How to prove $\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$ I want to prove the following and wanted to ask, if my proof is correct.

$$\lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$$

Remark (i): For $n \in \mathbb{N}$ and $n > 1$ holds: $2 < (1+\frac{1}{n})^n$
Define $b_n := \left\{\frac{2^n n!}{n^{n}}\right\}_{n \in \mathbb{N}}$.
Then $\left| \frac{b_{n+1}}{b_n} \right| = \left| \frac{\frac{2^{n+1} (n+1)!}{(n+1)^{n+1}}}{\frac{2^n n!}{n^{n}}} \right| = 2 \left(\frac{n}{n+1}\right)^n = 2 \left(1-\frac{1}{n+1}\right)^n = \frac{2 \left(1-\frac{1}{n+1}\right)^n \cdot (1+\frac{1}{n})^n}{(1+\frac{1}{n})^n} = \frac{2}{(1+\frac{1}{n})^n} \overbrace{<}^\text{(i)} 1$ holds for $n \in \mathbb{N}$ and $n > 1$
Hence: $\lim \limits_{n \rightarrow \infty} \left|\frac{b_{n+1}}{b_n}\right| < 1$, which yields: $\lim \limits_{n \rightarrow \infty} b_n = 0$
Define $a_n := \left\{\frac{2^n n!}{n^{n+1}}\right\}_{n \in \mathbb{N}}$.
Obviously $0 \leq a_n \leq b_n$ holds for all $n \in \mathbb{N}$.
Hence with the sandwich-theorem: $\lim \limits_{n \rightarrow \infty} 0 = \lim \limits_{n \rightarrow \infty} b_n = \lim \limits_{n \rightarrow \infty} a_n = \lim \limits_{n \rightarrow \infty} \frac{2^n n!}{n^{n+1}} = 0$
 A: At $n\to\infty$, $n!$ can be written as $\sqrt{2\pi n}\left(\dfrac ne\right)^n$. Now limit can be written as
$$\lim_{n\to\infty}{\dfrac{2^n\sqrt{2\pi n}\left(\dfrac ne\right)^n}{n^{n+1}}}$$
Now we need to simplify it.
$$e^{\lim_{n\to\infty}{\left((\ln2-1)n-\dfrac12\ln n+\dfrac12\ln2\pi\right)}}$$
So, we need to prove that $\lim_{n\to\infty}{\left((\ln2-1)n-\dfrac12\ln n\right)=-\infty}$
Function $(\ln2-1)n$ at $n\to\infty$ is $-\infty$ and $-\dfrac12\ln n$ at $n\to\infty$ is $-\infty$, so sum of this functions must be $-\infty$.
A: If $b_n\rightarrow 0$ then naturally $b_n/n\rightarrow 0$ by distributing the limits. You might want to also mention why $b_{n+1}/b_n$ has a limit in the first place (even though it seems implicit in your assertion that the ratio is <1). Basically, the proof is fine.
A: It is correct, but I find this a bit faster:
$$\frac{a_{n+1}}{a_n}=\frac{2n}{n+1}\left(\frac n{n+1}\right)^{n+1}\to\frac 2e<1$$
Nevertheless, yours is more elementary since it does not use the limit $\lim\left(1+\frac1n\right)^n=e$
A: I present an elementary solution.
Write $$\ell = \lim_{n \to \infty} \frac {2^n n!} {n^{n + 1}} = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n} \frac {2k} {n}.$$ For $k < n/2$, write $2k/n = 1 - a_k$, where $0 < a_k < 1$. Then, obviously, for $k > n/2$ we'll have $2k/n = 1 + a_k$. Hence, $$\ell = \lim_{n \to \infty} \frac {1} {n} \prod_{k = 1}^{n/2} (1 - a_k^2) = \lim_{n \to \infty} \frac {1} {n} \cdot \lim_{n \to \infty} \prod_{k = 1}^{n/2} (1 - a_k^2) = 0 \cdot 0 = 0.$$
