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I have already evaluated this sum:

\begin{equation*} \sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42 \end{equation*} using the identity $\displaystyle\frac{1}{1-x^2}\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)x^{2n-1}$

but kind of lengthy. any other approaches?

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    $\begingroup$ @Henry Lee H_n is the harmonic number =1+1/2+1/3+....+1/n $\endgroup$ – Ali Shather Apr 25 at 19:44
  • $\begingroup$ Ah thanks, sorry for the mistake $\endgroup$ – Henry Lee Apr 25 at 20:11
  • $\begingroup$ I guess there isn't any short route. I hope I'm wrong. $\endgroup$ – Felix Marin Apr 25 at 21:19
  • $\begingroup$ @Felix Martin I guess so too. I'll provide my solution soon. $\endgroup$ – Ali Shather Apr 26 at 22:10
  • $\begingroup$ @AliShather $\Huge\left(\bullet\quad\bullet \atop {\mid \atop {\Huge\smile}}\right)$. $\endgroup$ – Felix Marin Apr 26 at 22:14
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using the following identity $$\frac{1}{1-z^2}\ln\left(\frac{1-z}{1+z}\right)=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)z^{2n-1}$$ that can be proved as follows: using Cauchy product rule, $$\frac{1}{1-z^2}\ln\left(\frac{1-z}{1+z}\right)=-2\frac{1}{1-z^2}\tanh^{-1}z=-2\left(\sum_{n=1}^\infty z^{2n-2}\right)\left(\sum_{n=1}^\infty\frac{z^{2n-1}}{2n-1}\right)\\=-2\sum_{n=1}^\infty z^{2n-1}\left(\sum_{k=1}^n\frac1{2k-1}\right)=-2\sum_{n=1}^\infty z^{2n-1}\left(H_{2n}-\frac12H_n\right)\\=\sum_{n=1}^{\infty}\left(H_n-2H_{2n}\right)z^{2n-1}$$ integrate both sides from $z=0$ to $x$, we get $$\ln^2\left(\frac{1-x}{1+x}\right)=-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}x^{2n}$$ multiply both sides by $\frac{\ln(1-x)}{x}$ and integrate from $0$ to $1$, we get $$-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\int_0^1x^{2n-1}\ln(1-x)\ dx =-2\sum_{n=1}^\infty\frac{H_n-2H_{2n}}{n}\left(-\frac{H_{2n}}{2n}\right)\\ =\int_0^1\frac{\ln(1-x)}{x}\ln^2\left(\frac{1-x}{1+x}\right)\ dx$$ or \begin{align} \sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}&=8\sum_{n=1}^\infty\left(\frac{H_{2n}}{2n}\right)^2+\int_0^1\frac{\ln(1-x)}{x}\ln^2\left(\frac{1-x}{1+x}\right)\ dx\\ &=4\sum_{n=1}^\infty\frac{H_n^2}{n^2}+4\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}+\int_0^1\frac{\ln(1-x)}{x}\ln^2\left(\frac{1-x}{1+x}\right)\ dx\tag{1} \end{align} lets evaluate the integral and after expanding the log, we get

$$I=\int_0^1\frac{\ln^3(1-x)}{x}\ dx-2\int_0^1\frac{\ln^2(1-x)\ln(1+x)}{x}\ dx+\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx$$ letting $1-x=y$ for the first and second integral and using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)$ for the third integral, we get \begin{align} I&=\int_0^1\frac{\ln^3x}{1-x}-2\int_0^1\frac{\ln^2x\ln(2-x)}{1-x}+2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln(1-x)\\ &=-6\zeta(4)-2\ln(2)\int_0^1\frac{\ln^2x}{1-x}\ dx-2\int_0^1\frac{\ln^2x\ln(1-x/2)}{1-x}\ dx\\ &\quad+2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(-\frac{H_n}{n}\right)\\ &=-6\zeta(4)-4\ln2\zeta(3)+2\sum_{n=1}^\infty\frac{1}{n2^n}\int_0^1\frac{x^n\ln^2x}{1-x}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\\ &=-6\zeta(4)-4\ln2\zeta(3)+2\sum_{n=1}^\infty\frac{2\zeta(3)-2H_n^{(3)}}{n2^n}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\\ &=-6\zeta(4)-4\ln2\zeta(3)+4\ln2\zeta(3)-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}\tag{2} \end{align} Plugging $(2)$ in $(1)$, we get $$\boxed{\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}=4\sum_{n=1}^\infty\frac{H_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-4\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}-6\zeta(4)}$$

I managed to prove the following sums $$\sum_{n=1}^{\infty}\frac{H_n^2}{n^2}=\frac{17}4\zeta(4)$$ $$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$ $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$ $$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$ combining these sums, we get $$\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left(\frac12\right)+\frac{13}8\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac{1}{6}\ln^42$$

the proof of the first, second and third sum can be found here, here and here respectively.

as for the forth sum, it can be found using the easy-to-prove generating function $\sum_{n=1}^\infty\frac{x^nH_n^{(3)}}{n}=\operatorname{Li_4}(x)-\ln(1-x)\operatorname{Li_3}(x)-\frac12\operatorname{Li_2}^2(x)$.

