Double sum to infinity

We consider a more general setting. Let $z > 0$ . Then $\begin{aligned} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty}\frac{1}{2^n k+ z} &= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n k + z - 1} \, \mathrm{d}x \\ &= \int_{0}^{1} \left( \sum_{n=0}^{\infty} \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} x^{2^n k} \right) x^{z - 1} \, \mathrm{d}x \\ &= \int_{0}^{1} \left( \sum_{n=0}^{\infty} \log(1 + x^{2^n}) \right) x^{z - 1} \, \mathrm{d}x, \end{aligned}$ where in the last step we utilized the Taylor expansion $\log(1+x) = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} x^k$ . Now, from the following famous (and easy-to-prove) identity $\frac{1}{1-x} = \prod_{n=0}^{\infty} (1 + x^{2^n}),$ the last step simplifies to $\begin{aligned} &= \int_{0}^{1} x^{z - 1} \log\left(\frac{1}{1-x}\right) \, \mathrm{d}x \\ &= \left[ \frac{x^z - 1}{z} \log\left(\frac{1}{1-x}\right) \right]_{0}^{1} + \frac{1}{z} \int_{0}^{1} \frac{1 - x^z}{1 - x} \, \mathrm{d}x \\ &= \frac{H(z)}{z}, \end{aligned}$ where $H(z)$ is the harmonic number. Plugging $z = 5$ , we end up with $\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k} \sum_{n=0}^{\infty}\frac{1}{2^n k+ 5} = \frac{H(5)}{5} = \frac{1}{5}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right) = \frac{137}{300},$ and multiplying 300 to both sides yields the answer $137$ .