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

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 $S = 2\log(2)=\log(4)$.

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