I hope there is a simpler solution than mine that does not reply much on the series I used.


Addendum:

Different proof for the identity above.

\begin{align} \ln^2\left(\frac{1-x}{1+x}\right)&=2\ln^2(1-x)+2\ln^2(1+x)-\ln^2(1-x^2)\tag1\\ &=4\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n+4\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}x^n-2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^{2n}\tag2\\ &=4\sum_{n=1}^\infty(1+(-1)^n)\frac{H_{n-1}}{n}x^n-2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^{2n}\tag3\\ &=8\sum_{n=1}^\infty\frac{H_{2n-1}}{2n}x^{2n}-2\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^{2n}\tag4\\ &=2\sum_{n=1}^\infty\frac{2H_{n}-H_n}{n}x^{2n}\tag5 \end{align}


Explanation:

$(1)$ $(a-b)^2=2a^2+2b^2-(a+b)^2$

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

$(3)$ Factoring out

$(4)$ $\sum_{n=1}^\infty (1+(-1)^n)a_{n}=2\sum_{n=1}^\infty a_{2n}$

$(5)$ $H_{n-1}=H_n-\frac1n$

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Outstanding solution due to Cornel Valean. Recall the generating function $\displaystyle-\ln(1+x)\ln(1-x)=\sum_{n=1}^\infty x^{2n}\frac{H_{2n}-H_n}{n}+\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2},$ where if we multiply both sides by $\ln(1+x)/x$ and use the simple fact $\displaystyle\int_0^1x^{2n-1}\ln(1+x)\ dx=\frac{H_{2n}-H_n}{2n}$ then the Au-Yeung eries result, $\displaystyle\sum_{n=1}^\infty\left(\frac{H_n}{n}\right)^2=\frac{17}{4}\zeta(4),$ and $\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$, we have $$\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}-2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}-2\sum_{n=1}^\infty\frac{H_n}{(2n)^3}=\frac{29}{16}\zeta(4)-\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=\frac{23}{16}\zeta(4)\tag{1}$$

where $\displaystyle\int_0^1\frac{\ln(1-x)\ln^2(1+x)}{x}\ dx=-3/8\zeta(4)$ is an already famous integral elementary to evaluate using the algebraic identity, $\displaystyle6a^2b=(a+b)^3-(a-b)^3-2b^3$. since $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n}{n^3}=\frac{11}4\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$ and $\displaystyle\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{n^2}=\frac{41}{16}\zeta(4)-\frac74\ln(2)\zeta(3)+\frac12\ln^22\zeta(2)-\frac1{12}\ln^42-2\operatorname{Li}_4\left(\frac12\right)$, if using for the last two series in (1) that $2\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_n-\sum_{n=1}^\infty (-1)^{n-1}a_n$, we conclude that $\displaystyle\sum_{n=1}^\infty\frac{H_nH_{2n}}{n^2}=\frac{13}8\zeta(4)+\frac72\ln(2)\zeta(3)-\ln^22\zeta(2)+\frac1{6}\ln^42+4\operatorname{Li}_4\left(\frac12\right)$ and the solution is complete.

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Again, here is a slightly different approach. Unfortunately, like your solution, it is quite (very) lengthy. I have tried to make my solution as self-contained as possible. What this means is that while many of the integrals which I evaluate can be found elsewhere on this site, I just go ahead and evaluate each as they appear.

Recalling $$-\frac{H_{2n}}{2n} = \int_0^1 x^{2n - 1} \ln (1 - x) \, dx,$$ the sum can be written as \begin{align} \sum_{n = 1}^\infty \frac{H_n H_{2n}}{n^2} &= 2\sum_{n = 1}^\infty \frac{H_n}{n} \cdot \frac{H_{2n}}{2n}\\ &= -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{H_n x^{2n}}{n} \, dx\tag1 \end{align} Making use of the following well-known generating function for the harmonic numbers $$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$ on replacing $x$ with $x^2$ we have $$\sum_{n = 1}^\infty \frac{H_n x^{2n}}{n} = \frac{1}{2} \ln^2 (1 - x^2) + \operatorname{Li}_2 (x^2).$$ Substituting the above result into (1) gives \begin{align} \sum_{n = 1}^\infty \frac{H_n H_{2n}}{n^2} &= - \int_0^1 \frac{\ln (1 -x) \ln^2 (1 - x^2)}{x} \, dx - 2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (x^2)}{x} \, dx\\ &= -I_1 - 2 I_2.\tag2 \end{align}


