If \(R_1 \neq R_2\), prove that the radius of convergence, \(R\), of the power series \(\sum_{n=0}^\infty (a_n+b_n)x^n\) is \(\min\{R_1, R_2\}\). What can be said about \(R\) when \(R_1 = R_2\)?
Solution
Here \(R_1\) and \(R_2\) are the radii of convergence of \(\sum a_n x^n\) and \(\sum b_n x^n\).
1. \(R \ge \min\{R_1, R_2\}\).
For \(|x| < \min\{R_1, R_2\}\) both \(\sum a_n x^n\) and \(\sum b_n x^n\) converge absolutely, and \(|(a_n + b_n)x^n| \le |a_nx^n| + |b_nx^n|\).
2. If \(R_1 < R_2\), then \(R \le R_1\).
Let \(R_1 < |x| < R_2\). Then \(\sum a_n x^n\) diverges and \(\sum b_n x^n\) converges. If \(\sum (a_n + b_n)x^n\) converged, then \(\sum a_n x^n\) would converge as the difference of two convergent series. So \(\sum (a_n + b_n)x^n\) diverges at points with \(|x| > R_1\) arbitrarily close to \(R_1\), and \(R \le R_1\).
3. If \(R_1 = R_2\), then \(R \ge R_1\), and both \(R = R_1\) and \(R > R_1\) occur.
1. With \(b_n = a_n\), the coefficients \(2a_n\) give \(R = R_1\). With \(b_n = -a_n\), the series is \(0\) and \(R = \infty\), which exceeds \(R_1\) whenever \(R_1 < \infty\).
4. Q.E.D.
If \(R_1 \neq R_2\), steps 1 and 2, with the roles of the series exchanged if \(R_2 < R_1\), give \(R = \min\{R_1, R_2\}\). Step 3 is the case \(R_1 = R_2\).