Sum of series with binary parity in the numerator I'm now stuck with this question, and I don't even know where to start:
Find sum of series$$\sum_1^\infty \frac{f(n)}{n(n+1)}$$, where f(n) - number of ones in binary representation of n.
I wish I could post some moves, that I've tried but I don't know what to do.
Thanks!
 A: Code:
double sum=0;
for(int i=1;i<9999999;i++){
    double s2=Integer.bitCount(i);
    sum+=(s2)/(i*(i+1));
}

Output Data
$$\begin{array}{r|l}
n&\sum\\\hline
9&1.065079365079365\\
99&1.3394382621894894\\
999&1.3800972409478014\\
9999&1.3854974129587205\\
99999&1.3852676077956714\\&(\text{limit of data type, therefore decreased value})
\end{array}$$
A: Maybe there is a quicker way, but here is one. Let the sum be $S$. We have
$$\frac{f(n)}{n(n+1)} = \frac{f(n)}{n}-\frac{f(n)}{n+1}$$
$$\implies \frac{f(2n)}{(2n)(2n+1)}+\frac{f(2n+1)}{(2n+1)(2n+2)} = \frac{f(2n)}{2n}-\frac{f(2n)-f(2n+1)}{2n+1}-\frac{f(2n+1)}{2n+2}$$
Now $f(2n+1) = f(2n)+1, \; f(2n) = f(n)$, so we can write:
$$\frac{f(2n)}{(2n)(2n+1)}+\frac{f(2n+1)}{(2n+1)(2n+2)} = \frac{f(2n)}{2n}+\frac1{2n+1}-\frac{f(2n)+1}{2n+2} \\ = \frac12\left(\frac{f(n)}n -\frac{f(n)}{n+1}\right)+\left(\frac1{2n+1}-\frac1{2n+2}\right)$$
$$\implies S = \frac12+ \frac12\sum_{n=1}^\infty \frac{f(n)}{n(n+1)}+\sum_{n=1}^\infty \left(\frac1{2n+1}-\frac1{2n+2}\right) $$
$$\implies 2S = 1 + S + 2\log 2 -1 \implies S = 2\log 2 \approx 1.386$$
