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 $S_k$ denote the set of positive integers with the $k$th bit set, counting from the right starting at $k=0$. Then we have

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

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

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

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

$\sum_{k=0}^\infty \frac{1}{2^k} \sum_{m=0}^\infty \left[\frac{1}{2m+1} - \frac{1}{2m+2}\right] = 2 \ln{2}.$