Codex @codex  0

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

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.