TITLE: Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$
QUESTION [57 upvotes]: There is a known asymptotic expansion of harmonic numbers $H_n$ for $n\to\infty$:
$$\begin{align}H_n&=\gamma+\ln n+\sum_{k=1}^\infty\left(-\frac{B_k}{k\cdot n^k}\right)\\
&=\gamma+\ln n+\frac1{2n}-\frac1{12n^2}+\frac1{120n^4}-\frac1{252n^6}\,+\,\dots,\end{align}\tag1$$
where $B_k$ are Bernoulli numbers.
We can take a linear combination of harmonic numbers to cancel constant and logarithmic terms, compensate for $O(n^{-1})$ term, and get the following series that is possible to evaluate in a closed form (e.g. using generating function):
$$\sum_{k=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}-\frac1{8n}\right)=\frac18-\frac\pi{16}.\tag2$$
Rather than compensating for $O(n^{-1})$ term, we can take a series with alternating signs, that is also possible to evaluate in a closed form:
$$\sum_{n=1}^\infty\,(-1)^n\,\Big(H_n-\,2H_{2n}+H_{4n}\Big)=\frac{3\pi}{16}-\frac\pi{4\sqrt2}-\frac{\ln2}8.\tag3$$
Generalizing, we can consider two families of series:
$$\mathcal A_m=\sum_{n=1}^\infty\,(-1)^n\,\Big(H_n-\,2H_{2n}+H_{4n}\Big)^m,\tag4$$
$$\mathcal B_m=\sum_{n=1}^\infty\Big(H_n-\,2H_{2n}+H_{4n}\Big)^m,\tag5$$
and try to evaluate them in a closed form.

So far I have the following conjectured result:

$$\large\sum_{n=1}^\infty\Big(H_n-\,2H_{2n}+H_{4n}\Big)^2\stackrel{\normalsize\color{gray}?}=\frac\pi8-\frac\pi{16}\,\ln2-\frac{\pi^2}{96}+\frac3{16}\,\ln^22-\frac{G}{4},\tag{$\diamond$}$$

where $G$ is the Catalan constant.
Could you please help me to prove this result and, possibly, find other values of $\mathcal A_m,\mathcal B_m$?

REPLY [5 votes]: I am going to use @nospoon's note in his solution above
$$H_n-2H_{2n}+H_{4n}=\overline{H}_{4n}-\overline{H}_{2n}$$
where $\overline{H}_n=\sum_{k=1}^n \frac{(-1)^{k-1}}{k}=\int_0^1\frac{1-(-x)^{n}}{1+x}\ dx$ is the alternating harmonic number
so 
$$S=\sum_{n=1}^\infty(H_n-2H_{2n}+H_{4n})^2=\sum_{n=1}^\infty(\overline{H}_{4n}-\overline{H}_{2n})^2$$
now lets use $$\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)+\sum_{n=1}^\infty (-1)^n f(n)$$
$$\Longrightarrow S=\frac12\sum_{n=1}^\infty(\overline{H}_{2n}-\overline{H}_{n})^2+\frac12\sum_{n=1}^\infty (-1)^n(\overline{H}_{2n}-\overline{H}_{n})^2$$
$$=\frac12S_1+\frac12S_2$$
From the integral representation of the alternating harmonic number, we have
$$(\overline{H}_{2n}-\overline{H}_{n})^2=\int_0^1\int_0^1\frac{((-x)^n-x^{2n})((-y)^n-y^{2n})}{(1+x)(1+y)}\ dxdy$$ 
$$\Longrightarrow S_1=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)}\left(\sum_{n=1}^\infty ((-x)^n-x^{2n})((-y)^n-y^{2n})\right)$$
$$=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)}\left(\frac32\frac{1}{1-xy}+\frac12\frac{1}{1+xy}-\frac{1}{1+x^2y}-\frac{1}{1+xy^2}\right)$$
$$=\frac32\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1-xy)}+\frac12\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+xy)}$$
$$-2\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+x^2y)}$$
$$=\frac32(\ln2)+\frac12\left(\frac14\zeta(2)\right)-2\left(-\frac{\pi}{8}+\frac{3}{16}\zeta(2)+\frac34\ln2\right)$$
$$=\frac{\pi}{4}-\frac14\zeta(2)$$
Similarly
$$S_2=\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+xy)}-2\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1-x^2y)}$$
$$+\int_0^1\int_0^1\frac{dxdy}{(1+x)(1+y)(1+x^2y^2)}$$
$$=\frac14\zeta(2)-2\left(\frac{G}{2}+\frac18\zeta(2)-\frac18\ln^22\right)+\left(\frac{G}{2}+\frac18\ln^22-\frac{\pi}{8}\ln2+\frac18\zeta(2)\right)$$
$$=\frac38\ln^22-\frac{\pi}{8}\ln2+\frac18\zeta(2)-\frac{G}{2}$$
Combine the results of $S_1$ and $S_2$ we get 
$$S=\frac3{16}\ln^22-\frac{\pi}{16}\ln2+\frac{\pi}{8}-\frac1{16}\zeta(2)-\frac{G}{4}$$
Note: All these sub-integrals are trivial and can be found by integrating with respect to $y$ first then with respect to $x$ or the other way around.