# How do I prove that $2^n=O(n!)$?

How do I prove that $2^n=O(n!)$?

Is this a valid argument?

 2<=2,
2<3,
2<4,....
2<n if n>2

therefore 2.2.2....n times < 1.2. ... n
so,2^n <n!


#### Solutions Collecting From Web of "How do I prove that $2^n=O(n!)$?"

Since $e^2=\sum_{n=0}^{+\infty}\frac{2^n}{n!}$ converges (by ratio test if you want), the general term tends to $0$, whence, actually, $2^n=o(n!)$ and not only $O(n!)$.

The argument is very informal and has a small hole in it, but the basic idea is correct. The hole lies in the fact that $2^n$ is actually larger than $n!$ for $n=1,2,3$. Properly you should show by induction on $n$ that $2^n\le n!$ for $n\ge 4$ and conclude immediately that $2^n$ is $O(n!)$.

\begin{align*}&(1)\;\;\text{Look at the positive series}\;\sum_{n=1}^\infty\frac{2^n}{n!}\\{}\\ &(2)\;\;\frac{a_{n+1}}{a_n}=\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^n}=\frac2{n+1}\xrightarrow[n\to\infty]{}0 \;,\;\text{thus}\\{}\\ &(3)\;\;\text{The series in (1) converges}\\{}\\ &(4)\;\;a_n=\frac{2^n}{n!}\xrightarrow[n\to\infty]{}0\end{align*}

Thus, for some

$$N\in\Bbb N\;\;\text{and}\;\;\forall n>N\;,\;\;\frac{2^n}{n!}<1\implies 2^n=\mathcal O(n!)$$

You can use induction for $n > 3$.

Assume $p(k): 2 ^ k < k!, k > 3$

Prove that $p(k +1 ): 2 ^ {k + 1} < (k + 1)!$

$2 ^ {k + 1} = 2 ^ k * 2$

$(k + 1)! = k! * k$

$k > 2 \implies 2 ^ {k + 1} < (k + 1)!$