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