= Solution
Write the formal <Maclaurin series> as $\sum_{n\geq0}a_nx^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the <ratio test> gives
$$
\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|
=|x|\left|\frac{f^{(n+1)}(0)}{(n+1)f^{(n)}(0)}\right|\longrightarrow\frac{|x|}{R},
$$
so the series converges absolutely for $|x|<R$, where
$$
\boxed{R=\lim_{n\to\infty}\left|\frac{(n+1)f^{(n)}(0)}{f^{(n+1)}(0)}\right|.}
$$
If that ratio does not exist, the <Cauchy-Hadamard theorem> gives the general formula $R^{-1}=\limsup|a_n|^{1/n}$. 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,
$$
\ln(2+x)=\ln2+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n2^n},\qquad |x|<2.
$$
The absolute successive-term ratio tends to $|x|/2$, so \b[the interval of <absolute convergence> is $(-2,2)$]. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln2$, giving the full sum $\ln4$. 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 \b[$(-2,2]$].
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