Published October 28, 20202 minute read

Here’s a nice solution of Putnam 1981 B5 that I haven’t seen anywhere else (so far). The main idea is to sum `bitwise,' rather than `termwise.'

Let SkS_k denote the set of positive integers with the kkth bit set, counting from the right starting at k=0k=0. Then we have

n=1B(n)n2+n=k=0nSk1n2+n.\sum_{n=1}^\infty \frac{B(n)}{n^2 + n} = \sum_{k=0}^\infty \sum_{n \in S_k} \frac{1}{n^2 + n}.

A bit of thinking shows that nSkn \in S_k if and only if 2kn=2m+1\lfloor 2^{-k} n\rfloor = 2m+1 is odd. Thus the sum becomes

k=0m=0n=2k(2m+1)2k(2m+2)11n2+n.\sum_{k=0}^\infty \sum_{m = 0}^\infty \sum_{n = 2^k \, (2m+1)}^{2^k \, (2m+2) - 1} \frac{1}{n^2 + n}.

Since (n2+n)1=n1(n+1)1(n^2 + n)^{-1} = n^{-1} - (n+1)^{-1}, the inner sum telescopes, yielding

k=012km=0[12m+112m+2]=2ln2.\sum_{k=0}^\infty \frac{1}{2^k} \sum_{m=0}^\infty \left[\frac{1}{2m+1} - \frac{1}{2m+2}\right] = 2 \ln{2}.