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As we know, using $$\frac{{{{\ln }^2}\left( {1 - x} \right)}}{{1 - x}} = \sum\limits_{n = 1}^\infty {\left( {H_n^2 - {\zeta _n}\left( 2 \right)} \right){x^n}} = \sum\limits_{n = 1}^\infty {\left( {\sum\limits_{k = 1}^n {\frac{{{H_k}}}{{n - k + 1}}} } \right){x^{n + 1}}} $$ we obtain $$\sum\limits_{k = 1}^{n - 1} {\frac{{{H_k}}}{{n - k}} = H_n^2 - {\zeta _n}\left( 2 \right)} $$ where $${\zeta _n}\left( p \right) = \sum\limits_{k = 1}^n {\frac{1}{{{k^p}}}} ,{H_n} = {\zeta _n}\left( 1 \right) = \sum\limits_{k = 1}^n {\frac{1}{k}} .$$ Similarly, how to solve the following harmonic sum $$\sum\limits_{k = 1}^{n - 1} {\frac{{{H_k}}}{{n - k}}{{\left( { - 1} \right)}^{k - 1}} = ?} .$$

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This is not an answer but it is too long for a comment.

You can't use the identical way to evalutate the sum. The first equality that you use is a particular form of this

Theorem: For $m,r\in\mathbb{N} $ holds $$\frac{\left(-\log\left(1-x\right)\right)^{r}}{\left(1-x\right)^{m+1}}=\sum_{k\geq m}P_{r}\left(H_{k}^{\left(1\right)}-H_{m}^{\left(1\right)},\dots,H_{k}^{\left(r\right)}-H_{m}^{\left(r\right)}\right)\dbinom{k}{m}x^{k-m}\tag{1}$$ where $H_{k}^{\left(r\right)}=\sum_{j=1}^{k}\frac{1}{j^{r}}$ is the generalized harmonic number and $$P_{r}\left(x_{1},\dots,x_{r}\right)=\left(-1\right)^{r}Y_{r}\left(-0!x_{1},\dots,-\left(r-1\right)!x_{r}\right)$$ and $Y_{r}$ is the Bell polynomial (for a reference see here page $840$). Note that in our case we have $r=2$, $m=0$ and so $P_{r}\left(H_{k}^{\left(1\right)},H_{k}^{\left(2\right)}\right)=H_{k}^{2}-H_{k}^{\left(2\right)}$. From this we note that we can't use $(1)$ because to get our sum we have to consider $$\frac{\log\left(1-x\right)\log\left(1+x\right)}{1-x}=\sum_{n\geq1}\left(\sum_{k=1}^{n}\frac{H_{k}\left(-1\right)^{n-k+1}}{n-k+1}\right)x^{n+1}. $$

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