TITLE: How find this $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$
QUESTION [25 upvotes]: How Find this sum
$$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My idea: since
$$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$
so
$$I=\sum_{n=1}^{\infty}H^3_{n}\int_{0}^{-1}x^ndx$$
then I can't.Thank you 
This problem is not Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

REPLY [4 votes]: We proved here  that 
$$\small{\sum_{n=1}^\infty x^nH_n^3=\frac1{1-x}\left(\frac32\ln x\ln^2(1-x)-3\zeta(2)\ln(1-x)-\ln^3(1-x)+\operatorname{Li}_3(x)+3\operatorname{Li}_3(1-x)-3\zeta(3)\right)}$$
Replace $x$ with $-x$ then integrate from $x=0$ to $1$, we get
\begin{align}
S&=\sum_{n=1}^\infty(-1)^n\frac{H_n^3}{n+1}\\
&=\frac32\int_0^1\frac{\ln (-x)\ln^2(1+x)}{1+x}\ dx-3\zeta(2)\int_0^1\frac{\ln(1+x)}{1+x}\ dx-\int_0^1\frac{\ln^3(1+x)}{1+x}\ dx\\
&\quad+\int_0^1\frac{\operatorname{Li}_3(-x)}{1+x}\ dx+3\int_0^1\frac{\operatorname{Li}_3(1+x)}{1+x}\ dx-3\zeta(3)\int_0^1\frac{1}{1+x}\ dx\\
&=\underbrace{\frac32\int_0^1\frac{\ln (-x)\ln^2(1+x)}{1+x}\ dx}_{\large I_1}-\frac32\zeta(2)\ln^22-\frac14\ln^42\\
&\quad+\underbrace{\int_0^1\frac{\operatorname{Li}_3(-x)}{1+x}\ dx}_{\large I_2}+3\operatorname{Li}_4(2)-3\zeta(4)-3\zeta(3)\ln2\tag{1}
\end{align}
For $I_1$, use $\ln(-x)=\ln x+i\pi$
\begin{align}
I_1&=\frac32\int_0^1\frac{\ln (x)\ln^2(1+x)}{1+x}\ dx+\frac32i\pi\int_0^1\frac{\ln^2(1+x)}{1+x}dx\\
&\overset{IBP}{=}-\frac12\int_0^1\frac{\ln^3(1+x)}{x}\ dx+i\frac{\pi}{2}\ln^32\\
&\overset{x=\frac{1-y}{y}}{=}\frac12\int_{1/2}^1\frac{\ln^3x}{x(1-x)}\ dx+i\frac{\pi}{2}\ln^32\\
&=\frac12\int_{1/2}^1\frac{\ln^3x}{x}\ dx+\frac12\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx+i\frac{\pi}{2}\ln^32\\
&=-\frac18\ln^42+\frac12\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^3x \ dx+i\frac{\pi}{2}\ln^32\\
&=-\frac18\ln^42+\frac12\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)+i\frac{\pi}{2}\ln^32\\
&=-\frac18\ln^42+\frac12\ln^42+\frac32\ln^22\operatorname{Li}_2\left(\frac12\right)+3\ln2\operatorname{Li}_3\left(\frac12\right)+3\operatorname{Li}_4\left(\frac12\right)-3\zeta(4)+i\frac{\pi}{2}\ln^32\\
&=\boxed{3\operatorname{Li}_4\left(\frac12\right)-3\zeta(4)+\frac{21}{8}\ln2\zeta(3)-\frac34\ln^22\zeta(2)+\frac18\ln^42+i\frac{\pi}{2}\ln^32\quad}
\end{align}
as for the second integral:
\begin{align}
I_2&\overset{IBP}{=}\ln2\operatorname{Li}_3(-1)-\int_0^1\frac{\ln(1+x)\operatorname{Li}_2(-x)}{x}\ dx\\
&=-\frac34\ln2\zeta(3)+\frac12\left.\operatorname{Li}_2^2(-x)\right|_0^1=\boxed{-\frac34\ln2\zeta(3)+\frac5{16}\zeta(4)}
\end{align}
To find $\operatorname{Li}_4(2) $, we use the following polylogarithmic identity:
$$\operatorname{Li}_4(x)=-\frac74\zeta(4)-\frac12\ln^2(-x)\zeta(2)-\frac1{24}\ln^4(-x)-\operatorname{Li}_4(1/x)$$
set $x=2$, we get
$$\boxed{\operatorname{Li}_4(2)=2\zeta(4)+\ln^22\zeta(2)-\frac1{24}\ln^42-\operatorname{Li}_4\left(\frac12\right)-i\frac{\pi}{6}\ln^32\quad}$$
Plugging the boxed results in $(1)$, we get

$$\sum_{n=1}^\infty(-1)^n\frac{H_n^3}{n+1}=\frac5{16}\zeta(4)-\frac98\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac14\ln^42\approx -0.0641$$

but the numerical value for the sum given by wolfram $\approx -0.05526$. I checked all my steps many times and did not spot any mistake, I even checked the integrals and the real value of polylorarithm and they all matched numerically but when we collect them, the final numerical value does not match. I do not know where is the mistake but I am sure that sometimes wolfram gives the wrong numerical value.