An Infinite Series Involving a Sideways Sum

I found a recent question on Mathematics Stack Exchange quite interesting. It simply asked

How does one easily calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$ ?

Here $\mathrm{pop}(n)$ denotes the “population count” or “sideways sum”, which is the number of 1s in the binary representation of $n$ (A000120).

The user achille hui provided a very nice answer which I would like to describe in some detail here. First, he introduces the function

$$\theta_k(n) = \begin{cases}1,&\text{ if the $k$th bit of $n$ is set,}\\0,&\text{ otherwise.}\end{cases}$$

which makes it possible to write the series as

$$\sum_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)} = \sum_{n=1}^\infty \sum_{k=0}^\infty \frac{\theta_k(n)}{n(n+1)} = \sum_{k=0}^\infty \sum_{n=1}^\infty \frac{\theta_k(n)}{n(n+1)}.$$

After reversing the order of summation (which requires justification), he asks: For fixed $k$, which values of $n$ has $\theta_k(n)=1$? Note here that $n$ has the $k$th bit set if and only if $\lfloor n/2^k \rfloor$ has the zeroth bit set. And a number has the zeroth bit set if and only if that number is odd. So $\theta_k(n)=1$ if and only if $\lfloor n/2^k \rfloor = 2 l + 1$ for some $l = 0, 1, 2, \ldots$. This means

$$\begin{aligned} \theta_k(n)=1 \quad&\Leftrightarrow\quad 2 l + 1 \leq n/2^k < 2 l + 2 \\ &\Leftrightarrow\quad (2 l + 1) 2^k \leq n \leq (2 l + 2) 2^k - 1, \end{aligned}$$

for some $l = 0, 1, 2, \ldots$. This provides us with the intervals of $n$ that we are interested in, so we have

$$\begin{aligned} \sum_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)} &= \sum_{k=0}^\infty \sum_{n=1}^\infty \frac{\theta_k(n)}{n(n+1)} \\ &= \sum_{k=0}^\infty \sum_{l=0}^\infty \sum_{n=(2 l + 1) 2^k}^{(2 l + 2) 2^k - 1} \frac{1}{n(n+1)} \\ &= \sum_{k=0}^\infty \sum_{l=0}^\infty \sum_{n=(2 l + 1) 2^k}^{(2 l + 2) 2^k - 1} \left( \frac{1}{n} - \frac{1}{n+1} \right) \\ &= \sum_{k=0}^\infty \sum_{l=0}^\infty \left( \frac{1}{(2 l + 1) 2^k} - \frac{1}{(2 l + 2) 2^k} \right) \\ &= \left( \sum_{k=0}^\infty 2^{-k} \right) \left( \sum_{m=1}^\infty (-1)^{m+1} \frac{1}{m} \right) \\ &= 2 \ln 2 \end{aligned}$$

Here, both a telescoping sum, a geometric progression sum (see Nice Proof of a Geometric Progression Sum with $r=1/2$) and the power series for the natural logarithm of 2 occurs.

Very nice.