Help on $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}$

From this post:

The sum: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n+1}=-\frac{9}{8}\zeta(3)\ln(2)+\frac{\pi^4}{288}-\frac{\ln^4(2)}{4}+\frac{\pi^2}{8}\ln^2(2)$$

by moving the $$n+1$$ back to $$n$$

What is the sum of: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}?$$

I guess it could be easy, but I cannot see it.

Can anyone please with this sum? I needs it to solve one of my current problem I am working on.

I will start completing Frank's answer: since $$\int_{0}^{1}x^n\log(1-x)\,dx = -\frac{H_{n+1}}{n+1}$$, the linear, alternating Euler sum with weight four $$\sum_{n\geq 1}\frac{(-1)^{n-1} H_n}{n^3}$$ can be computed from the integral $$\int_{0}^{1}\text{Li}_2(-x)\log(1-x)\frac{dx}{x}$$ or from the integral (equivalent by $$\text{IBP}$$) $$\int_{0}^{1}\text{Li}_2(x)\log(1+x)\frac{dx}{x}$$. Up to a $$\eta(4)=\frac{7\pi^4}{720}$$ term, this is the same as computing $$\int_{0}^{1}\log^2(1+x)\log^2(x)\frac{dx}{x}\quad\text{or}\quad\int_{0}^{1}\log(1+x)\log^3(x)\frac{dx}{1+x}$$ since $$\log^2(1-x)=\sum_{n\geq 1}\frac{2H_{n-1}}{n}x^n$$. In terms of the notation of Flajolet and Salvy, we are tackling $$S_{1,3}^{+-}$$, which is given by $$\mu_1=\frac{1}{2}\int_{0}^{1}\frac{\log^2(z)\log(1+z)}{z(1+z)}\,dz$$ or $$\mu_1=\frac{11 \pi ^4}{360}+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log(2)^4-2\,\text{Li}_4\left(\frac{1}{2}\right)-\frac{7}{4}\log(2)\zeta(3)$$ (De Doelder 1991). It remains to crack the quadratic, alternating Euler sum with weight four $$\sum_{n\geq 1}\frac{(-1)^{n+1}H_n^2}{n^2}$$ where $$\sum_{n\geq 0}(-1)^n\frac{H_n^{(2)}}{(n+1)^2} = \int_{0}^{1}\frac{\text{Li}_2(-x)}{1+x}(-\log(x))\,dx$$ is related to $$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\sum_{m\geq 1}\frac{1}{m^2}$$ and $$\pi^4$$ via symmetry and summation by parts, while $$\sum_{n\geq 1}\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}x^n = -3\log^3(1-x)$$ by the structure of the Stirling numbers of the first kind. Therefore the computation can be finished by noticing that $$\int_{0}^{1}\log^3(1-x)\frac{dx}{x} = -\frac{\pi^4}{15}.$$ Mathematica's command $$\text{FindIntegerNullVector}$$ returns $$\sum_{n\geq 1}\frac{(-1)^{n+1}H_n^3}{n}=\\=\frac{1}{90} (68 -\pi^4+ 12\pi^2\log^2(2)- 197\log(2)^4+ 47\,\text{Li}_4(1/2)+ 22\log(2)\zeta(3))$$ as a plausible identity.

• Thank you @Jack D'Aurizio. – user550260 Mar 25 '19 at 6:23
• @Downvoter: please explain your downvote. – Jack D'Aurizio Apr 28 '19 at 21:03
• The correct version should be: $$-\frac{9}{8} \zeta (3) \log (2)+\frac{\pi ^4}{144}-\frac{1}{4} \log ^4(2)+\frac{1}{8} \pi ^2 \log ^2(2)$$ – Hypergeometric Sep 2 at 14:39

Here might be a start. Observe that\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\frac {H_1^3}2-\frac {H_2^3}3+\frac {H_3^3}4-\cdots\\ & =\sum\limits_{n\geq2}(-1)^n\frac {H_{n-1}^3}n\end{align*}

Now use the fact that $$H_n=H_{n-1}+\tfrac 1n$$. Therefore, the original sum becomes\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\sum\limits_{n\geq2}(-1)^n\frac {1}n\left[H_n-\frac 1n\right]^3\\ & =\sum\limits_{n\geq2}(-1)^n\frac 1n\left[H_n^3-\frac {3H_n^2}n+\frac {3H_n}{n^2}-\frac 1{n^3}\right]\\ & =\sum\limits_{n\geq2}(-1)^{n}\frac {H_n^3}{n}-3\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\\ & =-\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^3}n+3\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\end{align*}

The second sum can be found with the help of this question. Reindex the identity to start the sum from two to get

$$\sum\limits_{n\geq2}(-1)^n\frac {H_n}{n^3}\color{red}{=1+2\operatorname{Li}_4\left(\frac 12\right)+\frac {7\log 2}4\zeta(3)-\frac {11\pi^4}{360}+\frac {\log^42}{12}-\frac {\pi^2\log^22}{12}}$$

In a similar fashion, some digging on MSE gives this question which might be a big help in tackling the second sum:$$\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}$$

• Thank you @Frank W – user550260 Mar 24 '19 at 21:30
• @coffeee No problem. I will try to finish up this problem/answer when I get some time! – Frank W Mar 24 '19 at 21:31