So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $\displaystyle H_n-2H_{2n}+H_{4n}=\int_0^1 \frac{x^{2n}-x^{4n}}{1+x}dx$.
I noticed that $H_n-2H_{2n}+H_{4n}=H_{4n}-H_{2n}-(H_{2n}-H_n)=H_{4n^{-}}-H_{{2n}^{-}}$, where $H_{n^{-}}=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$ is called
a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work.
If I make any unexplainable substitution, it's most likely $t=\frac{1-x}{1+x}$.
Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted:
$ -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} ,-\ln\cos x=\ln2+\sum_{n=1}^{\infty} \frac{(-1)^n\cos(2nx)}{n} \tag{1}$
$$ \int_0^{\frac{\pi}{2}} \cos x \cos(nx)dx=\begin{cases} \frac{\pi}{4} &n=1\\0 &n \,\,\text{odd}\\ \frac{(-1)^{1+n/2}}{n^2-1} &n \,\,\text{even} \end{cases} \tag{2}$$
$$ \int_0^1 \frac{\ln(1-x)}{a+x}dx=-\operatorname{Li_2}\left(\frac1{a+1}\right)\tag{3}$$
Starting,
$$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=\sum_{n=0}^{\infty}\int_0^1\int_0^1\frac{(x^{2n}-x^{4n})(u^{2n}-u^{4n})}{(1+x)(1+u)}dxdu
\\=\small\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}-2\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^4)}+\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^4)}
\\=I_{22}-2I_{24}+I_{44}$$
Computing $I_{22}$.
Substitute $u=\frac{y}{x}$ ,change the order of integration, evaluate the inner integral, and substitue $t=\frac{1-x}{1+x}$ to get
$$\begin{align} I_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}=\int_0^1\int_0^x\frac{dydx}{(1+x)(x+y)(1-y^2)}
\\=\int_0^1 \frac1{1-y^2}\int_1^y \frac{dx}{(1+x)(x+y)} dy=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1+x)(1-x^2)}\,dx
\\=\frac{-1}{4}\int_0^1 \frac{(1+t)}{t^2}\ln(1-t^2)dt=-\frac14\int_0^1\frac{\ln(1-t^2)}{t^2}dt-\frac14\int_0^1\frac{\ln(1-t^2)}{t}dt
\\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+1)}+\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+2)}=\frac{\ln2}{2}+\frac{\pi^2}{48}.\end{align}$$
Computing $I_{44}$.
Start the same as with $I_{22}$ to get $\displaystyle I_{44}=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1-x)(1-x^4)}\,dx=\frac{-1}{8}\int_0^1 \frac{\ln(1-t^2)}{t^2(1+t^2)}(1+t)^3dt$.
We can calculate these integrals:
$$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx+\int_0^1 \frac{\ln(1-x)}{1+x^2}dx \tag{4}
\\=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx +\int_0^1 \frac{\ln\left(\frac{2t}{1+t}\right)}{1+t^2}dt
\\=\frac{\pi}{4}\ln2+\sum_{n=0}^{\infty} (-1)^n\int_0^1\ln(t) t^{2n}dt=\frac{\pi}{4}\ln2-G. \end{align}$$
$$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x^2}dx-\int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx \tag{5}
\\=-\sum_{n=0}^{\infty} \frac1{n+1}\int_0^1 x^{2n}dx-\frac{\pi}{4}\ln2+G=G-\frac{\pi}{4}\ln2-2\ln2.\end{align}$$
$$\begin{align} \int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx=\frac12\int_0^1 \frac{\ln(1-x)}{1+x}dx \tag{6}
\\=-\frac12 \operatorname{Li_2}\left(\frac12\right)=\frac{\ln^2 2}{4}-\frac{\pi^2}{24}.\end{align}$$
$$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x}dx-\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx \tag{7}
\\=-\sum_{n=0}^{\infty}\frac1{n+1}\int_0^1 x^{2n+1}dx-\frac{\ln^2 2}{4}+\frac{\pi^2}{24}=-\frac{\pi^2}{24}-\frac{\ln^2 2}{4}.\end{align}$$
Altogether,
$$I_{44}=\frac{-1}{8}\int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}(1+3x+3x^2+x^3)dx
\\=-\frac{\pi}{16}\ln2+\frac{\ln2}{4}+\frac{\ln^2 2}{16}+\frac{\pi^2}{48}+\frac{G}{4}.$$
Computing $I_{24}$.
