ai and bi are real, and ∑1∞ ai2 and ∑1∞ bi2 converge. Prove that ∑1∞ (ai - bi)p converges for p ≥ 2.
Notice first that it is sufficient to prove the result for p = 2. For that is equivalent to the statement that ∑ |ai - bi|2 converges. Hence for sufficiently large i, |ai - bi| < 1, and hence |ai - bi|p ≤ |ai - bi|2. So ∑ (ai - bi)p is absolutely convergent and hence convergent.
(ai - bi)2 = ai2 - 2aibi + bi2. The only tricky part is the middle term. It may be positive, so we cannot simply argue that 0 ≤ (ai - bi)2 ≤ ai2 + bi2. However, it is true that 0 ≤ (ai - bi)2 = 2ai2 + 2bi2 - (ai + bi)2 ≤ 2ai2 + 2bi2. That suffices, since ∑ ai2 and ∑ bi2 are absolutely convergent, hence also ∑ (2ai2 + 2bi2).
3rd Putnam 1940
© John Scholes
15 Sep 1999