Let $B(n)$ be the number of ones in the base 2 expression for the positive integer n.
Determine whether or not $$\exp\left(\sum_{n=1}^{\infty}\frac{B(n)}{n(n+1)}\right)$$ is a rational number.
Attempt:
I tried to make the sum into something that resembles the power series of log, that way it would be easier to determine whether this number is rational. But I have no idea how to deal with $B(n)$.
Thanks in advance!
For $B(n)$ we have the following properties: \begin{align} & B(2k) = B(k) & \text{if }n = 2k+1 \\ & B(2k + 1) = B(k) + 1 & \text{if }n = 2k \end{align}
Hence,
$$S = \sum\limits_{n=1}^{+\infty} \dfrac{B(n)}{n(n+1)} = \sum\limits_{k=0}^{+\infty} \dfrac{B(2k+1)}{(2k+1)(2k+2)} + \sum\limits_{k=0}^{+\infty} \dfrac{B(2k + 2)}{(2k + 2)(2k+3)} = \sum\limits_{k=0}^{+\infty} \dfrac{B(k) + 1}{(2k+1)(2k+2)} + \sum\limits_{k=0}^{+\infty} \dfrac{B(k + 1)}{(2k + 2)(2k+3)} = \sum\limits_{k=0}^{+\infty} \dfrac{1}{(2k+1)(2k+2)} + \sum\limits_{k=0}^{+\infty} B(k + 1)\left(\dfrac{1}{(2k + 2)(2k+3)} + \dfrac{1}{(2k + 3)(2k+4)}\right) = \ln{2} + \sum\limits_{k=1}^{+\infty} B(k)\left(\dfrac{4k + 2}{2k(2k+1)(2k+2)}\right) = \ln{2} + \dfrac{1}{2}\sum\limits_{n=1}^{+\infty} \dfrac{B(k)}{k(k+1)} = \ln{2} + \dfrac{1}{2}S \Rightarrow S = 2\ln{2} = \ln{4}$$
Thus, $\exp\left\{\sum\limits_{n=1}^{+\infty} \dfrac{B(n)}{n(n+1)} \right\} = 4$, which is a rational number.
Note that manipulations with series above are legal, because $B(n) \le 1 + [\log_2(n)]$, so $\dfrac{B(n)}{n(n+1)} \le \dfrac{1}{n^{3/2}}$ for large enough $n$, which means that the series in question converges.
