How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$? How does one [easily] calculate $\sum\limits_{n=1}^\infty\frac{\mathrm{pop}(n)}{n(n+1)}$, where $\mathrm{pop}(n)$ counts the number of bits '1' in the binary representation of $n$?
Is there any trick to calculate the sum? From what I already have, it definitely converges.

There is one more way to calculate this expression. Consider the following recurrence: $a_0 = 0, a_1 = 1, a_2 = 1$ and $$a_{2n} = a_n, \  a_{2n+1} = a_n + 1, \ n \ge 1.$$ 
It is easy to see that $a_n$ counts the number of ones in the binary expansion of $n$. We wish to calculate 
\begin{align*}
\sum_{n = 1}^{\infty} \frac{a_n}{n(n+1)} &= \sum_{n=1}^{\infty} \frac{a_{2n}}{2n(2n+1)} + \sum_{n = 0}^{\infty} \frac{a_{2n+1}}{(2n+1)(2n+2)} \\
 &= \sum_{n=1}^{\infty} \frac{a_{n}}{2n(2n+1)} + \sum_{n = 0}^{\infty} \frac{a_{n} + 1}{(2n+1)(2n+2)} \\
&= \frac{1}2 + \frac{1}2 \sum_{n = 1}^{\infty} \frac{a_n}{n(2n+1)} + \frac{1}2 \sum_{n=1}^{\infty} \frac{a_n}{(n+1)(2n+1)} + \frac{1}2 \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} \\
&= \frac{1}2 + \frac{1}2 \sum_{n=1}^{\infty} a_n \left( \frac{1}{n(2n+1)} + \frac{1}{(n+1)(2n+1)} \right) + \frac{1}2 \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} \\
&= \frac{1}2+ \frac{1}2 \sum_{n=1}^{\infty} \frac{a_n}{n(n+1)} + \frac{1}2\sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)}.
\end{align*}
Thus, 
$$\sum_{n=1}^{\infty} \frac{a_n}{n(n+1)} = 1 + \sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)}.$$
Now 
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+1)} &= \sum_{n=1}^{\infty}  \int_0^1 \frac{x^{2n}}{n+1} \ dx \\
&= \int_0^1 \sum_{n=1}^{\infty} \frac{x^{2n}}{n+1} \ dx  \\
&= \int_0^1 -\frac{ \log(1-x^2)}{x^2} \ dx -1\\
&= \log(4) - 1. 
\end{align*}
Therefore, our desired value is 
$$1 + \log(4) - 1 = 2 \log(2).$$
