Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$

Question

Prove that

$$S=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$ where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the n$th$ generalized harmonic number of order $m$ and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel Valean and can be found here.

Here is how I managed to find the integral representation:

We have $\int_0^1 x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$ , then we can write

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=-\frac12\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}(x^2)^n\ dx\\=\small{-\frac12\int_0^1\frac{\ln(1-x)}{x}\left(\operatorname{Li}_3(x^2)+2\operatorname{Li}_3(1-x^2)-\ln(1-x^2)\operatorname{Li}_2(1-x^2)-\zeta(2)\ln(1-x^2)-2\zeta(3)\right)\ dx}$$

So any idea how to crack this integral or different approach?

Thanks.

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UPDATE:

This result was mentioned by @nospoon here in equation $(3)$. He didn't post the solution but he provided the idea.

Answer 1

We will look into the integral \begin{align*} I = &\int_{0}^{1}\arcsin^4 x\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \end{align*} taking the @nospoon's novel approach presented here. Using the MacLaurin series of $\arcsin^4 x$ $$ \arcsin^4 x =\frac 3 2 \sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}x^{2n} $$ and the fact that $$ \small\operatorname{B}(n+\tfrac 1 2,\tfrac 1 2) = \int_0^1 x^{n-1/2}(1-x)^{-1/2}\ \mathrm dx = 2\int_0^{\frac\pi 2} \sin^{2n}\theta\ \mathrm d\theta = \frac{\pi}{4^n}{2n \choose n},\tag{$\small x\mapsto \sin^2\theta$} $$ \begin{align*}\small \psi(n+\tfrac 12 ) -\psi(n+1) =&\small \sum_{k=1}^\infty \frac 1{\scriptsize k+n} - \frac 1{\scriptsize k+n-\tfrac 1 2} \\ =&\small\sum_{k=1}^\infty \left(\frac 1{\scriptsize k} - \frac 1{\scriptsize k-\tfrac 1 2}\right)-\sum_{k=1}^n\frac 1 {\scriptsize k} + \sum_{k=1}^n\frac 1{\scriptsize k-\tfrac 1 2}\\ =&\small-2\ln 2 -H_n +2(H_{2n}-\tfrac 1 2H_n)\\ =&\small 2(H_{2n}-H_n-\ln 2), \end{align*} \begin{align*} \Longrightarrow \ {\int_{ 0}^{1 }x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx} = & \frac 1 4\int_{0 }^{1 }x^{n-1/2} { \ln x \over \sqrt{1-x}}\ \mathrm dx\tag{$\small x^2\mapsto x$}\\ =& \frac 1 4 \left[\frac{\partial }{\partial x}\operatorname{B}(x,y) \right]_{x=n+1/2,y=1/2}\\ =&\frac 1 4\Big[ \operatorname{B}(x,y)\big[\psi(x) -\psi(x+y) \big]\Big]_{x=n+1/2,y=1/2}\\ =& \frac 1 4 \operatorname{B}(n+\tfrac 1 2,\tfrac 1 2)\big[\psi(n+\tfrac 12 ) -\psi(n+1) \big]\\ =& \frac{\pi}2\frac{{2n \choose n}}{4^{n}} \left(H_{2n} - H_n -\ln 2\right), \end{align*} where $\operatorname{B}(x,y)$ and $\psi(x)$ are the Beta and digamma function, respectively, we have \begin{align*} I = &\frac 3 2\sum_{n=1}^\infty \frac{4^{n}H_{n-1}^{(2)}}{n^2{2n \choose n}}\int_{0}^{1}x^{2n}\frac{ \ln x}{\sqrt{1-x^2}}\ \mathrm dx \\ =&\frac {3\pi}4 \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}\left(H_{2n} - H_n -\ln 2\right) \\ =&\frac {3\pi}4\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{2n}}{n^2}-\frac {3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}}_{=-2\zeta(5) +2\zeta(2)\zeta(3)}-\frac {3\pi\ln 2}4\underbrace{\sum_{n=1}^\infty \frac{H^{(2)}_{n-1}}{n^2}}_{=\frac{3}4 \zeta(4)}\\ =&\frac{3\pi}{4} \sum_{n=1}^\infty \frac{H^{(2)}_{n}H_{2n}}{n^2} -\frac{3\pi}4\underbrace{\sum_{n=1}^\infty \frac{H_{2n}}{n^4}}_{=\frac{37}{4}\zeta(5)-4\zeta(2)\zeta(3)} +\frac{3\pi}2 \zeta(5) -\frac{\pi^3}4\zeta(3) -\frac{\pi^5\ln 2}{160}\\ =&\boxed{3\pi S -\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}} \end{align*} where $S = \sum_{n=1}^\infty \frac{H_{2n}H^{(2)}_{n}}{4n^2}$ is the sum in question, and the known values of several Euler sums $$ \sum_{n=1}^\infty \frac{H^{(2)}_{n-1}H_{n}}{n^2}=-2\zeta(5) +2\zeta(2)\zeta(3),\tag{1} $$ $$\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}=\frac{7}4 \zeta(4),\tag{2} $$ \begin{align*}\sum_{n=1}^\infty \frac{H_{2n}}{n^4} =& 8\sum_{n=1}^\infty \frac{H_{n}}{n^4}-8\sum_{n=1}^\infty \frac{(-1)^{n-1} H_{n}}{n^4}\\ =&8\big(3\zeta(5)-\zeta(2)\zeta(3)\big)-8\left(\frac{59}{32}\zeta(5)-\frac 1 2\zeta(2)\zeta(3)\right)\\ =&\frac{37}4\zeta(5) - 4\zeta(2)\zeta(3)\tag{3} \end{align*} are used.

