Prove or Disprove - power series

Question

Let $f(x)$ be a differentiable function for all order $n$ in $\Bbb R$.

Let $R_n(x)$ be McLaurin's remainder of order $n$ of $f(x)$.

Prove or disprove the following statement:

if $$\lim\limits_{x \to 0} \frac{R_n(x)}{x^n}=0$$ for all $n$ then the radius of convergence of the power series $$\Sigma_{n=0}^\infty \frac{f^{(n)} (0)}{n!}\cdot x^n$$ is greater than $\frac{1}{2}$

I think the statement is correct, but I have no idea how to prove it. Couldn't find a counter example.

Thanks!

Answer 1

Fix $n\in\Bbb N.$ We have $$R_n(x)=\frac{f^{(n)}(c_x)}{n!}x^n,$$ so $$\frac{R_n(x)}{x^n}=\frac{f^{(n)}(c_x)}{n!}.$$ Now let $x\to 0.$ Since $0\le c_x\le|x|$ we infer that $c_x\to 0$. All derivatives $f^{(k)}$ are continuous, then $f^{(n)}(c_x)\to f^{(n)}(0).$ Dividing by $n!$ we obtain that $f^{(n)}(0)=0$ for all $n$. Now our function is constant under assupmtion that $f$ is equal to its McLaurin's expansion.