Old ghosts from the past are catching up to me since I have either seem to completely forgotten how to connect convergence of the Taylor series of a function to the original function or I have never properly learned this in the first place. Take $f(x)=\sqrt{1 - x}$. Then $f(x)$ is defined on $\mathbb{R}$ and is smooth on $\mathbb{R}\setminus \{0\}$ with $f^{(k)}(x) = (-1)^{k+1}\left(\frac{1}{2}\right)^k\left(1 - x\right)^{\frac{1}{2} - k},k\geq 1$. Hence $\left|\frac{f^{(k)}(0)}{k!}\right| = \frac{1}{2^kk!}$ and
$$\lim_{k\to\infty}\left|\frac{f^{(k+1)}(0)x^{k+1}}{(k+1)!}\cdot \frac{k!}{f^{(k)}(0)x^{k}}\right| = \lim_{k\to\infty}\left|\frac{x}{2k}\right| = 0$$
showing that $g(x) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k$ has infinite radius of convergence. But clearly $g$ cannot converge to $f$ at $x = 1$ since $f(1) = 0$ but $g(1) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}1^k > 0$ as can be verified by computing first few terms. So what, if anything, is going on?
Your derivative is incorrect. It should be
$$f'(x)=-\frac12(1-x)^{-1/2},~~~~f^{(k)}(x)=-\frac{(2k-3)!!}{2^k}(1-x)^{-\frac{2k-1}2},~~k=2,3\dots$$
So you need to re-do everything.