A question says: True / False
There exists a power series about $a = 0$ with infinite convergence ratio, that converges to the function
$f(x) = \sqrt{1-x}$
I know the function is differentiable infinity times at $a=0$, but how can you find the convergence ratio or how can you know it's not infinity?
Power series with a positive convergence radius give holomorphic functions. If a power series (centered at the origin) converges to $\sqrt{1-x}$ over a neighbourhood of the origin, it has to be the Maclaurin series of $\sqrt{1-x}$, namely $$ \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(1-2n)}x^n $$ by the principle of analytic continuation. On the other hand the radius of convergence of such power series is one, either from the fact that $$\frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta \sim \frac{1}{\sqrt{\pi n}}, $$ or from the fact that the previous series is a $\phantom{}_2 F_1\left(\ldots,x\right)$, or from the fact that $\sqrt{1-z}$ has a singularity (a branch point) at $z=1$. In particular no entire function equals $\sqrt{1-x}$ over a neighbourhood of the origin.