im working on a problem where im confronted with this fraction $\frac{3 \cdot 5\cdot9\cdot15\cdot17\cdot23\cdot27\cdot29 …}{1\cdot7\cdot11\cdot13\cdot19\cdot21\cdot25\cdot31 …}$ .
I found that the product in the numerator is the product over all odd evil numbers and the product in the denominator is the product over all odd odious numbers. Im especially interested in the sequence $\frac{3}{1}=3$ , $\frac{3\cdot5}{1\cdot7}=\frac{15}{7}$ , $\frac{3\cdot5\cdot9\cdot15}{1\cdot7\cdot11\cdot13}=\frac{2025}{1001}$ , … where $2^n$ factors are put together. Its easy to see that this series converges monotonically decreasing to 2 but i have no idea how to proof that. Can somebody help me ?
Note: an odious number is a nonnegative integer which has an odd number of $1$s in its binary expansion. An evil number is one which has an even number of $1$s.
Edit:
Because some people are a bit suspicious about odious and evil numbers i want to give an alternative formulation of my problem. Consider the following recursive function: $f(n)=(-1)\cdot f(l)$ where $n=2^k+l$ and $l>0$ is the unique representation of n as the maximal power of 2 with a remainder $0\le l \lt 2^k$ and $f(1)=(-1)^k$ where $n=2^k$ which is defined on non negative odd integers. Im looking for the value of the product $\lim_{N->\infty}\prod_{n=1}^{2^N}{(2\cdot n-1)^{-f(n)}}$ .
This is the way i came to this fraction given above.
Edit2: I think i found another representation without a proof that these are equal
$\frac{3 \cdot 5\cdot9\cdot15\cdot17\cdot23\cdot27\cdot29 …}{1\cdot7\cdot11\cdot13\cdot19\cdot21\cdot25\cdot31 …}=(1+\frac{2}{1})\cdot(1-\frac{2}{7})\cdot(1-\frac{2}{11})\cdot(1+\frac{2}{13})\cdot(1-\frac{2}{19})\cdot(1+\frac{2}{21})\cdot(1+\frac{2}{25})\cdot(1-\frac{2}{31})…$
It seems to me that the rule of building this product is $\prod_{n=1}^{\infty}{(1+\frac{(-1)^{m_n}\cdot2}{o_n})}$ where $m_n$ is the n-th element of the Thue–Morse sequence and $o_n$ is the n-th element of the odd odious number sequence. Maybe this looks familiar to somebody
Best Answer
This follows easily from corollary 2.4(b) of "More Infinite Products: Thue-Morse and the Gamma function" (arXiv:1709.03398) by Jean-Paul Allouche, Samin Riasat, and Jeffrey Shallit. The statement there is
$$ \prod_{n \ge 0} \left( {4n+1 \over 4n+3} \right)^{(-1)^{t_n}} = {1 \over 2} $$
where $t_n$ is the Thue-Morse sequence, i. e. $t_n$ is the sum of the bits of the binary expansion of $n$. Your desired result follows by taking the reciprocal of both sides to get
$$ \prod_{n \ge 0} \left( {4n+3 \over 4n+1} \right)^{(-1)^{t_n}} = 2$$
and some rewriting on the left-hand side to get a product over odd numbers.