limits - How to find $lim_{n to infty} frac {2^n n!}{n^n}$

$$\lim_{n \to \infty} \frac {2^n n!}{n^n}$$

How can I find this limit? I have tried to use the ratio test and then I have got the the following limit: $$\lim_{n \to \infty} 2 \frac{ n^{n+1}}{(n+1)^{n+1}}$$ I have tried for a while but I can't figure out this limit either. How can I find this limit?

Answer

You should clarify: do you want to

1) find $\lim \limits_{n\to\infty} \frac{2^nn!}{n^n}$ or

2) test the convergence of the series $\sum_{n=1}^{\infty} \frac{2^nn!}{n^n}$ (for which the ratio test is used)?

I assume the question stated is the first, which is calculated by the Stirling's formula:

$\lim \limits_{n\to\infty} \frac{2^nn!}{n^n} = \lim \limits_{n\to\infty} \frac{2^n\cdot\sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n}{n^n}=\lim \limits_{n\to\infty} \frac{\sqrt{2\pi}\cdot \sqrt{n}}{\left(\frac{e}{2}\right)^n} \stackrel{l'H}=\lim \limits_{n\to\infty} \frac{\sqrt{2\pi}}{2\left(\frac{e}{2}\right)^n\cdot \ln{\left(\frac{e}{2}\right)}\cdot\sqrt{n}}=0.$

Note: Another direction: $\frac{2^nn!}{n^n}=\frac{2\cdot1}{n}\cdot\frac{2\cdot2}{n}\cdot\cdot\cdot\frac{2\cdot n}{n}$