Let Bn(x) = 1x + 2x + ... + nx and let f(n) = Bn(logn2) / (n log2n)2. Does f(2) + f(3) + f(4) + ... converge?
This looks much worse than it is. A crude estimate for Bn(logn2) is sufficient. We know that ∑ 1/n only just fails to converge. So it is likely that ∑ 1/n 1/log2n will converge. In that case we need only prove that Bn(logn2) ≤ O(n). But it has n terms, the biggest of which is 2. So certainly Bn(logn2) < 2n.
We can easily check that ∑ 1/n 1/log2n converges by the integral test (the indefinite integral is -1/log n).
43rd Putnam 1982
© John Scholes
16 Jan 2001