= Solution
Let $R_f,R_g$ be the two <radii of convergence>. First take $|x|<t<R_f$, with $t>0$. Absolute convergence at $t$ bounds the terms: $|a_nt^n|\leq M$ for all $n$. Thus
$$
|(n+1)a_{n+1}x^n|\leq\frac Mt(n+1)(|x|/t)^n.
$$
At $x=0$ the right-hand series is finite. Otherwise it converges by the <ratio test>, because its limiting ratio is $|x|/t<1$. So the differentiated <power series> converges absolutely at every interior point of $f$, proving $R_g\geq R_f$. This proof also covers $R_f=\infty$ by choosing a finite $t$ above any given $|x|$; if $R_f=0$, that inequality is automatic.
Conversely, if $0<|x|<R_g$, absolute convergence gives
$$
\sum_{n=1}^\infty|a_n||x|^n
=|x|\sum_{m=0}^\infty\frac{|(m+1)a_{m+1}||x|^m}{m+1}
\leq |x|\sum_{m=0}^\infty|(m+1)a_{m+1}||x|^m<\infty.
$$
The constant term does not affect convergence, and both series converge at zero. Hence $R_f\geq R_g$, including the zero and infinite cases. Therefore \b[$\boxed{R_f=R_g}$]. This proof uses coefficient estimates, rather than assuming the <termwise differentiation of a power series> result whose radius assertion is being established.
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