<s>[INST] Catalan numbers formula derivation I'm trying to follow a proof of the Catalan numbers being equal to $\frac{1}{n+1} {2n \choose n}$ from the recurrence relation $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-2}C_{1}+C_{n-1}C_0$
Now it's seen that the generating function satisfies $xf^2-f+1=0$ so $f=\frac{1-\sqrt{1-4x}}{2x}$ since the other root has a pole at 0 but I'm struggling to see how $f=\frac{1-\sqrt{1-4x}}{2x}$ can be expanded to obtain the necessary power series
I've seen Wikipedia's proof but I don't see how ${ \frac{1}{2} \choose n} = \frac{(-1)^{n+1}}{4^n(2n-1)} {2n \choose n}$ nor how this gets lost into ${2n \choose n}$ via plugging y=-4x and putting it into the expression $f=\frac{1-\sqrt{1-4x}}{2x}$
 [/INST] Hint: By binomial theorem, we have $$(1 + z)^\alpha = \sum\limits_{k \geq 0} \binom{\alpha}{k} z^k.$$
Use this for $\alpha = \frac{1}{2}$ on the square root term, and equate coefficients.

Edit: Here, we define $$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)\cdots(\alpha - (k-1))}{k!}.$$
Note that this agrees with our usual definition when $\alpha$ is an integer.

Edit 2: We have \begin{align}
\frac{1 + \sum_{n \geq 0} \binom{2n}{n} \frac{x^n}{2n - 1}}{2x} &= \frac{1 + (-1) + \sum_{n \geq 1} \binom{2n}{n} \frac{x^n}{2n - 1}}{2x} \\
&= \sum\limits_{n \geq 1}\binom{2n}{n} \frac{x^{n-1}}{(2n-1)2} \\
&= \sum\limits_{n \geq 0} \binom{2n+2}{n+1} \frac{x^n}{2(2n + 1)} \\
&= \sum\limits_{n \geq 0} \frac{(2n + 2)!}{(n+1)!(n+1)!} \frac{x^n}{2(2n + 1)} \\
&=\sum\limits_{n \geq 0} \frac{(2n + 2)(2n+1)(2n)!}{(n+1)(n+1)(n!)(n!)} \frac{x^n}{2(2n + 1)}\\
&= \sum\limits_{n \geq 0} \binom{2n}{n}\frac{x^n}{n + 1}
\end{align}
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