Find $\sum_{n=1}^\infty {\frac {f(n)} {n(n+1)}}$ where $f(n)$ is the number of $1$s in $n$'s binary expansion

Question

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?

Answer 1

Let $S$ be the sum (which exists as already noted in the comments). Use that $f(2n)=f(n)$ and $f(2n+1)=f(n)+1$. Then splitting the sum in $n$ even and $n$ odd and grouping terms results in the recursion $S= \log(2)+\tfrac{1}{2}S$ so $S = 2\log(2)=\log(4)$.

$$ \begin{eqnarray} S&=&\sum_{n=1}^\infty \frac{f(n)}{n(n+1)}\\ &=&\sum_{n=0}^\infty \frac{1+f(n)}{(2n+1)(2n+2)} + \sum_{n=1}^\infty \frac{f(n)}{2n(2n+1)}\\ &=&\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)+\sum_{n=1}^\infty f(n)\left(\frac{1}{(2n+1)(2n+2)}+\frac{1}{2n(2n+1)}\right)\\ &=&\log(2) +\sum_{n=1}^\infty\frac{f(n)}{2n(n+1)}\\ &=&\log(2)+\tfrac{1}{2}S \end{eqnarray} $$