Substitute $u=\frac{y}{x^2}$, change the order of integration, let $y\to y^2$, evaluate the inner integral,and substitue $t=\frac{1-x}{1+x}$:
$$\begin{align*} I_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^2)}=\int_0^1\int_0^{x^2} \frac{dydx}{(1+x)(x^2+y)(1-y^2)}
\\=\int_0^1 \frac1{1-y^2}\int_{\sqrt{y}}^1\frac{dx}{(1+x)(x^2+y)}dy=2\int_0^1\frac{y}{1-y^4}\int_{y}^1\frac{dx}{(1+x)(x^2+y^2)}dy
\\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1-x^4)}dx
=I_{241}-I_{242}. \end{align*}$$
Evaulation of $I_{241}$.
Substitute $t=\frac{1-x}{1+x}$ to get $\displaystyle I_{241}=\frac14\int_0^1 \frac{\tan^{-1}(t)}{t(1+t^2)^2}(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=\tan\theta$:
$$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^2(\theta)d\theta=\frac{\pi^2}{64}+\frac{\pi}{16}-\frac18.\tag{8}\end{align}$$
$$\begin{align} \int_0^1 \frac{x\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\tan\theta\cos^2{\theta}d\theta=\frac12\int_0^{\frac{\pi}{4}}\theta\sin(2\theta)d\theta=\frac18.\tag{9}\end{align}$$
$$\begin{align} \int_0^1 \frac{x^2\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^2(\theta)d\theta=\frac{\pi^2}{64}-\frac{\pi}{16}+\frac18.\tag{10}\end{align}$$
$$\begin{align} \int_0^1 \frac{x^3\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^3(\theta)\sec{\theta}\,d\theta \tag{11}
\\=\int_0^{\frac{\pi}{4}}\theta\tan\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=\frac{\pi}{8}\ln2-\frac18+\int_0^{\frac{\pi}{4}}\ln\cos\theta \,d\theta
\\=\frac{\pi}{8}\ln2-\frac18-\int_0^{\frac{\pi}{4}}\ln2 \,d\theta-\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta
\\=-\frac{\pi}{8}\ln2-\frac18+\frac12\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\sin(\frac{\pi n}{2})=\frac{G}{2}-\frac{\pi}{8}\ln2-\frac18.\end{align}$$
$$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^3(\theta)\csc{\theta}\,d\theta \tag{12}
\\=\int_0^{\frac{\pi}{4}}\theta\cot\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=-\frac18-\frac{\pi}{8}\ln2-\int_0^{\frac{\pi}{4}}\ln\sin\theta \,d\theta
\\=-\frac18-\frac{\pi}{8}\ln2+\int_0^{\frac{\pi}{4}}\ln2 \,d\theta+\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta
\\=-\frac18+\frac{\pi}{8}\ln2+\frac12\sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n^2}=\frac{G}{2}+\frac{\pi}{8}\ln2-\frac18.\end{align}$$
Altogether,
$$I_{241}=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx= \frac14\int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}(1+4x+6x^2+4x^3+x^4)dx
\\=\frac{\pi^2}{32}+\frac18+\frac{G}{4}$$
Evaulation of $I_{242}$.
