Let $A_n$ be a sequence of monotonically increasing, positive integers. How can one construct another sequence $a_n$, again positive integers, such that $$\lim_{n \to \infty} \frac{A_n}{\displaystyle \prod_{k=1}^n a_k} = 1$$
Is it even always possible to construct such a sequence?
The obvious thing is to choose $a_{n+1}$ to be $\frac{A_{n+1}}{A_n}$, but its not always guaranteed that this will be an integer. I tried to use $a_{n+1} = \lfloor \frac{A_{n+1}}{A_n} \rfloor$, but I cant seem to work it out. For sequences like $A_n = k^n$ it is very easy to construct $a_n$ if $k$ is an integer itself. What about $A_n =\lfloor \alpha^n \rfloor $ if $\alpha$ is any number greater than 1?
It is not always possible.
Let $A_n=n$ and $p(n)=a_1.a_2....a_n$. Then suppose that, for $n\ge N$, $$0.9<\frac{n}{p(n)}< 1.1$$
Then $\frac{2n}{p(2n)}= \frac{2}{A}\frac{n}{p(n)}$ for some integer $A$. For this to be in the assumed range we require $A=2$ but then $$\frac{n}{p(n)},\frac{2n}{p(2n)},\frac{4n}{p(4n)},...$$ are all constant and so do not tend to $1$.
The only remaining possibility is that, for all sufficiently large $n$, $\frac{n}{p(n)}=1$. But then $p(n)$ does not divide $p(n+1)$.