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How to elegantly prove that

$$S=\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=3\operatorname{Li}_4\left(\frac12\right)-\frac{29}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac18\ln^42$$

where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the skew harmonic number and $H_n=\sum_{k=1}^n\frac{1}{k}$ is the harmonic number.

I managed to prove the equality above using the same strategy here but too many harmonic series were involved and some of these series are advanced, so I am looking for a simpler more independent solution.

Thank you,


Edit

My closed form gives $-0.973154$ but Mathematica gives $-0.972344$. I think my closed form is right because $Mathematica$ also said that "The general form of the sequence could not be determined, and the result may be incorrect." as attached numerical value

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  • $\begingroup$ have you tried wolframalpha.com? I can verify the identity upto 2 decimal places $\endgroup$ – Sandeep Silwal Jan 30 at 3:34
  • $\begingroup$ @Sandeep Silwal Actually I faced the same problem but I think my result is right because Mathematica added along with the numerical answer the following sentence "The general form of the sequence could not be determined, and the \ result may be incorrect." $\endgroup$ – Ali Shather Jan 30 at 3:42
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Another approach

Using the same strategy of @omegadot,

from this paper page $105$ we have

$$\overline{H}_n=\ln2-\int_0^1\frac{(-x)^n}{1+x}\ dx$$

multiply both sides by $\frac{(-1)^nH_n}{n^2}$ then $\sum_{n=1}^\infty$ we get

$$S=\ln2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\underbrace{\int_0^1\frac{1}{1+x}\sum_{n=1}^\infty\frac{H_nx^n}{n^2}\ dx}_{\large \mathcal{I}}\tag1$$

From here we have

$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

$$\Longrightarrow \mathcal{I}=\underbrace{\int_0^1\frac{\operatorname{Li}_3(x)}{1+x}\ dx}_{\large \mathcal{I}_1}-\underbrace{\int_0^1\frac{\operatorname{Li}_3(1-x)}{1+x}\ dx}_{\large \mathcal{I}_2}+\underbrace{\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{1+x}\ dx}_{\large \mathcal{I}_3}$$ $$+\underbrace{\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx}_{\large \mathcal{I}_4}+\zeta(3)\underbrace{\int_0^1\frac{1}{1+x}\ dx}_{\ln2}$$


$$\mathcal{I}_1=\int_0^1\frac{\operatorname{Li}_3(x)}{1+x}\ dx=-\sum_{n=1}^\infty(-1)^n\int_0^1 x^{n-1}\operatorname{Li}_3(x)\ dx$$ $$=-\sum_{n=1}^\infty(-1)^n\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)$$

$$=\ln2\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$


$$\mathcal{I}_2=\int_0^1\frac{\operatorname{Li}_3(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\int_0^1\frac{\operatorname{Li}_3(x)}{2-x}\ dx$$ $$=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\operatorname{Li}_3(x)\ dx =\sum_{n=1}^\infty\frac1{2^n}\left(\frac{\zeta(3)}{n}-\frac{\zeta(2)}{n^2}+\frac{H_n}{n^3}\right)$$

$$=\ln2\zeta(3)-\zeta(2)\operatorname{Li}_2\left(\frac12\right)+\sum_{n=1}^\infty\frac{H_n}{2^nn^3}$$


$$\mathcal{I}_3=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{2-x}\ dx$$

$$=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln x\operatorname{Li}_2(x) \ dx=\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2H_n}{n^3}+\frac{H_n^{(2)}}{n^2}-\frac{2\zeta(2)}{n^2}\right)$$

$$=2\sum_{n=1}^\infty\frac{H_n}{2^nn^3}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^nn^2}-2\zeta(2)\operatorname{Li}_2\left(\frac12\right)$$


$$\mathcal{I}_4=\frac12\int_0^1\frac{\ln x\ln^2(1-x)}{1+x}\ dx\overset{1-x\to x}{=}\frac12\int_0^1\frac{\ln(1-x)\ln^2x}{2-x}\ dx$$

$$=\frac12\sum_{n=1}^\infty\frac1{2^n}\int_0^1 x^{n-1}\ln(1-x)\ln^2x \ dx$$ $$=\frac12\sum_{n=1}^\infty\frac1{2^n}\left(\frac{2\zeta(3)}{n}+\frac{2\zeta(2)}{n^2}-\frac{2H_n}{n^3}-\frac{2H_n^{(2)}}{n^2}-\frac{2H_n^{(3)}}{n}\right)$$

$$=\ln2\zeta(3)+\zeta(2)\operatorname{Li}_2\left(\frac12\right)-\sum_{n=1}^\infty\frac{H_n}{2^nn^3}-\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^nn^2}-\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}$$


Combine the results of $\mathcal{I}_1$, $\mathcal{I}_2$, $\mathcal{I}_3$ and $\mathcal{I}_4$

$$\Longrightarrow \mathcal{I}=2\ln2\zeta(3)-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}$$

now plug this result in $(1)$

$$ \Longrightarrow S=\frac54\zeta(4)-2\ln2\zeta(3)+\ln2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$

Finally, substitute

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

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{2^nn}=\operatorname{Li}_4\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42\tag{ii}$$

$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li}_4\left(\frac12\right)-\frac{11}{4}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42\tag{iii}$$

we obtain

$$S=3\operatorname{Li}_4\left(\frac12\right)-\frac{29}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac18\ln^42$$


Note that the results of $(i)$ and $(ii)$ follow from using the the generating functions

$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$

$$\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}x^n=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}_2^2(x).$$

As for $(iii)$, its already calculated here.


The interesting thing about this approach is that some tough series got cancelled and we used only well-known results of harmonic series.

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I think this is an easier approach

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

$$\Longrightarrow \sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=-\color{blue}{\sum_{n=1}^\infty\frac{\overline{H}_{2n-1}H_{2n-1}}{(2n-1)^2}}+\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{2n}}{4n^2}\tag1$$

Similarly

$$\sum_{n=1}^\infty\frac{\overline{H}_nH_n}{n^2}=\color{blue}{\sum_{n=1}^\infty\frac{\overline{H}_{2n-1}H_{2n-1}}{(2n-1)^2}}+\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{2n}}{4n^2}\tag2$$

By combining $(1)$ and $(2)$, the blue sum nicely cancels out

$$\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=\frac12\color{orange}{\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{2n}}{n^2}}-\color{red}{\sum_{n=1}^\infty\frac{\overline{H}_nH_n}{n^2}}\tag3$$

The red sum was elegantly evaluated by @omegadot here

$$\color{red}{\sum_{n = 1}^\infty \frac{H_n \overline{H}_n}{n^2}} = - 3 \operatorname{Li}_4 \left (\frac{1}{2} \right )+\frac{43}{16} \zeta (4) + \frac{3}{4} \ln^2 2\zeta (2) - \frac{1}{8} \ln^4 2$$

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

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

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

$$=\boxed{4\operatorname{Li}_4\left(\frac12\right)+\frac{27}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac1{6}\ln^42}$$

where we used

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

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

and from here we have

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

Combine the boxed results

$$\Longrightarrow \color{orange}{\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{2n}}{n^2}}=\frac74\zeta(4)$$

now substitute the results of the red and orange sums in $(3)$ we get

$$\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=3\operatorname{Li}_4\left(\frac12\right)-\frac{29}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac18\ln^42$$

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