Substitute $t=\frac{1-x}{1+x}$ to get
$$ I_{242}=\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)^2}(1-t^2)(1+t)^2dt
\\=\frac12\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{(1+t^2)^2}(1-t^2)dt+\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)}(1-t^2)dt
\\=\frac12\int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx-\frac14\int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx\\+\frac18\int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx-\frac14\int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx
$$
Calculating these integrals:
$$\begin{align} \int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx=-\frac12\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x}(1+x)}\frac{1-x}{1+x}dx \tag{13}
\\=-\frac12\int_0^1\frac{t\ln t}{\sqrt{1-t^2}}dt=-\frac18\int_0^1\frac{\ln t}{\sqrt{1-t}}dt=-\frac18\int_0^1 t^{-1/2}\ln(1-t)dt
\\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+3)}=\frac12-\frac{\ln2}{2}.\end{align}$$
$$\begin{align} \int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1+x)(1-x)}{x(1+x)}dx \tag{14}
\\=\frac12\int_0^1\frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx
\\=\frac12\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}\int_0^1 x^n dx-\frac12\ln^2(1+x)\bigg{|}_0^1=\frac{\pi^2}{24}-\frac{\ln^2 2}{2}.\end{align}$$
$$\begin{align} \int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1-x)}{x}dx-\int_0^1\frac{\ln(1-x)}{1+x}dx \tag{15}
\\=-\frac{\pi^2}{12}-\left(\frac{\ln^2 2}{2}-\frac{\pi^2}{12}\right)=-\frac{\ln^2 2}{2}.\end{align}$$
$$
\int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx=-2\int_0^{\frac{\pi}{4}}\cos^2(\theta)(1-\tan^2\theta)\ln\cos\theta\,d\theta \tag{16}
\\=-2\int_0^{\frac{\pi}{4}}\cos(2\theta)\ln\cos\theta\,d\theta=2\ln2\int_0^{\frac{\pi}{4}}\cos(2\theta)d\theta+\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{2}}\cos\theta \cos(n\theta)d\theta
\\=\ln2-\frac{\pi}{4}+\sum_{n=1}^{\infty} \frac{(-1)^{2n}}{2n}\frac{(-1)^{n+1}}{(2n)^2-1}=\frac{\ln2}{2}-\frac{\pi}{4}+\frac12.$$
Altogether, $\displaystyle I_{242}=\frac{\pi^2}{192}+\frac{\pi}{16}+\frac{\ln^2 2}{16}-\frac{3\ln2}{8}+\frac18$,
leading to $\displaystyle I_{24}=I_{241}-I_{242}=\frac{5\pi^2}{192}-\frac{\pi}{16}-\frac{\ln^2 2}{16}+\frac{3\ln2}{8}+\frac{G}{4}$,
and finally, confirming the conjecture,
$$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=I_{22}-2I_{24}+I_{44}=\frac{\pi}{8}-\frac{\pi}{16}\ln2-\frac{\pi^2}{96}+\frac{3\ln^2 2}{16}-\frac{G}{4}.$$
I don't know about higher powers. I guess the case $\mathcal A_2$ can also be done. If we start the same as with $\mathcal B_2$, writing $\mathcal A_2=J_{22}-2J_{24}+J_{44}$
we can find that $\displaystyle J_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^2u^2)}=2I_{44}-I_{22}=-\frac{\pi}{8}\ln2+\frac{G}{2}+\frac{\pi^2}{48}+\frac{\ln^2 2}{8}$
$J_{44}$ can be reduced to $\displaystyle =-\frac12\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx$.
already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2\sqrt{2})(x+3-2\sqrt{2}).$
I can get $\displaystyle \int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}dx=-\frac{\pi^2}{12}+\frac{4-3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{4}\right)+\frac{4+3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{4}\right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain
$$\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx=\frac{1+2\sqrt{2}}{4}\pi\ln2-\frac{\pi^2}{24}-\frac14\ln\left(\frac{2+\sqrt{2}}{4}\right)\ln\left(\frac{2-\sqrt{2}}{4}\right)
-\frac{\sqrt{2}+1}{2}\Im\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)-\frac{\sqrt{2}-1}{2}\Im\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)$$
I obtained it by calculating $\displaystyle \int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2\ln\left(\frac{a+1}{a-1}\right)+\operatorname{Li_2}\left(\frac2{1-a}\right)-\operatorname{Li_2}\left(\frac1{1-a}\right)$,
which together with $(3)$ can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x+a}dx$, which in turn, through partial fractions, can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x^2+a^2}dx$.
