Putnam 1966/B3 solution

27th Putnam 1966

Problem B3

a_n is a sequence of positive reals such that ∑ 1/a_n converges. Let s_n = ∑₁ⁿ a_i. Prove that ∑ n²a_n/s_n² converges.

Solution

Let A = (∑1/a_n)^1/2 and B_N = ∑₁^N n²a_n/s_n².

Since all a_n are positive, s_n-1 < s_n and hence B_N < ∑₁^N n²a_n/(s_ns_n-1). But a_n = s_n - s_n-1, so for n > 1 we may write the summand as n²(1/s_n-1 - 1/s_n). Hence B_N < 1/a₁ + (4/s₁ - 4/s₂) + (9/s₂ - 9/s₃) + ... + (N²/s_N-1 - N²/s_N) < 5/a₁ + 5/s₂ + 7/s₃ + 9/s₄ + ... + (2N-1)/s_N-1 - N²/s_N < 2/a₁ + 3(1/s₁ + 2/s₂ + 3/s₃ + ... + N/s_N).

Now ∑₁^N n/s_n= ∑ (1/√a_n) (n(√a_n)/s_n ) <= (∑ 1/a_n)^1/2 (∑ n²a_n/s_n²)^1/2 < A B_N^1/2. So we have B_N < 2/a₁ + 3A B_N^1/2. That implies that B_N is bounded above. For example, we certainly have B_N < (1 + 3/a₁ + 3A)². But any increasing sequence which is bounded above must converge.

27th Putnam 1966