TITLE: which is bigger $2^{n^{1.001}}$ or $n!$
QUESTION [5 upvotes]: for a big enough n, How to detrmine which is bigger?
$2^{n^{1.001}}$ or $n!$
I have tried to make a series:  $a_n = \frac{2^{n^{1.001}}}{n!}$
and then try finding the limit of $\frac{a_{n+1}}{a_n}$
and see if its bigger or smaller then one, also tried to use induction but didnt really got my anywhere
btw I know the answer is that $2^{n^{1.001}}$ is bigger

REPLY [5 votes]: Why didn't it lead you anywhere?
\begin{aligned}
\frac{a_{n+1}}{a_n} &= \frac{\frac{2^{(n+1)^{1.001}}}{(n+1)!}}{\frac{2^{n^{1.001}}}{n!}} \\
                    &= \frac{2^{(n+1)^{1.001}}}{(n+1)2^{n^{1.001}}}
\end{aligned}
Take the log of both sides.
\begin{aligned}
\log\left(\frac{a_{n+1}}{a_n}\right) &= \log\left(\frac{2^{(n+1)^{1.001}}}{(n+1)2^{n^{1.001}}}\right) \\
&= \log\left(2^{(n+1)^{1.001}}\right) -\log\left(2^{n^{1.001}}\right) - \log\left(n+1\right) \\
&= (n+1)^{1.001}\log\left(2\right) - n^{1.001}\log\left(2\right) - \log(n + 1)
\end{aligned}
Divide both sides by $\log(2)$.
\begin{aligned}
\log_2\left(\frac{a_{n+1}}{a_n}\right) &= (n+1)^{1.001} - n^{1.001} - \log_2(n + 1)
\end{aligned}
Now, we have a new function $f(n)$ for which the following is true:
$$
f(n) = (n+1)^{1.001} - n^{1.001} - \log_2(n + 1)
$$
$$
\lim_{n \to \infty}f(n) > 0 \implies 2^{n^{1.001}} > n!\\
\lim_{n \to \infty}f(n) < 0 \implies 2^{n^{1.001}} < n!
$$
Let's break up this function into two parts:
$$
f(n) = (n+1)^{1.001} - \left(n^{1.001} + \log_2(n + 1)\right)
$$
Now, let's repeat the process:
$$
\lim_{n \to \infty} \frac{(n+1)^{1.001}}{\left(n^{1.001} + \log_2(n + 1)\right)} > 1 \implies \lim_{n \to \infty}f(n) > 0
$$
Here, we can simply say that the highest degree term of the numerator is larger than that of the denominator, so the limit goes to $\infty$. Therefore, following the chain of implications back up, we end up with $2^{n^{1.001}} > n!$. If you followed that train of thought, you could indeed come to the answer.