Generating functions for $\log^3(1-x)$ of $\log^3(x)$ I am trying to find generating functions which will give me a power logarithm. 
I am trying to find generating sums in the form
$$\sum_{n=1}^{\infty} a_n\,x^n = -\frac{\log^2(1-x)}{1-x}$$
or 
$$\sum_{n=1}^{\infty} a_n\,x^n = \frac{\log^2(x)}{x}.$$
Something, which will return $\log^3$ in the end.     
Help is required! 
Thanks
 A: We have:
$$-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$$
for any $x$ such that $|x|<1$, hence:
$$-\frac{\log(1-x)}{1-x}=\sum_{n\geq 1}H_n x^n.$$
and since $\frac{d}{dx}\log^2(1-x) = -2\frac{\log(1-x)}{1-x}$ we have:
$$\log^2(1-x) = 2\sum_{n\geq 1}\frac{H_n}{n+1}x^{n+1}\tag{1}.$$
Since, by partial summation:
$$\sum_{n=1}^{N}\frac{H_n}{n}= H_N^2-\sum_{n=1}^{N-1}\frac{H_n}{n+1} = H_N^2-\sum_{n=1}^{N}\frac{H_n}{n}+\sum_{n=1}^{N}\frac{1}{n^2}$$
we have:
$$ \sum_{n=1}^{N}\frac{H_n}{n}=\frac{H_N^2+H_N^{(2)}}{2}, \quad \sum_{n=1}^{N}\frac{H_{n-1}}{n}=\frac{H_N^2-H_N^{(2)}}{2}$$
so:

$$\frac{\log^2(1-x)}{1-x}=\sum_{n\geq 1}(H_n^2-H_n^{(2)})\,x^{n}\tag{2}$$

and:

$$\log^3(1-x)=-3\sum_{n\geq 1}\frac{H_n^2-H_n^{(2)}}{n+1}\,x^{n+1}\tag{3}$$

for any $x$ such that $|x|<1$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{\verts{x}\ <\ 1}$:

\begin{align}
\pars{1 - x}^{\mu}&=\sum_{n\ =\ 0}^{\infty}{\mu \choose n}x^{n}\quad\imp\quad
\left\{\begin{array}{lcl}
\pars{1 - x}^{\mu}\ln^{2}\pars{1 - x}
&=&\sum_{n\ =\ 0}^{\infty}\partiald[2]{{\mu \choose n}}{\mu}\,x^{n}
\\
\pars{1 - x}^{\mu}\ln^{3}\pars{1 - x}
&=&\sum_{n\ =\ 0}^{\infty}\partiald[3]{{\mu \choose n}}{\mu}\,x^{n}
\end{array}\right.
\end{align}

It leads to:
\begin{align}
{\ln^{2}\pars{1 - x} \over 1 - x}
&=\sum_{n\ =\ 0}^{\infty}\bracks{\lim_{\mu\ \to\ -1}\partiald[2]{{\mu \choose n}}{\mu}}x^{n}
=\sum_{n\ =\ 0}^{\infty}\bracks{\color{#00f}{\pars{-1}^{n}\lim_{\mu\ \to\ 0}
\partiald[2]{{\mu + n\choose n}}{\mu}}}x^{n}
\\[5mm]
\ln^{3}\pars{1 - x}
&=\sum_{n\ =\ 0}^{\infty}\bracks{\dsc{\lim_{\mu\ \to\ 0}\partiald[3]{{\mu \choose n}}{\mu}}}x^{n}
\end{align}

Limits can be expressed in terms of ${\sf Gamma}$ and ${\sf PolyGammas}$ functions.

