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I ma trying to prove

$$S=\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$

where $\overline{H}_n$ is the alternating harmonic number and $H_n$ is the harmonic number.

I need this sum to complete my solution here.

Here is my trial,

Following @user97357329's note in the comments of the same link above

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

Giving us

$$S=\underbrace{\sum_{n=1}^\infty\frac{\overline{H}_{2n-1}H_{n-1/2}}{(2n-1)^2}}_{\large S_1}+\frac14\underbrace{\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{n}}{n^2}}_{\large S_2}$$

I managed to evaluate $S_2$ using $\overline{H}_{2n}=H_{2n}-H_n$.

Regarding $S_1$, I used $\overline{H}_{2n-1}=H_{2n}-H_n+\frac1{2n}$ and $H_{n-1/2}=2H_{2n}-H_n-2\ln2$

therefore

$$S_1=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n-1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n-1)^2}}-2\ln2\sum_{n=1}^\infty\frac{H_{2n}}{(2n-1)^2}+\color{red}{\sum_{n=1}^\infty\frac{2H_{2n}-H_n-2\ln2}{2n(2n-1)^2}}$$

and I am stuck with the blue and red sums, any idea? Thank you.

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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.

$$\zeta(4)$$ $$=\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{ \left(H_{2 n}\right)^2}{ (2 n+1)^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^3}$$ $$-\frac{8}{5}\sum _{n=1}^{ \infty } \frac{\left(H_{2 n}\right){}^2}{ n^2}-\frac{32}{5}\underbrace{\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^2}}_{\text{The series you need}}-\frac{64}{5}\log(2)\sum _{n=1}^{ \infty } \frac{H_{2 n}}{(2 n+1)^2}-\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^2}.$$

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.

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    $\begingroup$ Thank you. (+1). What about about the red sum ? $\endgroup$ – Ali Shather Jan 19 at 16:23
  • $\begingroup$ @AliShather Welcome. For some reasons I missed the red part. In a few minutes I'll type the way to go. $\endgroup$ – user97357329 Jan 19 at 16:49
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    $\begingroup$ thank you for the awesome work. $\endgroup$ – Ali Shather Jan 19 at 16:51
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    $\begingroup$ @AliShather Let me know if you're fine with the answer provided for the red sum. $\endgroup$ – user97357329 Jan 19 at 17:14
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    $\begingroup$ It's all good :) $\endgroup$ – Ali Shather Jan 19 at 17:27
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$$S=\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=H_{1/2}+\sum_{n=2}^\infty\frac{\overline{H}_nH_{n/2}}{n^2},\quad H_{1/2}=2-2\ln2$$

notice that

$$\sum_{n=2}^\infty f(n)=\sum_{n=1}^\infty f(2n)+\sum_{n=1}^\infty f(2n+1)$$

therefore

$$S=H_{1/2}+\frac14\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{n}}{n^2}+\sum_{n=1}^\infty\frac{\overline{H}_{2n+1}H_{n+1/2}}{(2n+1)^2}$$

$$S=2-2\ln2+\frac14S_1+S_2\tag{*}$$


For $S_1$, use $\overline{H}_{2n}=H_{2n}-H_n$

$$\Longrightarrow S_1=\sum_{n=1}^\infty\frac{{H}_{2n}H_{n}}{n^2}-\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}$$


For $S_2$, use: $$\overline{H}_{2n+1}=H_{2n+1}-H_n$$

$$H_{n+1/2}=2H_{2n}-H_n+\frac2{2n+1}-2\ln2$$

so

$$\overline{H}_{2n+1}H_{n+1/2}\\=2H_{2n}^2+H_n^2-3H_{2n}H_n-2\ln2H_{2n}+2\ln2H_n+\frac{4H_{2n}}{2n+1}-\frac{3H_n}{2n+1}-\frac{2\ln2}{2n+1}+\frac{2}{(2n+1)^2}$$

$$\Longrightarrow S_2=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^2}+\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^2}-3\color{orange}{\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}}$$ $$-2\ln2\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}+2\ln2\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^2}+4\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}$$ $$-3\sum_{n=1}^\infty\frac{H_{n}}{(2n+1)^3}-2\ln2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^3}}_{\large \frac{7}{8}\zeta(3)-1}+2\underbrace{\sum_{n=1}^\infty\frac{1}{(2n+1)^4}}_{\large \frac{15}{16}\zeta(4)-1}$$