The first integral $I_1$

Since $$\ln^2(1 - x^2) = \ln^2 (1 - x) + 2 \ln (1 - x) \ln (1 + x) + \ln^2 (1 + x),$$ the first integral can be written as \begin{align} I_1 &= \int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx + 2 \int_0^1 \frac{\ln^2 (1 - x) \ln (1 + x)}{x} \, dx + \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x)}{x} \, dx\\ &= I_a + 2 I_b + I_c \end{align}

Integral $I_a$

\begin{align} I_a &= \underbrace{\int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx}_{x \, \mapsto \, 1 - x} = \int_0^1 \frac{\ln^3 x}{1 - x} \, dx = \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\int_0^1 x^{n + s} \, dx \right ]_{s = 0}\\ &= \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \right ]_{s = 0} = -6 \underbrace{\sum_{n = 0}^\infty \frac{1}{(n + 1)^4}}_{n \, \mapsto n - 1} = -6 \sum_{n = 1}^\infty \frac{1}{n^4} = - 6 \zeta (4) \end{align}

Integrals $I_b$ and $I_c$

Note that $$a^2 b = \frac{1}{6} (a + b)^3 + \frac{1}{6} (a - b)^3 - \frac{1}{3} a^3,$$ and $$ab^2 = \frac{1}{6}(a + b)^3 + \frac{1}{6}(a - b)^3 - \frac{1}{3} a^3.$$

If we set $a = \ln (1 - x)$ and $b = \ln (1 + x)$, on applying the first of the above identities we see that \begin{align} I_b &= \frac{1}{6} \underbrace{\int_0^1 \frac{\ln^3 (1- x^2)}{x} \,dx}_{x \, \mapsto \, \sqrt{x}} - \frac{1}{6} \underbrace{\int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{x}}_{x \, \mapsto \, (1 - x)/(1 + x)}- \frac{1}{3} \int_0^1 \frac{\ln^3 (1 + x)}{x} \, dx\\ &= \frac{1}{12} \int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx - \frac{1}{3} \int_0^1 \frac{\ln^3 x}{1 - x^2} \, dx - \frac{1}{3} \int_0^1 \frac{\ln^3 (1 + x)}{x} \, dx\\ &= \frac{1}{12} I_a - \frac{1}{3} I_y - \frac{1}{3} I_z, \end{align} while on applying the second of the above identities we see that \begin{align} I_c &= \frac{1}{6} \underbrace{\int_0^1 \frac{\ln^3 (1- x^2)}{x} \,dx}_{x \, \mapsto \, \sqrt{x}} + \frac{1}{6} \underbrace{\int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{x}}_{x \, \mapsto \, (1 - x)/(1 + x)}- \frac{1}{3} \int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx\\ &= -\frac{1}{4} \int_0^1 \frac{\ln^3 (1 - x)}{x} \, dx + \frac{1}{3} \int_0^1 \frac{\ln^3 x}{1 - x^2} \, dx\\ &= -\frac{1}{4} I_a + \frac{1}{3} I_y. \end{align}

Integral $I_y$

\begin{align} I_y &= \int_0^1 \frac{\ln^3 x}{1 - x^2} \, dx = \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\int_0^1 x^{2n + s} \, dx \right ]_{s = 0} = \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\frac{1}{2n + s + 1} \right ]_{s = 0}\\ &= -6 \sum_{n = 0}^\infty \frac{1}{(2n + 1)^4} = -6 \lambda (4) = - 6 \left (1 - \frac{1}{2^4} \right ) \zeta (4) = - \frac{45}{8} \zeta (4). \end{align}

