A double Putnam 2016 series

Evaluate the series:

$$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\sum_{n=0}^{\infty}\frac{1}{k2^n+1}$$

(Putnam 2016)

Solution

We begin by the simple observation that the series converges absolutely. Why? Simply, because it is dominated by $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^2}\sum_{n=0}^{\infty}\frac{1}{2^n}$. (The last sum is equal to $2 \zeta(2)$.) Now in order to compute the sum we can interchange the summations. Thus:

\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \frac{1}{k2^n+1} &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \int_{0}^{1} x^{k 2^n } \, {\rm d}x\\
 &= \sum_{n=0}^{\infty} \int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^{2^n k} \, {\rm d}x\\
 &= \sum_{n=0}^{\infty} \int_{0}^{1} \log \left ( 1+x^{2^n} \right ) \, {\rm d}x\\
 &= \int_{0}^{1}\sum_{n=0}^{\infty} \log \left ( 1+ x^{2^n} \right ) \, {\rm d}x  \\
 &= \int_{0}^{1} \log \prod_{n=0}^{\infty} \left ( 1+x^{2^n} \right ) \, {\rm d}x \\
 &= - \int_{0}^{1} \log  \left ( 1-x \right ) \, {\rm d}x \\
 &=1
\end{align*}