Maclaurin series of $\frac{1}{1-x}\ln\frac{1}{1-x}$ Find the Maclaurin series expansion (power law series) of the function:
$$f(x) = \frac{1}{1-x}\ln\frac{1}{1-x}.$$
I could find the expansion terms, up to the 5-th term, and is:
$$F(x) = x + \frac{3}{2!}x^2 + \frac{11}{3!}x^3 + \frac{50}{4!}x^4 + \frac{274}{5!}x^5 + \dots$$ 
and so on...
But I could not find a closed formula for it. Does any exist?
I've tried manipulating the power law series for $\dfrac{1}{1-x}$ and $\ln(1+x)$, but without success...
 A: $$ 1/(1-x)ln(1/(1-x)) = $$
$$ (1+x+x^2+x^3+x^4+...)(x+x^2/2+x^3/3+x^4/4+...) = $$
$$ x+(1+1/2)x^2+(1+1/2+1/3)x^3+(1+1/2+1/3+1/4)x^4+...= $$
$$ x + 3x^2/2 + 11x^3/6+25x^4/12+...=$$
$$ \sum_{a=1..\infty } (\sum_{b=1..a} 1/b) x^a =$$
$$ \sum_{a=1..\infty} H_a x^a$$
Numerators of harmonic numbers $H_n$ : https://oeis.org/A001008
Denominators of harmonic numbers $H_n$ : https://oeis.org/A002805
https://en.wikipedia.org/wiki/Harmonic_number
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{{1 \over 1 - x}\,\ln\pars{1 \over 1 - x} =
-\,{\ln\pars{1 - x} \over 1 - x}}$ is the
  Harmonic Number Generating Function. 

Namely,
\begin{align}
{1 \over 1 - x}\,\ln\pars{1 \over 1 - x} =
-\,{\ln\pars{1 - x} \over 1 - x} = \sum_{n = 1}^{\infty}H_{n}\, x^{n}
\end{align}

$\ds{H_{z}}$ is a Harmonic Number.

$$
\begin{array}{|c|c|}\hline
\ds{\quad n\quad} & \ds{\quad H_{n}\quad}
\\ \hline
\ds{1} & \ds{1}
\\ \hline
\ds{2} & \ds{3 \over 2}
\\ \hline
\ds{3} & \ds{11 \over 6}
\\ \hline
\ds{4} & \ds{25 \over 12}
\\ \hline
\ds{5} & \ds{137 \over 60}
\\ \hline
\ds{6} & \ds{49 \over 20}
\\ \hline
\end{array}
$$
