Proof that taylor series converges to function using taylors inequality

I would like to proof that the function $f(x)=\frac{1}{\sqrt{1-x}}$ converges to its Maclaurin series $$Tf(x;0)=1+\sum_{n=1}^\infty{\frac{(2n)!}{(2^n n!)^2}}x^n$$ for $0<x<1$ by using taylors inequality $$|R_{n}|\leq M\frac{|x|^{n+1}}{(n+1)!}$$ In this case M (the upper bound for the magnitude of the n+1th derivative in the interval $[0,x]$) would be given by the n+1th derivative of f at x, which is $$f^{(n+1)}(x)=\frac{(2n+2)!}{2^{2n+2} (n+1)!}(1-x)^{-\frac{2n+3}{2}}$$ Plugging that into taylors inequality yields: $$|R_{n}|\leq \frac{(2n+2)!}{2^{2n+2} ((n+1)!)^2}(1-x)^{-\frac{2n+3}{2}}|x|^{n+1}$$ The taylor series converges if $\lim_{n\rightarrow +\infty}(|R_{n}|)=0$. Thus the following should hold true: $$\lim_{n\rightarrow +\infty}\frac{(2n+2)!}{2^{2n+2} ((n+1)!)^2}(1-x)^{-\frac{2n+3}{2}}|x|^{n+1}=0$$ I tried using the ratio test to evaluate this limit and the result is: $$\lim_{n\rightarrow +\infty}\left|\frac{a_{n+1}}{a_n}\right|=|\frac{x}{1-x}|$$ which is a problem, since for $x>1/2$ the limit does not exist and hence is not zero. But clearly the taylor series does converge, even for $1/2<x<1$.

So my question is whether there has to be an error in my calculation, which I should search for or the method to prove convergence involving taylors inequality does not work generally (or at least not in this case)?

So you want to prove that for any $x\in(0,1)$, $$\sum_{n\geq 0}a_n x^n=\sum_{n\geq 0}\frac{x^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-x}}\tag{1}$$ and you already know that the LHS is the Taylor series of the RHS in a neighbourhood of the origin. Well, the radius of convergence of the LHS is one, since: $$0\leq \frac{a_{n+1}}{a_n} = \frac{2n+1}{2n+2}<1\tag{2}$$ so for any $x\in(0,1)$: $$\left|\sum_{n\geq N}\frac{x^n}{4^n}\binom{2n}{n}\right|\leq\frac{x^N}{1-x}\tag{3}$$ but the RHS of $(3)$ converges to zero as $N\to +\infty$, proving pointwise convergence.