# Radius of convergence of a non-zero function with zero Taylor series

A classic example of a nonzero function with identically zero Taylor expansion is the following:

$$\begin{equation*} f(x)= \begin{cases} e^{-\frac{1}{x^2}}\quad &\text{if x\neq 0}\\ 0\quad &\text{if x=0} \end{cases} \end{equation*}$$ according to this post Maclaurin series expansion for $e^{-1/x^2}$, it is clear that the Taylor series is null, however I want to know what the radius of convergence is, I know that it is convergent in the neighborhood of zero, however using the definition

$$$$\alpha=\limsup_{n\to\infty}\sqrt[n]{|c_{n}|},\quad R=\frac{1}{\alpha}$$$$ I think that the radius of convergence is zero, however I don't know how to justify it, which help is well received

• All the coefficients are zero, so the radius of convergence is infinite ($\alpha=0$ so by convention, $R=\infty$). Now, just because the Taylor series has infinite radius of convergence, doesn't mean the function it sums to (in this case $0$) is equal to the function you started with (namely $f$). Jun 22 '21 at 5:57

The series itself is convergent everywhere, since $$\alpha = 0$$, by convention $$R = \infty$$. This, however, does not mean that the series converges to the function from which you computed those coefficients - you need to prove that separately. The function you are talking about is not analytic at $$x=0$$, meaning it cannot be expressed as a Taylor series about that point.