Note: $(1)$ is in @nospoon's answer here, $(2)$ can be found here, and for $(3)$ you can see Euler's formula and here.

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Evaluation of $I$: By making substitution $x = \sin \theta$ and using the Fourier series of $$ \ln (\sin\theta) = -\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}, $$ we get \begin{align*} I =& \int_{0}^{\frac\pi 2} \theta^4 \ln(\sin\theta)\ \mathrm d\theta\\ =&\int_{0}^{\frac\pi 2} \theta^4\left(-\ln 2 -\sum_{k=1}^\infty \frac{ \cos(2k \theta)}{k}\right)\ \mathrm d\theta\\ =& -\ln 2\int_0^{\frac \pi 2}\theta^4\ \mathrm d\theta-\sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_{0}^{\frac\pi 2}\theta^4 \cos(2k \theta) \ \mathrm d\theta}_{\text{IBP}\times 4}\\ =& -\frac{\pi^5\ln 2}{160}-\sum_{k=1}^\infty \frac{1}{k}\cdot\left(-\frac{\pi^3}{8}\frac{(-1)^{k-1}}{k^2} +\frac{3\pi}{4}\frac{(-1)^{k-1}}{k^4}\right)\\ =&-\frac{\pi^5\ln 2}{160}+\frac{\pi^3}8\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^3}}_{=\frac 3 4 \zeta(3)} - \frac{3\pi}4\underbrace{\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}}_{=\frac{15}{16}\zeta(5)}\\ =&\boxed{-\frac{\pi^5\ln 2}{160}+\frac{3\pi^3}{32}\zeta(3) -\frac{45\pi}{64}\zeta(5).} \end{align*}

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Combining these, we get the equation $$ 3\pi S-\frac{87\pi}{16} \zeta(5) +\frac{\pi^3}{4}\zeta(3) -\frac{\pi^5\ln 2}{160}=-\frac{\pi^5\ln 2}{160} +\frac{3\pi^3}{32}\zeta(3)-\frac{45\pi}{64}\zeta(5), $$hence it follows $$ \boxed{S = \frac{101}{64}\zeta(5) -\frac{5\pi^2}{96}\zeta(3).} $$

