How do I solve this? $\lim_{n\to\infty} \frac{2^n}{n!}$?

Question

Good day to everyone. So, basically, it's no big deal that factorial grows faster than $2^n$, that's quite obvious, but I need to prove it, and that's where the problem begins.

I've tried some standard ways — no result. Stirling's approximation — no result. I tend to think that I either don't know something minor yet helpful or do something wrong. Help me, please:з

Answer 1

Put $p_n=2^n/n!$ and notice that $$ \frac{p_{n+1}}{p_n}=\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \frac{2}{n+1} \to 0. $$ Hence the series $\sum_n p_n$ converges, and therefore $\lim_n p_n =0$.

Answer 2

$n!$ is, a product of $n$ factors, as is $2^n$. What can you say about the quotients of these factors?

Answer 3

Let $\epsilon >0$. Choose $n\in\mathbb{N}$ large enough so that $\epsilon>\frac{8}{n}$.

Notice that $$\frac{2^n}{1\cdot2\cdots n}\leqslant\frac{2^n}{1\cdot2^{n-2}\cdot n}=\frac{4}{n}<\frac{8}{n}<\epsilon$$

concluding that $$\lim _{n\to 0} \frac{2^n}{n!}=0$$

Answer 4

Use the inequality

$\frac{2^n}{n!}\leq(\frac{2}{3})^{n-1}$ for all $n\geq k$

Answer 5

Let $$A=\frac{2^n}{n!}$$ Take the logarithms of both sides so $$\log(A)=n\log(2)-\log(n!)$$ Use now the simplest approximation by Stirling $$\log(n!)\simeq\left(n+\frac{1}{2}\right) \log (n)+\frac{1}{2} \log (2 \pi ) $$ So $$\log(A)\simeq n\Big(\log(2)-\log(n)\Big) -\frac{1}{2} \log (n)-\frac{1}{2} \log (2 \pi ) $$

I am sure that you can take from here.