Prove that the radius of convergence, \(R\), of \(\sum_{n=0}^\infty a_nb_nx^n\) satisfies \(R \geq R_1R_2\). Show by means of example that this inequality can be strict.
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. For \(|x| < R_1 R_2\), \(\sum a_n b_n x^n\) converges absolutely.
1.1. There are \(u < R_1\) and \(v < R_2\) with \(|x| < uv\).
The set \(\theset{uv : 0 \le u < R_1,\ 0 \le v < R_2}\) is the interval \([0, R_1R_2)\), with \(R_1R_2 = \infty\) if either radius is infinite and the other positive.
1.2. \(|a_n|u^n \le 1\) and \(|b_n|v^n \le 1\) for all large \(n\).
\(\sum a_n u^n\) and \(\sum b_n v^n\) converge, since \(u < R_1\) and \(v < R_2\), so their terms tend to \(0\).
1.3. Q.E.D.
By step 2, \(|a_n b_n x^n| = (|a_n|u^n)(|b_n|v^n)\left|\frac{x}{uv}\right|^n \le \left|\frac{x}{uv}\right|^n\) for large \(n\), and \(\sum_n |x/(uv)|^n\) is a convergent geometric series by step 1.
2. \(R \ge R_1 R_2\).
Step 1.
3. For \(a_n = 1\) (\(n\) even), \(a_n = 0\) (\(n\) odd), \(b_n = 0\) (\(n\) even), \(b_n = 1\) (\(n\) odd), \(R_1 = R_2 = 1\) and \(R = \infty\).
\(\limsup |a_n|^{1/n} = 1\) along the even indices and \(\limsup|b_n|^{1/n} = 1\) along the odd indices, so the Cauchy–Hadamard formula gives \(R_1 = R_2 = 1\). Every \(a_nb_n = 0\), so \(\sum a_nb_nx^n\) is the zero series and \(R = \infty > 1 = R_1R_2\).
4. Q.E.D.
Step 2 is the inequality and step 3 shows that it can be strict.