How is this series in denominator converted to a series in numerator? How do we go from 
\begin{equation*}
\frac{1}{1+(\frac{x}{2})+(\frac{x^2}{3})+(\frac{x^3}{4})+\dots} 
\end{equation*}
to 
\begin{equation*}
1-\left(\frac{x}{2}\right)-\left(\frac{x^2}{12}\right)-\dots?
\end{equation*}
If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power $-1$, but I don't see the one in denominator in any known form other than converting it to $\log.$
 A: 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
$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$
A: Let $f(x) = \sum f_n x^n$ be a formal power series.  To compute the inverse of $1 - x f(x)$, write
$$\frac{1}{1 - x f(x)} = \sum_{n \ge 0} x^n f^n(x).$$
To compute the coefficient of $x^k$ it suffices to compute the contributions from the first $k$ terms on the RHS, so this results in a finite algorithm which will let you compute any particular coefficient of the inverse.  If $f$ has special properties (for example it is a polynomial or more generally meromorphic) then often one can even find a nice closed form.  
