A calculus problem by Rajyawardhan Singh

Before anything else, the sum is absolutely convergent since we have $\sum_{\substack{k \ge 1 \\ n \ge 0}} \frac{1}{k(k \cdot 2^n+1)} < \sum_{k \ge 1} k^{-2} \sum_{n \ge 0} \frac{1}{2^n} = \frac{\pi^2}{6} \cdot 2 < \infty.$ Thus we may swap the order of summation.

We use $d = k \cdot 2^n+1 \ge 2$ as the summation variable, so that the sum in question is $\sum_{d \ge 2} \frac 1d \sum_{\substack{k \\ \exists n : d-1 = k \cdot 2^n}} \frac{(-1)^{k-1}}{k}.$ Now we claim that the inner sum is exactly $\frac{1}{d-1}$ . Indeed, if $d - 1 = 2^r m + 1$ with $m$ odd, then the sum is $\frac{(-1)^{m-1}}{m} + \frac{(-1)^{2m-1}}{2m} + \dots + \frac{(-1)^{2^r m-1}}{2^r m}$ &= $\frac 1m \left( \frac11 - \frac12 - \dots - \frac{1}{2^r} \right)$ &= $\frac{1}{2^r m}$ &= $\frac{1}{d-1}$ . Consequently, the final answer is $\sum_{d \ge 2} \frac{1}{d(d-1)} = \sum_{d \ge 2} \left( \frac{1}{d-1} - \frac 1d \right) = \boxed{1}.$