Fortunately, things didn't get too ugly as both $3+2\sqrt{2}$ and $3-2\sqrt{2}$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_{24}$. Starting similarly as with $I_{24}$,
we have:
$$J_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^4u^2)}
\\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1+x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1+x^4)}dx
\\=J_{241}-J_{242}$$
Through $t=\frac{1-x}{1+x}$, $J_{241}$ turns to $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)(x^4+6x^2+1)}(1+x)^4\,dx$. I don't have any idea about that yet. \Edit 1.
Here is a sketch of Cornel's way.
We use that $\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+2n+1)}=\frac{H_{2n}^2+H_{2n}^{(2)}}{4n}$ and then multiply all by $1/n^2$ and consider the summation from $n=1$ to $\infty$. Later in the process we use another critical step, $\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2k+2n+1) 2n}=\frac{1}{(2k+1)^2}+\frac{H_{2k}}{2k+1}-\frac{H_k}{2(2k+1)}-\frac{\log(2)}{2k+1}$.
Essentially, these are almost exactly the steps presented in section 6.59, pages $530$-$532$, in the book (Almost) Impossible Integrals, Sums, and Series.
So, after simple calculations and rearrangements we arrive at
$$\frac{1}{4} \sum _{n=1}^{\infty } \frac{\left(H_{2 n}\right){}^2}{n^3}+\frac{1}{4} \sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^3}$$
$$=\frac{1}{8}\sum _{n=1}^{\infty }\frac{H_n}{n^4}-\frac{1}{4}\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}+\frac{\pi^2}{24}\sum _{n=1}^{\infty }\frac{H_{2 n}}{n^2}-4\sum _{n=1}^{\infty }\frac{H_{2 n+1}^2}{(2 n+1)^3}+\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^2}\\+4\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^4}+2\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{(2 n+1)^3}+4 \log (2)\sum _{n=1}^{\infty }\frac{ H_{2 n+1}}{(2 n+1)^3}-\frac{\pi^2}{48}\sum _{n=1}^{\infty }\frac{1}{n^3}-\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^3}-4 \log (2)\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^4}.$$
Since all the series are known except the desired one, the extraction is immediately achieved.
For example, a solution to the challenging series $\displaystyle \sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}$ is presented in https://math.stackexchange.com/q/3345138.
The way to go also appears in details in the preprint On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$ by Cornel Ioan Valean
Best Answer
The blue sum, with the denominator rearranged, comes immediately from the result given in Section 4.59, page $313$, from the book (Almost) Impossible Integrals, Sums, and Series.
In fact, in the book the author nicely exploits the fact that for the linear Euler sum of the type $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^m}$, with $m=3$, we arrive at $5/4 \zeta(4)$ which allows us to express the $\zeta(4)$ value in terms of a sum of seven series. You might not need this precise representation, but almost all the steps presented in the solution. It is precisely the same strategy as for the weight $5$ case which is given in On the calculation of two essential harmonic series with a weight 5 structure, involving harmonic numbers of the type $H_{2n}$. In this case we play with weight $4$ series. Observe that all the other series above are known or easily reducible to known series.
A note: In this question Two very advanced harmonic series of weight $5$, if you take a look at the second and third series you may see how they look like when having $2n-1$ and $2n+1$ in denominator (the latter version looks way better in terms of a closed-form). Well, like our case, except that there we are on the realm of weight $5$ series.
What about the red part? We want a clever rearrangement of the initial series, that is $$\sum _{n=1}^{\infty } \frac{2 H_{2 n}-H_n-2 \log (2)}{2 n (2 n-1)^2}$$ $$=2\sum _{n=1}^{\infty } \frac{H_{2 n-1}+1/(2n)}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{2 n (2 n-1)}-2 \log (2)\sum _{n=1}^{\infty } \frac{1}{(2 n-1)^2}$$ $$+2 \sum _{n=1}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{2 n (2 n-1)}.$$
Both the first and the second series are done by using the results from this paper A new powerful strategy of calculating a classof alternating Euler sums by Cornel Ioan Valean, particularly the main theorem and lemma $4$. Then, the third and the fourth sums are trivial.
Finally, there is a nice thing to observe about the fifth sum, that is if we reindex it and start from $n=0$, we can simply use the series from the second step in this answer Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$, which is finalized elementarily.
End of story.