If R_1,R_2 are the radii of convergence of sum a_nx^n
If $R_1,R_2$ are the radii of convergence of $\sum a_nx^n$ and $\sum b_nx^n$, what can be said about the radii of $1^{\circ}\ \sum(a_n+b_n)x^n$ and $2^{\circ}\ \sum a_nb_nx^n$?
Explanation of Method
For the sum, absolute convergence of both parent series on $|x|<\min(R_1,R_2)$ guarantees convergence of the combined series there, giving a clean lower bound — though cancellation can push the true radius higher. For the coefficient-wise product, the root test gives the general (Hadamard) lower bound $R_1R_2$, but no exact formula exists, since the product coefficients can behave very differently from either parent sequence.
- Bound the radius of $\sum(a_n+b_n)x^n$ below by $\min(R_1,R_2)$, and show equality can fail.
- Bound the radius of $\sum a_nb_nx^n$ below by $R_1R_2$ (Hadamard's theorem).
- Illustrate with examples that no exact formula exists for the product case.
Step 1 · Part $1^{\circ}$: the sum $\sum(a_n+b_n)x^n$
For $|x|<\min(R_1,R_2)$, both $\sum a_nx^n$ and $\sum b_nx^n$ converge absolutely, so their sum converges absolutely too. Hence the radius $R$ of $\sum(a_n+b_n)x^n$ satisfies $$ R \ge \min(R_1,R_2). $$ Equality typically holds, but cancellation can make $R$ strictly larger — even infinite. Example: $a_n=1$, $b_n=-1$ for all $n$ (so $R_1=R_2=1$); then $a_n+b_n=0$ for every $n$, giving $R=\infty$.
Step 2 · Part $2^{\circ}$: the coefficient-wise product $\sum a_nb_nx^n$
This is the Hadamard product of the two series. Using the root test, since $\limsup(u_nv_n)\le(\limsup u_n)(\limsup v_n)$ for non-negative sequences $u_n=|a_n|^{1/n}$, $v_n=|b_n|^{1/n}$: $$ \limsup_n|a_nb_n|^{1/n} \le \limsup_n|a_n|^{1/n}\cdot\limsup_n|b_n|^{1/n}, $$ which translates to the radius $R'$ of $\sum a_nb_nx^n$ satisfying (Hadamard's multiplication theorem) $$ R' \ge R_1R_2. $$ Unlike Part $1^{\circ}$, there is no general upper bound or exact formula: $R'$ can equal $R_1R_2$, or be strictly larger, depending entirely on the specific coefficients.
Step 3 · Examples for Part $2^{\circ}$
Equality: $a_n=n$ ($R_1=1$), $b_n=1/n^2$ ($R_2=1$): $a_nb_n=1/n$, giving $\sum x^n/n$ with $R'=1=R_1R_2$.
Strictly larger: $a_n=n!$ ($R_1=0$), $b_n=1/n!$ ($R_2=\infty$): $a_nb_n=1$, giving $\sum x^n$ with $R'=1$ — finite, despite $R_1=0$ making the "bound" $R_1R_2$ ill-defined as a naive product.
Infinite: $a_n=2^n$ ($R_1=1/2$), $b_n=1/(2^nn!)$ ($R_2=\infty$): $a_nb_n=1/n!$, giving $\sum x^n/n!=e^x-1$ with $R'=\infty$.
Summary Table
| Series | Radius bound |
|---|---|
| $\sum(a_n+b_n)x^n$ | $R\ge\min(R_1,R_2)$, can be larger |
| $\sum a_nb_nx^n$ (Hadamard product) | $R'\ge R_1R_2$, no exact formula |
Additional Notes
- The sum case mirrors the elementary fact that a sum of two convergent series converges — power series add termwise, so their regions of absolute convergence simply intersect, with the only subtlety being possible cancellation enlarging the region.
- The product case (Hadamard product) is genuinely richer: it can be computed via a classical contour-integral formula, and its radius is famously not determined by $R_1,R_2$ alone — the third example above (factorial times reciprocal factorial) is the standard textbook illustration that $R'$ can exceed any naive expectation from $R_1$ or $R_2$ individually.
- This theoretical question sits apart from the concrete convergence computations of Problems 223–235, but its tools (root test, limsup inequalities) are the same machinery used throughout — for instance, Problem 233's root-test radius $1/e$ relies on exactly the same limsup framework used here.