Finding $\rho$ for which $(\frac k\rho)^k<\prod\limits_{i=1}^kp_i<((\frac k\rho)+k\varepsilon)^k$ It is known and provable (for instance, it is mentioned here), that $$\left(\frac{n}{e}\right)^{n} < n! < \left(\left(\frac{n}{e}\right) + n\varepsilon)^{n}\right)$$ for an arbitrarily small $\varepsilon$ and all $n\geq N$, where $N$ depends on $\varepsilon$.
As a first question, it would be great some reference or proof showing that $$\lim_{n\to\infty}\frac{(\frac{n}{e})^{n}}{n!}=1$$
As a second question, I would like to know if there is known some $\rho$ such that
$$\left(\frac{k}{\rho}\right)^k<\prod_1^{k}{p_{k}}<\left(\left(\frac{k}{\rho}\right)+k\varepsilon \right)^k$$ where $\prod_1^{k}{p_{k}}$ is the product of the first $k$ prime numbers, for an arbitrarily small $\varepsilon$ and all $k\geq N$, where $N$ depends on $\varepsilon$.
I have found no reference to this second question, so any reference would be welcomed.
Thanks!
 A: I will only comment on the first question regarding the limit.
Your limit is incorrect. We have
$$
\lim_{n\to\infty} \frac{ (\frac{n}{e})^n}{n!} = 0.
$$
This property follows from
$$
\lim_{n\to\infty} \frac{ (\frac{n}{e})^n\sqrt{2\pi n}}{n!} = 1
$$
which is given by Stirling's approximation.
A: First, by Stirling's formula
$$
n! = \left( {\frac{n}{\mathrm{e}}} \right)^n \sqrt {2\pi n} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right).
$$
This shows that
$$
\frac{{\left( {\frac{n}{\mathrm{e}}} \right)^n }}{{n!}} = \frac{1}{{\sqrt {2\pi n} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) \to 0 \ne 1
$$
as $n\to +\infty$. It follows that $$
\frac{{\sqrt[n]{{n!}}}}{n} = \frac{1}{\mathrm{e}}\sqrt[{2n}]{{2\pi n}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right)^{1/n}  = \frac{1}{\mathrm{e}}\left( {1 + \mathcal{O}\!\left( {\frac{{\log n}}{n}} \right)} \right) \to \frac{1}{\mathrm{e}}
$$ as $n\to +\infty$, but this does not imply $\frac{n!}{n^n}\sim \mathrm{e}^{-n}$ at all (it cannot be true by Stirling's result).
Regarding the primes,
$$
\log (p_1 p_k  \ldots p_k ) = \vartheta (p_k )
$$
where $\vartheta$ is the first Chebyshev function. From this paper, we have
$$
\vartheta (p_k ) = p_k (1 + \mathcal{O}(\exp ( - c\sqrt {\log p_k } ))
$$
with a suitable $c>0$. Cipolla (La determinazione asintotica dell' $n$-esimo numero primo, Napoli. Rend., 8 ($1902$), pp. $132$–$166$) proved more than a century ago that
$$
\log p_k  = \log k + \log \log k + \frac{{\log \log k - 1}}{{\log k}} + \mathcal{O}\!\left( {\left( {\frac{{\log \log k}}{{\log k}}} \right)^2 } \right),
$$
$$
\frac{{p_k }}{k} = \log k + \log \log k - 1 + \frac{{\log \log k - 2}}{{\log k}} + \mathcal{O}\!\left( {\left( {\frac{{\log \log k}}{{\log k}}} \right)^2 } \right).
$$
Consequently,
$$
\frac{{p_k }}{k} = \log p_k  - 1 + \mathcal{O}\!\left( {\frac{1}{{\log k}}} \right) = \log \left( {\frac{{p_k }}{{\rm e}}} \right) + \mathcal{O}\!\left( {\frac{1}{{\log k}}} \right) = \log \left( {\frac{{p_k }}{{\rm e}}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log ^2 k}}} \right)} \right).
$$
Thus,
\begin{align*}
\frac{1}{k}\log (p_1 p_k  \ldots p_k ) &= \log \left( {\frac{{p_k }}{{\rm e}}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log ^2 k}}} \right)} \right)(1 +\mathcal{O}(\exp ( - c\sqrt {\log p_k } )) \\ &= \log \left( {\frac{{p_k }}{{\rm e}}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log ^2 k}}} \right)} \right)=\log \left( {\frac{{p_k }}{{\rm e}}} \right)+\mathcal{O}\!\left( {\frac{1}{{\log k}}} \right),
\end{align*}
i.e.,
$$
\sqrt[k]{{p_1 p_k  \ldots p_k }} = \frac{{p_k }}{{\rm e}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\log k}}} \right)} \right)
$$
where the error term cannot be improved. Since $p_k \sim k\log k$, there is no such $\rho$ you are looking for.
