A Conjecture : can anyone find a counter-example or otherwise disprove ?

[uart]

Hi, I have a conjecture and I am not sure whether it is true. I can't construct a counter example but perhaps someone more mathemetically resourceful than myself can do so (or perhaps even offer a direct proof or disproof).

Here's the conjecture.

Let X_n = r_1 \, r_2 \, r_3 \, ... \, r_n, be a product of n rational fractions (r_i) , such that, for each n in [1,2,3 …] the numerator of X_n has at least one prime factor (uncancelled of course) greater than n.

Conjecture : If the limit as n goes to infinity of X_n is finite then it (the limit) is irrational.


If you cant find a counter-example (or direct proof or disproof) then what does your mathematical "intuition" think about it, do you think it's probably true or probably false.

Thankyou. :)

[Hurkyl]

Start with \inline{r_1=2/1}.

For any \inline{n\geq 1}:
if \inline{X_n>1}, define \inline{r_{n+1}=p_{n+1}/(p_{n+1}+e_{n+1}} where \inline{p_n+1} is the \inline{n+1}-th prime number and \inline{e_{n+1}} is the least integer that doesn't have \inline{p_{n+1}} as a factor and makes \inline{X_{n+1}<1}.
if \inline{X_n<1}, define \inline{r_{n+1}=p_{n+1}/(p_{n+1}-e_{n+1}} where \inline{e_{n+1}} is the greatest integer less than \inline{p_{n+1}} and makes \inline{X_{n+1}>1}.

Then, \inline{\lim_{n \rightarrow \infty}X_n = 1}


This isn't rigorous, but I'm fairly convinced one can make a real proof out of this.

[uart]

Hey, I think you've convinced me. :)

The only thing that I would change is to stipulate that each of the e_{n+1} must be an odd number, so as to make each (p_{n+1} \pm e_{n+1}) in the denominator an even number. That way you can guarantee that none of the denominator terms can contain the same prime as the next numerator term in the product. This modification will make sure the "not cancelled" clause insn't violated and I dont think it will change the fact that the limit is 1.

Now I can finally put that attempted proof of Pi irrational based on an infinite product to rest.

Thanks. :)