MathDB
Putnam 2016 B6

Source:

December 4, 2016
PutnamPutnam 2016Putnam calculus

Problem Statement

Evaluate k=1(1)k1kn=01k2n+1.\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}.