How is the last "=" true? How can the last equality be true? 

$$
G(u)=\frac{g}{(1+u)^g-1}-\frac1u=\frac{g}{gu + \cdots + u^g} - \frac{1}{u}=\sum_{k=0}^\infty c_ku^k
$$


 A: One may recall that, as $x \to 0$, by the Taylor series expansion, we have
$$
\frac1{1+x}=1-x+O(x^2).
$$ We assume $g\neq0$ and $ u \to 0$. Then we have
$$
\begin{align}
\frac{g}{gu+\cdots+u^g}&=\frac1{u+u\left(1/g+\cdots+u^{g}/g\right)}
\\\\&=\frac1{u}\frac1{1+u\left(1/g+\cdots+u^{g-1}/g\right)}
\\\\&=\frac1{u}\left(1-u\left(1/g+\cdots+u^{g-1}/g\right)+O(u^2)\right)
\\\\&=\frac1{u}-1/g+O(u)
\end{align}
$$ with
$$
O(u)=\sum_{k=1}^\infty c_ku^k
$$ for some constants $c_k$ to be determined. Convergence may be ensured by the implicit function theorem.
A: If you look at the left-hadn side of the equals sign, you have
$$
\begin{align}
\frac{g}{gu + \cdots + u^g} - \frac{1}{u} 
&= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\
&= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
\end{align}
$$
If you then expand the first term as a geometric series, you will see that the terms with $\frac{1}{u}$ cancel out, leaving a series of the form $\sum_{k=0}^\infty c_ku^k$ for some undetermined coefficents $c_k$.
Edit: Let's expand on this a little bit. Note that we have (formally!) the expansion
$$
\frac{1}{1 + A} = 1 - A + A^2 - A^3 + \cdots
$$
for any $A$. In particular, if we let $A =  \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$, then we have
$$
\begin{align}
\frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
&=
\frac{1}{u}\frac{1}{1 + A} - \frac{1}{u} \\
&=
\frac{1}{u}(1 - A + A^2 - A^3 + \cdots) - \frac{1}{u} \\
&=
\frac{1}{u} - \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots - \frac{1}{u} \\
&=
- \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots
\end{align}
$$
Since $A = \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$ this now becomes
$$
- \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big) + \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big)^2 -  \cdots
$$
which is a series that only contains positive exponent terms in $u$.
A: $$\frac{g}{(1+u)^g-1}-\frac{1}{u}=\frac{gu-(1+u)^g+1}{u((1+u)^g-1)}=\frac{1+gu-(1+u)^g}{u((1+u)^g-1)}$$
$$=\frac{\sum_{k=2}^g \binom{g}{k}u^k}{\sum_{k=1}^g\binom{g}{k}u^{k+1}}\approx\frac{\binom{g}{2}u^2}{\binom{g}{1}u^2}=\frac{g-1}{2} \text{ as } u\to0$$
$$\implies G(u)=\frac{\sum_{k=2}^g \binom{g}{k}u^{k-2}}{\sum_{k=1}^g\binom{g}{k}u^{k-1}}=\frac{\binom{g}{2}+\binom{g}{3}u+\cdots+\binom{g}{g}u^{g-2}}{\binom{g}{1}+\binom{g}{2}u+\cdots+\binom{g}{g}u^{g-1}}$$
is a ratio of polynomials, with the denominator finite and nonzero at the origin. It can hence be developed as a convergent power series at the origin.
Some extra prodding to find the roots of the denominator should give that the radius of convergence of this series ought to be $\vert u \vert < 2\sin\frac{\pi}{g}$ for $g\ge2$.
