Using $-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$
We have that
$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$
Now the series for $\frac{x}{\log(1-x)}$ is well known.
See the question asked on this very site here: Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana
And the page here: http://en.wikipedia.org/wiki/Euler-Mascheroni_constant. (search the page for Gregory).
The series expansion is given by
$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$
The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.
So to answer your question, we get
$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$