Sum of $\sum \frac{a(n)}{n(n+1)}$, where $a(n)$ is binary weight. Consider a series : $A = \sum_{n \ge 1} \frac{a(n)}{n(n+1)}$, where $a(n)$ is binary weight of number. 
It's easy to see that $A = 2\log{2}$ with using $a(2n+1) = a(2n) + 1$ and $a(2n) = a(n)$. 
But does there some probabilistic proof ? It's looks like some expectation, but I don't know how to find a random variables with such expectation.
 A: Not sure this is what you're looking for...  While my "solution" is written using random variables, all the heavy lifting is actually in summations, which is perhaps how you found the "easy to see" answer in the first place.  :)
The random variable you seek is $a(X)$, and $A = E[a(X)]$, where $X$ is a random positive integer picked according to $P(X=n) = \frac{1}{n(n+1)}$.  Note that this is a valid probability distribution since $\sum_{n\ge 1}\frac{1}{n(n+1)} = \sum_{n \ge 1} (\frac{1}{n} - \frac{1}{n+1}) = 1$, a "telescoping" feature of this sequence.
Note that  $a(n) = \sum_{j \ge 1} bit_j(n)$ where $bit_j(n)$ is the $j$th bit of the binary expression of $n$, counting $j=1$ as the least significant bit, etc., so we just use linearity of expectations and $A = E[a(X)] = \sum_{j \ge 1} E[bit_j(X)]$.  
Now $E[bit_1(X)] = \sum_{odd\ n \ge 1} (\frac{1}{n} - \frac{1}{n+1}) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... = \log 2$, a known result.
In general the $j$th bit alternates between 0 and 1 for "blocks" of consecutive integers.  E.g. $bit_2(n) = 1$ for $n \in \{2, 3, 6, 7, 10, 11, ...\}$.  Because of the "telescoping" nature, only the "boundary" values appear in the sum:  
$$E[bit_2(X)] = \sum_{n = 2,3,6,7,10, 11, ...} (\frac{1}{n} - \frac{1}{n+1}) = \frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + ...$$
I.e. you add $\frac{1}{n}$ for each $n$ which starts a 1-block, and subtract $\frac{1}{n}$ for each $n$ which starts a 0-block.  Then note that this summation is term by term exactly $\frac{1}{2}$ of the summation in $E[bit_1(X)]$.
Allow me to hand-wave a bit and declare $E[bit_j(X)] = E[bit_1(X)] / 2^{j-1}$.  Then $A = E[a(X)] = (\log 2) (1 + \frac{1}{2} + \frac{1}{4} + ...) = 2 \log 2$.
Is this what you're looking for?  Or is this how you solved it in the first place (and found it "easy to see") and all I did was dressing things up in the trappings of linearity of expectations?
