I was trying to solve some problem from a question here on MSE by first trying to find something about simplified version, and, if I calculated correctly I obtained:
$$\lim_{n \to + \infty} (\sqrt{a})^{3^n} \cdot \prod_{k=2}^{n} (\dfrac {a}{k})^{3^{n-k+1}}=1$$
Even if my calculations are not right this is interesting in itself and what I would like to know is does such an $a$ exists? That is, even though I obtained this result and am trying to find closed form for $a$ (or an approximation) I really do not know even if there exists an $a$ that would satisfy this limit problem.
Does it exists? How to calculate it if it exists?
For any $a \gt 1$ the limit is $0$, not $1$.
Lets start with the numerator of the product. $$\prod_{k=2}^na^{3^{n-k+1}}=a^{\sum_{k=2}^n3^{n-k+1}}\\ \sum_{k=2}^n3^{n-k+1}=\frac{3^n-3^1}{3-1}\approx \frac {3^n}2 \\ \text {and all the } a\text{s including the leading factor give us }a^{3^n} $$ We will loosely bound the denominator from below, remembering that $n \gg a$ $$\prod_{k=2}^nk^{3^{n-k+1}}\gt \prod_{k=a^3}^nk^{3^{n-k+1}}\gt \prod_{k=a^3}^n(a^3)^{3^{n-k+1}}=(a^3)^{\frac{3^n-3^{a^3}}{3-1}}=a^{\frac 323^n}\cdot a^{-3^{a^3}}$$ The trailing $a^{-3^{a^3}}$ is just a constant. For any $a \gt 1$ the terms depending on $n$ will be less than $a^{\frac 123^n}$, which can be made as small as we want.