How to find the limit of $ {n! e^n}/{n^{n+1/2}} $?

Question

What is the value of this limit and how to find it? $$ \lim_{n \to \infty} \frac{n! e^n}{n^{n+\frac{1}{2}}} $$ Can we use L'Hospital rule here? I tried but failed that how to do it.

Answer 1

As said by 900 sit-ups a day, Stirling approximation is a good approach for this kind of problem. Since Stirling gives $$n!\simeq n^n \sqrt{2\pi n}e^{-n}$$ then $$\frac{n! e^n}{n^{n+\frac{1}{2}}}\simeq \sqrt{2\pi}$$ A more accurate representation can be obtained using a Taylor expansion of the expression for large values of $n$. As a result,$$\frac{n! e^n}{n^{n+\frac{1}{2}}}=\sqrt{2 \pi }+\frac{\sqrt{\frac{\pi }{2}}}{6 n}+\frac{\sqrt{\frac{\pi }{2}}}{144 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$

By the way, if you use as an improved approximation by Stirling $$n!\simeq n^n \sqrt{2\pi n}e^{-n}e^{\frac{1}{12n}}$$ you find that $$\frac{n! e^n}{n^{n+\frac{1}{2}}}\simeq \sqrt{2 \pi } e^{\frac{1}{12 n}}$$ If instead you use Gosper's approximation of the factorial $$n!\simeq \sqrt{\pi(2n+\frac{1}{3})}n^n e^{-n}$$ you should find $$\frac{n! e^n}{n^{n+\frac{1}{2}}}\simeq \sqrt{2\pi}\sqrt{1+\frac{1}{6n}}$$

Answer 2

Using Stirling's approximation $n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ and substituting, the result is immediate after some cancellations.