Prove that $\frac{1}{\sqrt{1-z}}=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\binom{2n}{n}z^{n}$ using Cauchy product need to prove using Cauchy product for series for all $\left|z\right|<1$ that $$\frac{1}{\sqrt{1-z}}=\sum_{n=0}^{\infty}\frac{1}{4^{n}}\binom{2n}{n}z^{n}$$ (with appropriate branch of the root function)
 A: Prove the series converges (absolutely) for $|z|<1$, then the Cauchy product formula says
$$\left(\sum_{n=0}^\infty\frac1{4^n}\binom{2n}nz^n\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n\frac1{4^n}\binom{2(n-k)}{n-k}\binom{2k}kz^n$$
Now prove
$$\sum_{k=0}^n\binom{2(n-k)}{n-k}\binom{2k}k=4^n$$
So the square of the sum is
$$\sum_{n=0}^\infty z^n=\frac1{1-z}$$
$\dfrac1{\sqrt{1-z}}$ is analytic on the disk $|z|<1$ and matches the series (which is also analytic) for $z\in\mathbb [0,1)$ by the previous calculations, as both are clearly the positive roots of $\dfrac1{1-z}$. Therefore they have to be the same for all $|z|<1$.

Details:
The  ratio test tells us the radius of convergence is
$$\lim_{n\to\infty}\frac{4^{n+1}}{4^n}\frac{\binom{2n}n}{\binom{2(n+1)}{n+1}}=4\lim_{n\to\infty}\frac{(2n)!}{(n!)^2}\frac{((n+1)!)^2}{(2(n+1))!}=4\lim_{n\to\infty}\frac{(n+1)^2}{(2n+1)(2n+2)}=1$$
The combinatorial identity is harder than I expected, for a bijective proof see this answer: Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$
I suspect it should also be possible to find a bijection to the subsets of $2n$ as there's $4^n$ of them, but I don't see how.