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Addendum: By considering MacLaurin series of \begin{align*} \ln(1-x)\ln(1+x) =&-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)x^{2k} \end{align*} and \begin{align*} \frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k} =& \frac{\partial }{\partial k}\left[-\frac{H_k}{k}\right]\\ =& \int_0^1 x^{k-1}\ln x\ln(1-x)\ \mathrm dx\\ =&4\int_0^1 x^{2k-1}\ln x \ln(1-x^2)\ \mathrm dx \end{align*} we have that \begin{align*} &\int_{0}^{1}\ln(1-x)\ln(1+x) \frac{\ln x\ln(1-x^2)}x \ \mathrm dx \\&=-\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\int_{0}^{1}x^{2k-1} \ln x \ln(1-x^2)\ \mathrm dx \\ &=-\frac 1 4\sum_{k=1}^\infty \left(\frac{ H_{2k}}k-\frac{H_k}{k} + \frac1{2k^2}\right)\left(\frac{H_k}{k^2} + \frac{H_k^{(2)}}{k} -\frac{\zeta(2)}{k}\right). \end{align*} The integral can be attacked by considering algebraic identity $$ ab(a+b) = \frac 1 3 (a+b)^3 - \frac {a^3}3 -\frac{b^3}3 $$ with $a=\ln(1-x)$ and $b=\ln(1+x)$, and extant results.

For the sum, after expanding the summand, the only tricky part is $$ \sum_{k=1}^\infty\frac{H_{2k}H_k}{k^3}, $$ which can be found here. Then, the sum $\sum_{k=1}^\infty \frac{H_{2k}H_k^{(2)}}{4k^2}$ can be evaluated by solving the equation obtained.

Answer 2

I am going to establish two relations as Cornel suggested and solve them by elimination.

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From here we have

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

Multiply both sides by $\frac{H_{2n}}{4n}$ then sum them from $n=1$ to $\infty$ we get

\begin{align} R_1&=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)=\frac12\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^nH_{2n}}{2n}\ dx\quad \color{red}{x\mapsto x^2}\\ &=\int_0^1\frac{\ln^2(1-x^2)}{x}\sum_{n=1}^\infty\frac{x^{2n}H_{2n}}{2n}\ dx, \quad \color{red}{2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}+\sum_{n=1}^\infty (-1)^na_{n}}\\ &=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\sum_{n=1}^\infty\frac{x^{n}H_{n}}{n}+\sum_{n=1}^\infty\frac{(-x)^{n}H_{n}}{n}\right)\ dx\\ &=\frac12\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\underbrace{\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)}_{\frac12\operatorname{Li}_2(x^2)}+\frac12\ln^2(1-x)+\frac12\ln^2(1+x)\right)\ dx\\ &=\small{\frac14\underbrace{\int_0^1\frac{\ln^2(1-x^2)\operatorname{Li}_2(x^2)}{x}\ dx}_{x^2\mapsto x}+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx}\\ &=\frac18\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x)}{x}\ dx+\frac14\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\ &=\frac18A+\frac14B\tag{1} \end{align}

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Let's start with $A$ and by seting $1-x\mapsto x$ we can write

\begin{align} A&=\int_0^1\frac{\ln^2x\color{blue}{\operatorname{Li}_2(1-x)}}{1-x}\ dx\\ &=\int_0^1\frac{\ln^2x}{1-x}\left[\color{blue}{\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)}\right]\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^2x}{1-x}\ dx-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx-\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\ dx-\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^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{A=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 $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$

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To evaluate $B$, we are going to use the key identity

$$(a+b)^2(a^2+b^2)=\frac23a^4+\frac23b^4+\frac5{12}(a+b)^4-\frac1{12}(a-b)^4$$

and by taking $a=\ln(1-x)$ and $b=\ln(1+x)$ , we get

$$B=\int_0^1\frac{\ln^2(1-x^2)}{x}\left(\ln^2(1-x)+\ln^2(1+x)\right)\ dx\\ =\small{\frac23\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac5{12}\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}-\frac1{12}\underbrace{\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}}\\ =\frac78\int_0^1\frac{\ln^4(1-x)}{x}\ dx+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\int_0^1\frac{\ln^4x}{1-x^2}\ dx\\ =\frac78(24\zeta(5))+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx-\frac16\left(\frac{93}{4}\zeta(5)\right)\\ =\frac{137}{8}\zeta(5)+\frac23\int_0^1\frac{\ln^4(1+x)}{x}\ dx$$

