I want to comprehend the first proof of $C_n = \frac{1}{n + 1}\binom{2n}{n}$ on Wikipedia. Link: here. I have problems with two steps:
- $$\sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}y^n = \sum_{n=0}^{\infty} \frac{(-1)^{n + 1} }{4^n(2n - 1)} \binom{2n}{n}y^n$$ I calculated $$\binom{\frac{1}{2}}{n} = \frac{(-1)^{n + 1}}{4^n} \frac{(n - 1.5) \cdot (n - 1.5 - 1) \cdot \ldots \cdot (n - 1.5 - (n - 3) + 1) }{n!}$$, but how to proceed?
- One shall insert $$\sum_{n = 0}^{\infty} \frac{(-1)^{n+1}}{4^n(2n-1)}\binom{2n}{n} x^n$$ into $$\frac{1 - \sqrt{1 - 4x}}{2x} = \frac{1}{2} \frac{1}{x}(1 - \sqrt{1 - 4x})$$ by using standard power series operations. But how do we handle the division of $1$ by $x$ in terms of $x^n$? The wikipedia entry for power series says that the first coeffizient of the division is calculated as $d_0 = \frac{a_0}{b_0}$, but for $x = \sum b_n x^n$, $b_0$ is 0, such that we get a division by zero.
Please note that we consider formal power series, so I do not know if and how we can somehow detour via an other development point.
Background: The problem occured during computing the moments of the semicircular distribution in free probability theory.
EDIT: While elaborating the solution (I admit that I never had found it by myself) given by @BrianM.Scott, the specific second question was solved too: One does not calculate $\frac{1}{x}$ directly, but splits the sum up and lets the dangerous constant terms (for $n = 0$) eliminate one another before dividing the remaining sum terms (for $n > 0$) by $x$.
