= Radius of a Maclaurin series from successive derivatives
{title2=$R=\lim_{n\to\infty}\left|(n+1)f^{(n)}(0)/f^{(n+1)}(0)\right|$}
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 $R^{-1}=\limsup_{n\to\infty}|a_n|^{1/n}$, 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 $x\ne0$, $f(0)=0$, has all derivatives zero at zero, so its Maclaurin series is identically zero despite the nonzero values away from zero.
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