Since \begin{align} \int_0^1\frac{\ln^4(1+x)}{x}&=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\ &=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\ &=\small{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)} \end{align}

Then

$$\boxed{B=\frac83\ln^32\zeta(2)-7\ln^22\zeta(3)+\frac{265}{8}\zeta(5)-\frac8{15}\ln^52-16\ln2\operatorname{Li}_4\left(\frac12\right)-16\operatorname{Li}_5\left(\frac12\right)}$$

Plugging the boxed results of $A$ and $B$ in (1) we get our first relation:

$$R_1=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}\left(H_n^2+H_n^{(2)}\right)\\ =\small{\frac23\ln^32\zeta(2)-\frac74\ln^22\zeta(3)+\frac14\zeta(2)\zeta(3)+\frac{261}{32}\zeta(5)-\frac2{15}\ln^52-4\ln2\operatorname{Li}_4\left(\frac12\right)-4\operatorname{Li}_5\left(\frac12\right)}$$

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We have

$$\frac{\ln^2(1-y)}{1-y}=\sum_{n=1}^\infty y^n(H_n^2-H_n^{(2)})$$

integrate both sides from $y=0$ to $y=x$ to get

$$-\frac13\ln^3(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^2-H_n^{(2)}\right)=\sum_{n=1}^\infty\frac{x^{n}}{n}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Now replace $x$ with $x^2$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and integrate from $x=0$ to $x=1$, also note that $\int_0^1 -x^{2n-1}\ln(1-x)\ dx=\frac{H_{2n}}{2n}$ we get

$$\frac13\underbrace{\int_0^1\frac{\ln^3(1-x^2)\ln(1-x)}{x}\ dx}_{\large C}=\sum_{n=1}^\infty\frac{H_{2n}}{2n^2}\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac{2}{n^2}\right)$$

Rearranging the terms to get

$$R_2=\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^2}(H_n^2-H_n^{(2)})=4\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n)^3}-8\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac16C\tag{2}$$

Cornel elegantly calculated the first sum here

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

For the second sum: $$\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac12\sum_{n=1}^\infty\frac{H_{n}}{n^4}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}$$

plugging the common results:

$$\sum_{n=1}^\infty\frac{H_{n}}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

$$\sum_{n=1}^\infty(-1)^n\frac{H_{n}}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$

we get

$$\boxed{\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}=\frac{37}{64}\zeta(5)-\frac14\zeta(2)\zeta(3)}$$

For the remaining integral $C$, we use the magical identity

$$(a+b)^3a=a^4-b^4+\frac12(a+b)^4-\frac12(a-b)^4-(a-b)^3b$$

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

$$C=\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1+x)}{x}\ dx+\frac12\underbrace{\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2\mapsto x}\\-\underbrace{\frac12\int_0^1\frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}-\underbrace{\int_0^1\frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}\ dx}_{\frac{1-x}{1+x}\mapsto x}$$

$$C=\frac54\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{4!\zeta(5)}-\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{K}-\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}+\underbrace{2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx}_{J}$$

$$C=\frac{27}{4}\zeta(5)-K+J\tag{3}$$

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

for $J$

$$J=2\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x^2}\ dx=\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx+\int_0^1\frac{\ln^3x\ln\left(\frac{1+x}{2}\right)}{1+x}\ dx$$

using the rule

$$\int_0^1\frac{\ln^ax\ln\left(\frac{1+x}{2}\right)}{1-x}\ dx=(-1)^aa!\sum_{n=1}^\infty\frac{(-1)^nH_n^{a+1}}{n}$$

