The real sequence a1, a2, a3, ... has the property that limn→∞ (an+2 - an) = 0. Prove that limn→∞ (an+1 - an)/n = 0.
Suppose we have any series an satisfying the condition. Take a new series bn defined by b2n+1 = 0 and b2n = a2n. Then bn also satisfies the condition, but (b2n+1 - b2n)/n = - a2n/n, so we must have the apparently stronger result that a2n/(2n) tends to zero (and similarly that a2n+1/(2n+1) tends to zero).
Thus we need to prove that if cn is any sequence such that cn+1 - cn tends to zero, then cn/n tends to zero.
Given any positive k, we have |cn+1 - cn| < k for all n > some N. Now for any M > N, we have cM = cN + (cN+1 - cN) + ... + (cM - cM-1). Hence |cM| ≤ |cN| + (M-N)k. If we take M sufficiently large that |bN|/M < k, then |cM|/M < 2k, which establishes that cn/n tends to zero.
31st Putnam 1970
© John Scholes
25 Aug 2001