expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?
 A: There are quite a few ways to prove it. 
You could note that $\frac1n\binom{2n-2}{n-1}$ is $C_{n-1}$, the $(n-1)$-st Catalan number, so that the formula can be written
$$\sqrt{1-4x}=1-2\sum_{n\ge 0}\frac1{n+1}\binom{2n}nx^{n+1}=1-2\sum_{n\ge 0}C_nx^{n+1}\;.$$
Now you can simply manipulate the known generating function for the Catalan numbers:
$$\sum_{n\ge 0}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}\;,$$
so
$$\begin{align*}
\sqrt{1-4x}&=1-2x\sum_{n\ge 0}C_nx^n\\
&=1-2\sum_{n\ge 0}C_nx^{n+1}\;,
\end{align*}$$
as desired.
Or you can use the generalized binomial theorem:
$$\begin{align*}
(1-4x)^{1/2}&=\sum_{n\ge 0}\binom{1/2}n(-4x)^n\\
&=1+\sum_{n\ge 1}\binom{1/2}n(-1)^n4^nx^n\\
&\overset{*}=1+\sum_{n\ge 1}\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1}(-1)^n4^nx^n\\
&=1-\sum_{n\ge 1}\frac{2^{2n}}{2^{2n-1}n}\binom{2n-2}{n-1}x^n\\
&=1-2\sum_{n\ge 1}\frac1n\binom{2n-2}{n-1}x^n\;,
\end{align*}$$
where the starred step is carried out in detail here.
A: Hint. One may apply the Taylor series expansion to $f(x):=\sqrt{1-4x}$,
$$
f(x)=f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n,\quad |x|<\frac14. \tag1
$$
Then one may prove by induction that
$$
\begin{align}
(\sqrt{u})'=& \frac12\cdot u^{1/2-1}
\\(\sqrt{u})^{(2)}=& \frac12\left(\frac12-1 \right)\cdot u^{1/2-2}
\\(\sqrt{u})^{(3)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdot u^{1/2-3}
\\\cdots =& \cdots
\\(\sqrt{u})^{(n)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)\cdot u^{1/2-n}
\end{align}
$$ that is

$$
\left.(\sqrt{u})^{(n)}\right|_{u=0}= \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right) \tag2
$$ 

From $(2)$ we deduce by the chain rule that, for $n\geq1$,
$$
\begin{align}
\frac{f^{(n)}(0)}{n!}=&\frac1{n!}\left.(\sqrt{1-4x})^{(n)}\right|_{x=0}
\\\\= &\frac{(-4)^n}{n!}\cdot\frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)
\\\\= &\frac{(-4)^n}{n!}\cdot\frac{1\times(-1)\times(-3)\cdots \times(-(2n-3))}{2^n}
\\\\= &2\cdot\frac{(-1)^n}{n!}\cdot\frac{(-1)^{n-1} 1\times2\times3\cdots \times(2n-3)\times(2n-2)}{1\times2\times3\cdots \times(n-1)}
\\\\= &-\frac2n\cdot\frac{(2n-2)!}{((n-1)!)^2}
\\\\= &-\frac2n\binom{2n-2}{n-1}
\end{align}
$$ which inserted in $(1)$ gives the announced identity.
A: Another approach. The claim is equivalent to:
$$ f(x)=\frac{1-\sqrt{1-4x}}{2x}\stackrel{\color{red}{?}}{=}\sum_{n\geq 0}\frac{1}{n+1}\binom{2n}{n}x^n=g(x) \tag{1}$$
but if we set $C_n = \frac{1}{n+1}\binom{2n}{n}$ we may check that:
$$ C_{n+1} = \frac{1}{n+2}\binom{2n+2}{n+1} = \frac{(2n+2)(2n+1)}{(n+1)^2 (n+2)}\binom{2n}{n}=\frac{2(2n+1)}{n+2} C_n\tag{2}$$
hence it follows that $h(x)=x\cdot g(x)$ is an analytic solution of the differential equation:
$$ (1-4x)\,h'(x)+2\,h(x)=1.\tag{3} $$
However, it is easy to check that $j(x)=x\cdot f(x)$ is an analytic solution of the same differential equation. Since $f(0)=g(0)$ and $f'(0)=g'(0)$, it follows that $\color{red}{f\equiv g}$ as wanted.
