$\displaystyle \sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty (-1)^{n_1+n_2+\cdots+n_k}\dfrac{H_{n_1+n_2+\cdots+n_k}}{n_1 n_2 n_3 \cdots n_k}$

$\displaystyle \begin{aligned} &\sum_{n_1=1}^\infty \sum_{n_2=1}^\infty \cdots \sum_{n_k=1}^\infty (-1)^{n_1+n_2+\cdots+n_k}\dfrac{H_{n_1+n_2+\cdots+n_k}}{n_1 n_2 n_3 \cdots n_k} \\ &= (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \left[(-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \right]\end{aligned}$

The above is the claim which is proved here after in this note.

Proof:

First note that,

$\displaystyle H_n=\int_0^1 \dfrac{1-x^n}{1-x}\; dx$

Substituting this into our sum $S$ we have,

$\displaystyle \begin{aligned} S&= \int_0^1 \dfrac{\sum_{n_1,n_2,\cdots,n_k=1}^\infty \dfrac{(-1)^{n_1+n_2+\cdots+n_k}}{n_1 n_2\cdots n_k}-\sum_{n_1,n_2,\cdots,n_k=1}^\infty \dfrac{(-1)^{n_1+n_2+\cdots+n_k}x^{n_1+n_2+\cdots+n_k}}{n_1 n_2\cdots n_k}}{1-x}\; dx \\ &= (-1)^k \int_0^1 \dfrac{\log^k 2-\log^k (1+x)}{1-x}\; dx \\ &= (-1)^k \int_{1/2}^1 \dfrac{\log^k 2-\log^k (2x)}{1-x}\; dx \\ &= (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \int_{1/2}^1 \dfrac{\log^n x}{1-x}\; dx \end{aligned}$

Now it remains to calculate ,

$\displaystyle \int_{1/2}^1 \frac{\log^n x}{1-x}\; dx$

To find this we evaluate a preliminary integral,

$\displaystyle \begin{aligned} & \int_{1/2}^1 x^m \; dx = \dfrac{1}{m+1}-\dfrac{2^{-m}}{1+m} \\ & \int_{1/2}^1 x^m \log^n x \; dx = \dfrac{d^n}{dm^n}\left[\dfrac{1}{m+1}-\dfrac{2^{-m}}{1+m}\right] \\ & \int_{1/2}^1 x^m \log^n x \; dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}+2^{-m}(-1)^{n-1} \sum_{r=0}^n \binom{n}{r}\dfrac{(\log 2)^{n-r}r!}{(m+1)^{r+1}}\end{aligned}$

Now coming back to the main integral,

$\displaystyle \begin{aligned} \int_{1/2}^1 \frac{\log^n x}{1-x}\; dx &= \sum_{m=0}^\infty \int_{1/2}^1 x^m \log^n x \; dx \\ &= \sum_{m=0}^\infty \left[ \dfrac{(-1)^n n!}{(m+1)^{n+1}}+2^{-m}(-1)^{n-1} \sum_{r=0}^n \binom{n}{r}\dfrac{(\log 2)^{n-r}r!}{(m+1)^{r+1}}\right] \\ &= (-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \end{aligned}$

Putting this back,

$\displaystyle S = (-1)^{k} \sum_{n=1}^{k} \binom{k}{n} (\log 2)^{k-n} \left[(-1)^n n! \zeta(n+1)+2(-1)^{n-1}\sum_{r=0}^n \binom{n}{r}r! (\log 2)^{n-r} {\rm Li}_{n+1}\left(\dfrac12\right) \right]$