Simple test of Convergence.

The sum is $\sum_{n=0}^\infty \sum_{k=1}^\infty \frac1{2^nk + 1} \frac{(-1)^{k-1}}{k}.$ Consider how many times a number $N$ appears as $2^n k.$ If $N = 2^b a$ with $a$ odd, then $\frac1{N+1}$ will appear in $b+1$ terms of the sum, when $k= a, 2a, 4a, \ldots, 2^ba$ and $n= b, b-1, b-2, \ldots, 0.$ That is, our sum equals $\begin{aligned} \sum_{N=1}^\infty \frac1{N+1} \sum_{2^nk = N} \frac{(-1)^{k-1}}{k} &= \sum_{N=1}^{\infty} \frac1{N+1} \left( \frac1{a} - \frac1{2a} - \frac1{4a} - \cdots - \frac1{2^ba} \right) \\ &= \sum_{N=1}^{\infty} \frac1{N+1} \frac1{2^ba} = \sum_{N=1}^{\infty} \frac1{N(N+1)}= \fbox{1} \end{aligned}$ (the sum telescopes ).

Putnam 2016, B6

The given sum can be re-written as:

$\large\ S = \displaystyle \sum _{ k=1 }^{ \infty }{ \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ k - 1 } }{ k } \displaystyle \int _{ 0 }^{ 1 }{ { x }^{ k{ 2 }^{ n } }dx } } } .$

By absolute convergence, we are free to permute the integral and the sums:

$\large\ S = \displaystyle \int _{ 0 }^{ 1 }{ dx\sum _{ k=1 }^{ \infty }{ \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ k-1 } }{ k } { x }^{ { k2 }^{ n } } } } } = \displaystyle -\int _{ 0 }^{ 1 }{ dx \displaystyle \sum _{ n=0 }^{ \infty }{ \log { \left( 1 + { x }^{ { 2 }^{ n } } \right) } } } .$

Which evaluates to $1$ .