 Radius of a Maclaurin series from successive derivatives

Write the Maclaurin series coefficients as $a_n=f^{(n)}(0)/n!$. When the successive ratios exist eventually and have a limit, the ratio test gives the displayed radius of convergence, including zero and infinity. The unconditional formula is $R^{-1}={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$, from the Cauchy-Hadamard theorem. Endpoints must be checked separately.

The existence and convergence of this power series do not by themselves imply that it equals $f$: equality requires analyticity or control of the Taylor remainder. The smooth function $f(x)=e^{-1/x^2}$ for $x\ne 0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.