allows us to write

\begin{align} J&=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}\ dx-\ln2\int_0^1\frac{\ln^3x}{1+x}\ dx\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-\sum_{n=1}^\infty(-1)^n H_n\int_0^1x^n\ln^3x\ dx-\ln2\left(-\frac{21}4\zeta(4)\right)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}+6\sum_{n=1}^\infty\frac{(-1)^n H_n}{(n+1)^4}+\frac{21}{4}\ln2 \zeta(4)\\ &=-6\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}-6\sum_{n=1}^\infty\frac{(-1)^n H_n}{n^4}-\frac{45}{8}\zeta(5)+\frac{21}{4}\ln2 \zeta(4) \end{align}

Plugging

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

we get

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

Plugging the results of $K$ and $J$ in (3) we get

$$\boxed{\small{C=24\operatorname{Li}_5\left(\frac12\right)+24\ln2\operatorname{Li}_4\left(\frac12\right)+\frac3{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{21}2\ln^22\zeta(3)-4\ln^32\zeta(2)+\frac45\ln^52}}$$

and by substituting the boxed results in (2) we get our second relation

$$R_2=\sum _{n=1}^{\infty } \frac{H_{2 n} }{(2 n)^2}(H_n^2-H_n^{(2)}) =\frac78\zeta(2)\zeta(3)+5\zeta (5)+\frac{2}{3} \ln ^3(2)\zeta (2) -\frac{7}{4} \ln ^2(2)\zeta (3)\\ -\frac{2}{15} \ln^5(2) -4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)$$

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Thus

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{R_1-R_2}{2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$

and as a bonus

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^2}{(2n)^2}=\frac{R_1+R_2}{2}\\ =\small{\frac{421 }{64}\zeta (5)+\frac{9 }{16}\zeta (2) \zeta (3)+\frac{2}{3} \ln ^32\zeta (2) -\frac{7}{4} \ln ^22\zeta (3) -\frac{2}{15} \ln^52-4 \ln2\operatorname{Li}_4\left(\frac{1}{2}\right) -4 \operatorname{Li}_5\left(\frac{1}{2}\right)}$$

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Note:

$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ can be found here and $\sum_{n=1}^\infty\frac{(-1)^nH_n^{(4)}}{n}$ can be found here.

Answer 3

Update: the details may be found in the preprint The evaluation of a special harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$

The magical way by Cornel Ioan Valean

By the Cauchy product, we have $\operatorname{Li}_2(x^2) \log(1-x^2)= 3\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^3}-2\sum _{n=1}^{\infty } x^{2n}\frac{H_n}{n^2}-\sum _{n=1}^{\infty } x^{2n}\frac{H_n^{(2)}}{n}$, and if we multiply both sides by $\log(1-x)/x$, and integrate from $x=0$ to $x=1$, using that $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, and doing all the reductions, we arrive at

$$2\sum _{n=1}^{\infty } \frac{H_{2 n} H_n^{(2)}}{(2 n)^2}-12\sum _{n=1}^{\infty } \frac{H_n}{n^4}+12\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{n^4}+\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log \left(1-x^2\right) \log (1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x}\textrm{d}x+2 \int_0^1 \frac{\text{Li}_2(-x) \log ^2(1-x)}{x} \textrm{d}x\\+2 \int_0^1 \frac{\text{Li}_2(x) \log ^2(1-x)}{x} \textrm{d}x$$ $$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \sum _{n=1}^{\infty } \frac{H_n^2}{n^3}-2 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}H_n^2}{n^3}+2 \sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}\\-2 \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n^{(2)}}{n^3},$$ where the last integral is given here Two very advanced harmonic series of weight $5$, and all the last resulting harmonic series are given in the book (Almost) Impossible Integrals, Sums, and Series. The reduction to the last series has been achieved by using the identity, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$. The series $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$ maybe found calculated in the paper On the calculation of two essential harmonicseries with a weight 5 structure, involving harmonic numbers of the type H_{2n} by Cornel Ioan Valean. Thus, we have

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3).$$

All the details will appear in a new paper.