(January 2006, 1) Let the power series series \(\sum_{n=0}^{\infty}a_nx^n\) and \(\sum_{n=0}^{\infty}b_nx^n\) have radii of convergence \(R_1\) and \(R_2\), respectively.
If \(R_1\ne R_2\), 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 \(R_1=R_2\)?
Prove that the radius of convergence, \(R\), of \(\sum_{n=0}^{\infty}a_nb_nx^n\) satisfies \(R\ge R_1R_2\). Show by means of an example that this inequality can be strict.
The source page literally says “the power series series” in its opening sentence; that wording is preserved here.
Solution
Goal: (a) If \(\sum a_nx^n\) and \(\sum b_nx^n\) have radii \(R_1, R_2\) with \(R_1 \neq R_2\), show \(\sum(a_n+b_n)x^n\) has radius \(\min\{R_1, R_2\}\); discuss \(R_1 = R_2\). (b) Show \(\sum a_nb_nx^n\) has radius \(R \ge R_1R_2\), with a strict example.
1. (a) \(R_1 \neq R_2\): the radius of \(\sum(a_n+b_n)x^n\) is \(\min\{R_1, R_2\}\).
1.1. Say \(R_1 < R_2\).For \(|x| < R_1\), both series converge absolutely, so their sum converges absolutely: the radius \(R\) of the sum satisfies \(R \ge R_1\).
1.2. For \(R_1 < |x| < R_2\): \(\sum(a_n+b_n)x^n\) diverges.Proof: if it converged, then \(\sum a_nx^n = \sum(a_n+b_n)x^n - \sum b_nx^n\) would converge (difference of two convergent series, since \(\sum b_nx^n\) converges for \(|x| < R_2\)), contradicting the divergence of \(\sum a_nx^n\) for \(|x| > R_1\).
1.3. \(R = R_1 = \min\{R_1, R_2\}\).Proof: 1 gives \(R \ge R_1\) and 2 gives \(R \le R_1\).
1.4. If \(R_1 = R_2 = R_0\), then \(R \ge R_0\), and \(R\) can be strictly larger.Proof: \(R \ge R_0\) since both series converge absolutely for \(|x| < R_0\).Strictness: \(a_n = 1\), \(b_n = -1\) gives \(a_n + b_n = 0\), radius \(\infty\); or \(a_n = 1\), \(b_n = -1 + 2^{-n}\) gives \(a_n + b_n = 2^{-n}\), radius \(2 > 1 = R_0\).Nothing more can be said in general.
1.5. Q.E.D. Proof: 3 and 4 answer (a).
2. (b) The radius \(R\) of \(\sum a_nb_nx^n\) satisfies \(R \ge R_1R_2\).
2.1. Fix \(|x| < R_1R_2\); choose \(r_1 < R_1\), \(r_2 < R_2\) with \(|x| < r_1r_2\).Proof: e.g. take \(r_1, r_2\) close to \(R_1, R_2\) with \(r_1r_2 > |x|\), possible since \(R_1R_2 > |x|\).
2.2. \(|a_n|r_1^n \to 0\) and \(|b_n|r_2^n \to 0\).Proof: \(\sum a_nr_1^n\) and \(\sum b_nr_2^n\) converge absolutely (inside the radius of convergence), so their terms tend to \(0\); in particular they are bounded: \(|a_n|r_1^n \le C_1\), \(|b_n|r_2^n \le C_2\).
2.3. \(|a_nb_n||x|^n = (|a_n|r_1^n)(|b_n|r_2^n)\left(\frac{|x|}{r_1r_2}\right)^n \le C\rho^n\) with \(\rho := |x|/(r_1r_2) < 1\).Proof: 2 and \(|x| < r_1r_2\).
2.4. \(\sum a_nb_nx^n\) converges absolutely for \(|x| < R_1R_2\), so \(R \ge R_1R_2\).Proof: comparison with the convergent geometric series \(\sum C\rho^n\) by
2.3.
2.5. Strictness example: \(a_{2k} = 1\), \(a_{2k+1} = 0\) and \(b_{2k} = 0\), \(b_{2k+1} = 1\).Proof: \(\sum a_nx^n = \sum_k x^{2k}\) has radius \(R_1 = 1\) and \(\sum b_nx^n = \sum_k x^{2k+1}\) has radius \(R_2 = 1\); but \(a_nb_n = 0\) for all \(n\), so \(\sum a_nb_nx^n = 0\) converges everywhere: \(R = \infty > 1 = R_1R_2\).
2.6. Q.E.D. Proof: 4 gives \(R \ge R_1R_2\); 5 exhibits strictness.