# The form of an infinite product that converges to 1/2 or 1/3

I have a hypothesis from some code that in the limit, the products of some variables become $$\displaystyle \lim_{n\to\infty}\prod_{i=1}^n a_i=\frac{1}{2}$$ $$\displaystyle \lim_{n\to\infty}\prod_{i=1}^n b_i=\frac{1}{3}$$

I find that the terms $$a_i$$ and $$b_i$$ all tend to 1 as $$n$$ becomes large, and the products tend towards the values above.

I am interest to know if there are any general forms of functions $$a_i(i)$$ or $$b_i(i)$$ that satisfy the properties above i.e can we work backwards from the convergent solution to some examples of functions which satisfy these convergence properties.

I think that it is easier to discuss sums than products: take your favorite converging series $$\sum\limits_{i=1}^\infty x_i =X$$ where $$X\neq 0$$, divide by X, multiply by $$\ln(0.5)$$ and voila, $$a_i=\exp(x_i\ln(0.5)/X)$$ is a possible choice: $$\lim\limits_{n\to\infty}\prod\limits_{i=1}^n a_i = \lim\limits_{n\to\infty}\prod\limits_{i=1}^n \exp(x_i\ln(0.5)/X)= \lim\limits_{n\to\infty}\exp\left(\sum\limits_{i=1}^n x_i\ln(0.5)/X\right)= \exp\left(\lim\limits_{n\to\infty}\sum\limits_{i=1}^n x_i\ln(0.5)/X\right)= \exp\left(X\ln(0.5)/X\right)=\tfrac{1}{2}$$
• Thanks for this. I also have the condition that $a_{i+1}>a_i$, so I'll shop around for series that satisfy this... hopefully there's something simple (say a rational function) that works. – Tom Waits Jun 12 '20 at 19:34
• Take $x_i=-1/2^i$ ($X=-1$, starting from $i=1$). Then $\ln(0.5)/2^i$ is increasing and so does $a_i$. – YJT Jun 12 '20 at 19:48