= Solution
Use these standard facts about a <power series>: it is absolutely convergent at every point strictly inside its <radius of convergence> and divergent at every point strictly outside it. Also, the <triangle inequality> implies that a sum or difference of two absolutely convergent <series> is absolutely convergent.
Let $T$ be the <radius of convergence> of the coefficientwise sum. For $|z|<\min(R,S)$,
$$
\sum_{n\ge0}|(a_n+b_n)z^n|
\le\sum_{n\ge0}|a_nz^n|+\sum_{n\ge0}|b_nz^n|<\infty.
$$
Hence $T\ge\min(R,S)$.
Suppose $R<S$. If $T>R$, choose a finite $\rho$ with $R<\rho<\min(T,S)$ and choose $z$ of <complex modulus> $\rho$. The sum series and the $b_n$ series both converge absolutely there, so
$$
\sum_{n\ge0}|a_nz^n|
\le\sum_{n\ge0}|(a_n+b_n)z^n|+\sum_{n\ge0}|b_nz^n|<\infty,
$$
contradicting $\rho>R$. Thus $T=R$. Interchanging the series covers $S<R$, proving the <sum of power series with unequal radii of convergence> rule:
$$
\boxed{T=\min(R,S).}
$$
The argument also covers a zero radius or an infinite larger radius. When the two radii are equal, cancellation can enlarge the radius, so the inequality of the radii matters.
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