The orange sum can be extracted from the equality provided by @user97357329 in his solution above

$$3\color{orange}{\sum_{n=1}^\infty\frac{H_{2n}H_n}{(2n+1)^2}}$$ $$\small{=\frac34\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}+6\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^2}+6\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}-\frac34\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}-\frac34\sum_{n=1}^\infty\frac{H_{2n}^{(2)}}{n^2}-6\ln2\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}}$$

plugging this result in $S_2$ gives

$$S_2=2\ln2\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}}-3\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}}$$

$$-4\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^2}}-2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}}+4\ln2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}}+\frac34\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}}+\frac34\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^{(2)}}{n^2}}$$

$$-\frac34\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}+\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^2}-\frac74\ln2\zeta(3)+\frac{15}8\zeta(4)+2\ln2-2$$

Now plug $S_1$ and $S_2$ in $(*)$ we reach

$$S=2\ln2\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}}-3\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}}$$

$$-4\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^2}}-2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}}+4\ln2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}}+\frac34\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}}+\frac34\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^{(2)}}{n^2}}$$

$$-\frac12\sum_{n=1}^\infty\frac{H_{2n}H_n}{n^2}+\sum_{n=1}^\infty\frac{H_{n}^2}{(2n+1)^2}-\frac14\sum_{n=1}^\infty\frac{H_n^2}{n^2}-\frac74\ln2\zeta(3)+\frac{15}8\zeta(4)$$


Lets start with the easy ones, the blue sums can be calculated using the following generalization proved by @Random Variable here

$$ \sum_{n=1}^\infty\frac{H_n}{(n+a)^2}=\left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2}$$

so

$$\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}}=\frac74\zeta(3)-\frac32\ln2\zeta(2)$$

$$\color{blue}{\sum_{n=1}^\infty\frac{H_n}{(2n+1)^3}}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$


The red ones can be evaluated using the fact that

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

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n+1)^2}}=\sum_{n=1}^\infty\frac{H_{n}^2}{(n+1)^2}+\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{(n+1)^2}$$

$$=\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}-\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}^2}{n^2},\quad H_{n-1}=H_n-\frac1n$$

$$=\sum_{n=1}^\infty\frac{H_n^2}{n^2}-2\sum_{n=1}^\infty\frac{H_n}{n^3}+\sum_{n=1}^\infty\frac{1}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\sum_{n=1}^\infty\frac{(-1)^n}{n^4}$$

Similarly

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}}=\sum_{n=1}^\infty\frac{H_n}{n^2}-\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}+\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$

$$2\color{red}{\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^3}}=\sum_{n=1}^\infty\frac{H_n}{n^3}-\sum_{n=1}^\infty\frac{1}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\sum_{n=1}^\infty\frac{(-1)^n}{n^4}$$

$$\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}}=4\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}=2\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}$$

$$\color{red}{\sum_{n=1}^\infty\frac{H_{2n}^{(2)}}{n^2}}=4\sum_{n=1}^\infty\frac{H_{2n}^{(2)}}{(2n)^2}=2\sum_{n=1}^\infty\frac{H_{n}^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^{(2)}}{n^2}$$


Evaluating $\displaystyle \sum_{n=1}^\infty\frac{H_n^2}{(2n+1)^2}$

Using the identity

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

replace $x$ with $x^2$, then multiply both sides by $-\ln x$ and integrate from $x=0$ to $1$ we get

$$\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{(2n+1)^2}=-\underbrace{\int_0^1\frac{\ln x\ln^2(1-x^2)}{1-x^2}\ dx}_{\text{beta function}}=-3\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{15}4\zeta(4)$$

From here we have

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)$$

$$\Longrightarrow \sum_{n=1}^\infty\frac{H_n^2}{(2n+1)^2}=\frac13\ln^42+\ln^22\zeta(2)-\frac{61}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)$$


substitute the following results

$$\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)\tag1$$

$$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)\tag2$$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac58\zeta(3)\tag3$$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42\tag4$$

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}=\frac{7}4\zeta(4)\tag5$$

$$\sum_{n=1}^\infty\frac{H_n^2}{n^2}=\frac{17}4\zeta(4)\tag6$$

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)+\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42\tag7$$

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42\tag8$$

$$\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42\tag9$$

we obtain the closed form of $S$.


References

$(1)$ and $(2)$ can be obtained using Euler identity, $(3)$ and $(4)$ can be found here, $(5)$ and $(6)$ can be found here, $(7)$ and $(8)$ can be found here and $(9)$ can be found here.

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