TITLE: 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 [14 upvotes]: 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?

REPLY [17 votes]: 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 $S = 2\log(2)=\log(4)$.
$$
\begin{eqnarray}
S&=&\sum_{n=1}^\infty \frac{f(n)}{n(n+1)}\\
&=&\sum_{n=0}^\infty \frac{1+f(n)}{(2n+1)(2n+2)} + \sum_{n=1}^\infty \frac{f(n)}{2n(2n+1)}\\
&=&\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)+\sum_{n=1}^\infty f(n)\left(\frac{1}{(2n+1)(2n+2)}+\frac{1}{2n(2n+1)}\right)\\
&=&\log(2) +\sum_{n=1}^\infty\frac{f(n)}{2n(n+1)}\\
&=&\log(2)+\tfrac{1}{2}S
\end{eqnarray}
$$