We are given the series $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$, where $f(n)$ is such a function that it equals the sum of 1's in the binary representation of n.
I'm obliged to find the sum of the series. First, I decided to explicitly prove the convergence of the series. Use for that converging harmonic series,
$\sum {\frac 1 {n^2}}$.
$\lim_{n \to \infty} \frac {f(n)* n^2} {n \cdot (n+1)} = \lim_{n \to \infty} f(n)$
Where $f(n)$ could be represented as $\sum_{i=0}^{\log_2 n} m_i$, if we define n as $\sum_{i=0}^{\log_2 n} m_i \cdot 2^i$ and $m_i$ is able to take values of 0 or 1.
I can get it intuitively that $f(n)$ doesn't yield $\infty$, when $n \to \infty$. A graph of the function would behave like wave, growing and then returning to 1. I wrote a script to get some sense of the data, which outputted:
For p= 10 series= 1.0873015873015872
For p= 100 series= 1.3398423025935298
For p= 1000 series= 1.3801052489558094
For p= 100000 series= 1.3861981879674006
This result looks for me like $ln(4)$, but it doesn't help much, since the solution is supposed to be purely analytical. How can I find a sum?