How to use Stirling's approximation in this limit?

Question

I want to compute $$L=\lim_{n\to\infty}{\prod_{k=1}^{n}{\left(\frac{k}{n}\right)^{1/n}}}.$$ Let $P_n$ denote the $n$th partial product. Then, $$P_n=\left(\frac{n!}{n^n}\right)^{1/n}.$$ Now we can apply Stirling's approximation, $n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ to obtain $$L=\frac{\sqrt{2\pi}}{e}\lim_{n\to\infty}{n^{\frac{1}{2n}}}=\frac{\sqrt{2\pi}}{e}.$$ However, the correct answer is apparently $L=e^{-1}$. Why am I off by a factor of $\sqrt{2\pi}$?

Answer 1

Note that the $\sqrt{2\pi}$ in Stirling's approximation is just some constant, unlike the $e$ raised to the $n^{\text{th}}$ power. So, when you take $(2\pi)^{1/n},$ and let $n\rightarrow\infty,$ this $2\pi$ constant will go to 1 and not end up contributing to the limit.

Answer 2

You would see the $2\pi$ appearing in the asymptotics but not in the limit $$P_n=\left(\frac{n!}{n^n}\right)^{1/n}=\frac{1}{e}+\frac{\log (2 \pi n)}{2 e n}+O\left(\frac{1}{n^2}\right)$$