It's easy to see that : $a_1x=U_1+x_1 , a_2x_1 = U_2+x_2, ... , a_{n+1}x_n=U_{n+1}+x_{n+1},...$, where $0\leq x_n < 1$ and $U_n$ is an integer such that ${0 \leq U_n\leq a_n-1}$ (1) All $x_n$ are irrational because $x$ is irrational , so $x_n>\frac{1}{a_{n+1}}$ or $0<x_n<\frac{1}{a_{n+1}}$ $\exists m\forall n \geq m , \ x_n<\frac{1}{a_{n+1}}\Longrightarrow \forall n \geq m,$$\ U_n =0 \Longrightarrow \forall n \geq m ,\ x_m =\frac{x_{n+1} }{a_{m+1}...a_{n+1}}\leq \frac 1{2^{n-m+1}} \Longrightarrow x_m=0$, manifestly false !! So $x_n>\frac{1}{a_{n+1}}$ for infinitely many $n$ (2) Let $x\in(0,1)$ be such that that $\forall n,\ x_n>\frac{1}{a_{n+1}}$, so $\forall n>1,\ 1 \leq U_n \leq a_n-1$ $x = \frac{U_1}{a_1}+\frac{x_1}{a_1} = \frac{U_1}{a_1}+\frac{U_2}{a_1a_2}+\frac{x_2}{a_1a_2}=...=\sum_{n=1}^{\infty}\frac{U_n}{a_1..a_n}$ So the asked sequences have a infinite sub-sequence $a_{p(n)}$ such that ${a_{p(n)} \geq 3}$

yes! this is correct!

Diogene wrote: So the asked sequences have a infinite sub-sequence $a_{p(n)}$ such that ${a_{p(n)}\geq 3}$ What does this mean?

Yeah, what does it mean? Can someone tell?

Somebody please help me