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...

$$ 1/(1-x)ln(1/(1-x)) = $$
$$ (1+x+x^2+x^3+x^4+...)(x+x^2/2+x^3/3+x^4/4+...) = $$
$$ x+(1+1/2)x^2+(1+1/2+1/3)x^3+(1+1/2+1/3+1/4)x^4+...= $$
$$ x + 3x^2/2 + 11x^3/6+25x^4/12+...=$$
$$ \sum_{a=1..\infty } (\sum_{b=1..a} 1/b) x^a =$$
$$ \sum_{a=1..\infty} H_a x^a$$
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
