an is a sequence of positive reals such that ∑ 1/an converges. Let sn = ∑1n ai. Prove that ∑ n2an/sn2 converges.
Let A = (∑1/an)1/2 and BN = ∑1N n2an/sn2.
Since all an are positive, sn-1 < sn and hence BN < ∑1N n2an/(snsn-1). But an = sn - sn-1, so for n > 1 we may write the summand as n2(1/sn-1 - 1/sn). Hence BN < 1/a1 + (4/s1 - 4/s2) + (9/s2 - 9/s3) + ... + (N2/sN-1 - N2/sN) < 5/a1 + 5/s2 + 7/s3 + 9/s4 + ... + (2N-1)/sN-1 - N2/sN < 2/a1 + 3(1/s1 + 2/s2 + 3/s3 + ... + N/sN).
Now ∑1N n/sn= ∑ (1/√an) (n(√an)/sn ) <= (∑ 1/an)1/2 (∑ n2an/sn2)1/2 < A BN1/2. So we have BN < 2/a1 + 3A BN1/2. That implies that BN is bounded above. For example, we certainly have BN < (1 + 3/a1 + 3A)2. But any increasing sequence which is bounded above must converge.
27th Putnam 1966
© John Scholes
25 Jan 2002