Finding value of $\displaystyle \frac{e^n\cdot n!}{n^n\sqrt{n}}$
Try: Using Stirling Approximation result $$n!\approx\bigg(\frac{n}{e}\bigg)^n\sqrt{2\pi n}$$
So $$\lim_{n\rightarrow \infty }\frac{e^n\cdot n!}{n^n\sqrt{n}}=\sqrt{2\pi}$$
Could anyone explain me How can i prove that result, Thanks
First of all, the expression $$ n! \simeq \left( \frac{n}{e} \right)^n \sqrt{2\pi n} $$ is valid for $n$ large. I know that you are probably aware of that, because I imagine that you are taking the limit when $n \rightarrow \infty$, but you should not use an equality on that expression anyway.
With this in mind, you shall do this in two different ways:
Not so rigorous way: plug the expression for $n!$ given by the approximation directly in $$ \lim_{n \rightarrow \infty} \frac{e^n n!}{n^n \sqrt{n}} $$ and simplify.
Rigorous way: use the definition of limit at infinity. Take $\varepsilon > 0$, and try to find $n_0 \in \mathbb{N}$ such that $$ \left| \frac{e^n n!}{n^n \sqrt{n}} - \sqrt{2\pi}\right| < \varepsilon $$ if $n \geq n_0$.
Hint: Notice that $n! \simeq \left( \frac{n}{e} \right)^n \sqrt{2\pi n}$ really means that for $\varepsilon > 0$ exists $n_1 \in \mathbb{N}$ so that $$ \left| \frac{n!}{\left(\frac{n}{e}\right)^n \sqrt{2\pi n}} - 1\right| < \varepsilon $$ if $n \geq n_1$.
Edit: proving Stirling's approximation is not direct. You shall find a proof of it Rudin, W: Principles of Mathematical Analysis. The proof is based on using Euler's Gamma function, for $\Gamma(n+1) = n!$.