# Conjectured value of a harmonic sum ∑∞n=1(Hn−2H2n+H4n)2\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2

There is a known asymptotic expansion of harmonic numbers $H_n$ for $n\to\infty$:

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):

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:

Generalizing, we can consider two families of series:

and try to evaluate them in a closed form.

So far I have the following conjectured result:

where $G$ is the Catalan constant.

Could you please help me to prove this result and, possibly, find other values of $\mathcal A_m,\mathcal B_m$?

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:
$-\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} ,-\ln\cos x=\ln2+\sum_{n=1}^{\infty} \frac{(-1)^n\cos(2nx)}{n} \tag{1}$

Starting,

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

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:

Altogether,

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 $t=\frac{1-x}{1+x}$:

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 $x=\tan\theta$:

Altogether,

Evaulation of $I_{242}$.

Substitute $t=\frac{1-x}{1+x}$ to get

Calculating these integrals:

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,

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

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:

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.