boxed{\small{\sum _{n=1}^{\infty } \frac{H_{2 n}H_n }{(2 n)^3}=\frac{307}{128}\zeta(5)-\frac{1}{16}\zeta (2) \zeta (3)+\frac{1}{3}\ln ^3(2)\zeta (2) -\frac{7}{8} \ln ^2(2)\zeta (3)-\frac{1}{15} \ln ^5(2) -2 \ln (2) \operatorname{Li}_4\left(\frac{1}{2}\right) -2 \operatorname{Li}_5\left(\frac{1}{2}\right)}}$$


Evaluation of the integral $A$:

By using the identity $$a(a+b)^2=\frac23a^3-\frac23b^3+\frac12(a+b)^3-\frac16(a-b)^3$$

and setting $a=\ln(1-x)$ and $b=\ln(1+x)$ we can write

$$A=\int_0^1\frac{\ln x\ln(1-x)\ln^2(1-x^2)}{x}\ dx\\=\small{\frac23\int_0^1\frac{\ln x\ln^3(1-x)}{x}-\frac23\int_0^1\frac{\ln x\ln^3(1+x)}{x}}+\frac12\underbrace{\int_0^1\frac{\ln x\ln^3(1-x^2)}{x}}_{x^2\mapsto x}-\frac16\underbrace{\int_0^1\frac{\ln x\ln^3\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}\\=\frac{19}{24}\int_0^1\frac{\ln x\ln^3(1-x)}{x}\ dx-\frac23\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx-\frac13\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3x}{1-x^2}\ dx$$

lets calculate the first integral:

\begin{align} \int_0^1\frac{\ln x\ln^3(1-x)}{x}\ dx&=\int_0^1\frac{\ln(1-x)\ln^3x}{1-x}\ dx\\ &=-\sum_{n=1}^\infty H_n\int_0^1 x^n\ln^3x\ dx\\ &=6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}\\ &=6\sum_{n=1}^\infty\frac{H_n}{n^4}-6\zeta(5)\\ &=12\zeta(4)-6\zeta(2)\zeta(3) \end{align}

The second integral was calculated here

\begin{align} \int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx&=-12\operatorname{Li}_5\left(\frac12\right)-12\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{99}{16}\zeta(5)+3\zeta(2)\zeta(3)\\ &\quad-\frac{21}4\ln^22\zeta(3)+2\ln^32\zeta(2)-\frac25\ln^52 \end{align}

For the third integral, by using the identity

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

we can write

\begin{align} \int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln^3x}{1-x^2}\ dx&=\sum_{n=1}^\infty(H_n-2H_{2n})\int_0^1 x^{2n-1}\ln^3x\ dx\\ &=12\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}-\frac38\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=6\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}+\frac{48}{5}\sum_{n=1}^\infty\frac{H_n}{n^4}\\ &=\frac{92}{16}\zeta(5)-\frac{21}8\zeta(2)\zeta(3) \end{align}

where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$ which can be found here.

Collecting these three integrals we get

$$\boxed{\small{A=8\operatorname{Li}_5\left(\frac12\right)+8\ln2\operatorname{Li}_4\left(\frac12\right)+\frac{55}{16}\zeta(5)-\frac{47}8\zeta(2)\zeta(3)+\frac72\ln^22\zeta(3)-\frac43\ln^32\zeta(2)+\frac4{15}\ln^52\ \ }}$$


Evaluation of the integral $B$:

\begin{align} B&=\int_0^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(x)}{x}\ dx\overset{IBP}{=}\frac12\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx\\ &=\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{n-1}\operatorname{Li}_2(x)\ dx=\frac12\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\ &=\frac12\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac{H_n}{n^4}=\boxed{\zeta(2)\zeta(3)-\frac32\zeta(5)} \end{align}


Evaluation of integral $C$:

\begin{align} C&=\int_0^1\frac{\ln x\ln(1-x)\operatorname{Li}_2(-x)}{x}\ dx\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln x\ln(1-x)\ dx\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\frac{H_n^{(2)}}{n}+\frac{H_n}{n^4}-\frac{\zeta(2)}{n}\right)\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}-\zeta(2)\underbrace{\sum_{n=1}^\infty\frac{(-1)^n}{n^3}}_{-\frac34\zeta(3)}\\ &=\boxed{\frac58\zeta(2)\zeta(3)-\frac32\zeta(5)} \end{align}

where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}=\frac{11}{32}\zeta(5)-\frac58\zeta(2)\zeta(3)$ which can be found here.


Substituting the boxed results in (1) we get

$$\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^{(2)}}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)-\frac{581}{128} \zeta (5)-\frac{2}{3}\ln ^32 \zeta (2)+\frac{7}{4} \ln^22\zeta (3)\\ +\frac{2}{15} \ln ^52 +4\ln2 \operatorname{Li}_4\left(\frac{1}{2}\right) +4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$


In the body, we proved

$$\sum_{n=1}^\infty\frac{H_n}{(2n)^2}(H_{2n}^2+H_{2n}^{(2)})=\frac{23}{16}\zeta (2) \zeta (3)+\frac{21}{8} \zeta (5)$$

then

$$\sum _{n=1}^{\infty } \frac{H_n H_{2 n}^2}{(2 n)^2} =\frac{23 }{32}\zeta (2) \zeta (3)+\frac{917 }{128}\zeta (5)+\frac{2}{3} \ln ^32\zeta (2)-\frac{7}{4} \ln ^22\zeta (3)\\-\frac{2}{15} \ln ^52 -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right)-4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$

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  • $\begingroup$ (+1) Great talent here. $\endgroup$ – user97357329 Sep 13 at 10:54
  • $\begingroup$ @user97357329 thank you :) $\endgroup$ – Ali Shather Sep 13 at 11:11

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