Integral $I_z$

\begin{align} I_z &= \underbrace{\int_0^1 \frac{\ln^3 (1 + x)}{x} \, dx}_{x \, \mapsto \, x/(x + 1)}\\ &= -\int_0^{\frac{1}{2}} \frac{\ln^3 (1 - x)}{1 - x} \, dx - \underbrace{\int_0^{\frac{1}{2}} \frac{\ln^3 (1 - x)}{x} \, dx}_{x \, \mapsto \, 1 - x}\\ &= \frac{1}{4} \ln^4 2 - \int_{\frac{1}{2}}^1 \frac{\ln^3 x}{1 - x} \, dx\\ &= \frac{1}{4} \ln^4 2 - \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\int_{\frac{1}{2}}^1 x^{n + s} \, dx \right ]_{s = 0}\\ &= \frac{1}{4} \ln^4 2 - \sum_{n = 0}^\infty \frac{d^3}{ds^3} \left [\frac{1}{n + s + 1} \left (1 - \frac{1}{2^{n + s + 1}} \right )\right ]_{s = 0}\\ &= \frac{1}{4} \ln^4 2 - \sum_{n = 0}^\infty \left [\frac{6}{(n + 1)^4} + \frac{1}{2^{n + 1} (n + 1)^4} + \frac{\ln^3 2}{2^{n + 1} (n + 1)} + \frac{\ln^2 2}{2^{n + 1} (n + 1)^2}\right.\\ & \qquad \qquad \qquad \qquad \left. + \frac{\ln^2 2}{2^n (n + 1)^2} + \frac{\ln 2}{2^{n - 1} (n + 1)^3} + \frac{\ln 2}{2^n (n + 1)^3} \right ]\\ &= \frac{1}{4} \ln^4 2 + 6 \sum_{n = 0}^\infty \frac{1}{n^4} - 6 \sum_{n = 0}^\infty \frac{1}{2^n n^4} - \ln^3 2 \sum_{n = 0}^\infty \frac{1}{2^n n} - 3 \ln^2 2 \sum_{n = 0}^\infty \frac{1}{2^n n^2} + 6 \ln 2 \sum_{n = 0}^\infty \frac{1}{2^n n^3}\\ &= -\frac{1}{4} \ln^4 2 - 6 \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{21}{4} \zeta (3) \ln 2 +\frac{3}{2} \zeta (2) \ln^2 2 + 6 \zeta (4). \end{align} Thus $$I_b = 2 \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{7}{4} \zeta (3) \ln 2 - \frac{1}{2} \zeta (2) \ln^2 2 + \frac{1}{12} \ln^4 2 - \frac{5}{8} \zeta (4),$$ and $$I_c = -\frac{3}{8} \zeta (4),$$ so that, finally $$I_1 = 4 \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{7}{2} \zeta (3) \ln 2 - \zeta (2) \ln^2 2 + \frac{1}{6} \ln^4 2 - \frac{61}{8} \zeta (4).$$


The second integral $I_2$

\begin{align} I_2 &= \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (x^2)}{x} \, dx}_{IBP}\\ &= -\zeta^2 (2) - 2 \int_0^1 \frac{\ln (1 - x^2) \operatorname{Li}_2 (x)}{x} \, dx\\ &= -\zeta^2 (2) - 2 \underbrace{\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (x)}{x} \, dx}_{IBP} - 2 \int_0^1 \frac{\ln (1 + x) \operatorname{Li}_2 (x)}{x} \, dx\\ &= -2 \int_0^1 \frac{\ln (1 + x) \operatorname{Li}_2 (x)}{x} \, dx\\ &= 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n} \int_0^1 x^{n - 1} \operatorname{Li}_2 (x) \, dx\\ &= 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n} \left [\frac{\zeta (2)}{n} + \frac{1}{n} \int_0^1 x^{n - 1} \ln (1 - x) \, dx \right ]\\ &= 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n} \left (\frac{\zeta (2)}{n} - \frac{H_n}{n^2} \right )\\ &= 2 \zeta (2) \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} - 2 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^3} \end{align} For the Euler sum that appears its value can be found from the following generating function \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}x^n &=2{\rm Li}_4(x)+{\rm Li}_4\left(\tfrac{x}{x-1}\right)-{\rm Li}_4(1-x)-{\rm Li}_3(x)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\tfrac{x}{x-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(x)\ln^2(1-x)+\frac{1}{2}{\rm Li}_2^2(x)+\frac{1}{6}\ln^4(1-x)-\frac{1}{6}\ln{x}\ln^3(1-x)\\ &+\frac{1}{2} \zeta (2) \ln^2(1-x)+\zeta(3)\ln(1-x)+\zeta (4),\tag3 \end{align} which is proved in this answer here.

Setting $x = -1$ in (3) gives \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}=2{\rm Li}_4\left(\frac{1}{2}\right)-\frac{11}{4} \zeta (4) + \frac{7}{4}\zeta(3)\ln{2} - \frac{1}{2} \zeta (2) \ln^2{2} + \frac{1}{12}\ln^4{2}, \end{align} Also, as $$\sum_{n = 1}^\infty \frac{(-1)^n}{n^2} = - \frac{1}{2} \zeta (2),$$ it follows that $$I_2 = - 4\operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{7}{2} \zeta (3) \ln 2 + \zeta (2) \ln^2 2 - \frac{1}{6} \ln^4 2 + 3 \zeta (4).$$


The main sum

On plugging the values for $I_1$ and $I_2$ into (2), the value for the sum becomes $$\sum_{n = 1}^\infty \frac{H_n H_{2n}}{n^2} = 4 \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{13}{8} \zeta (4) + \frac{7}{2} \zeta (3) \ln 2 - \zeta (2) \ln^2 2 + \frac{1}{6} \ln^4 2,$$ as required!

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