3rd Putnam 1940

a_i and b_i are real, and ∑₁^∞ a_i² and ∑₁^∞ b_i² converge. Prove that ∑₁^∞ (a_i - b_i)^p converges for p ≥ 2.

Solution

Moderately easy.

Notice first that it is sufficient to prove the result for p = 2. For that is equivalent to the statement that ∑ |a_i - b_i|² converges. Hence for sufficiently large i, |a_i - b_i| < 1, and hence |a_i - b_i|^p ≤ |a_i - b_i|². So ∑ (a_i - b_i)^p is absolutely convergent and hence convergent.

(a_i - b_i)² = a_i² - 2a_ib_i + b_i². The only tricky part is the middle term. It may be positive, so we cannot simply argue that 0 ≤ (a_i - b_i)² ≤ a_i² + b_i². However, it is true that 0 ≤ (a_i - b_i)² = 2a_i² + 2b_i² - (a_i + b_i)² ≤ 2a_i² + 2b_i². That suffices, since ∑ a_i² and ∑ b_i² are absolutely convergent, hence also ∑ (2a_i² + 2b_i²).  

3rd Putnam 1940

© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999