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


3 Answers 3


Hint: $$n!\le n^n=2^{n\log_2 n}$$ Spoiler:

$2^{n\log_2 n}<2^{n^{1.001}}\iff n\log_2 n<n^{1.001}\iff \log_2n<n^{0.001}$, which is true eventually as $\log_2 n$ grows slower than any $n^\alpha$ with $\alpha>0$. (e.g. by L'Hôpital's rule on $n^{\alpha}/\log_2 n$)


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.


Hint: Compare $$ n^{1.001}\log(2) $$ and the log of Stirling's Formula $$ n\log(n)-n+\frac12\log(2\pi n) $$

Note that since $x\gt\log(x)$ for all $x\gt0$, we have $$ \begin{align} n^{0.001} &=\color{#C00000}{n^{0.0005}}n^{0.0005}\\ &\ge\color{#C00000}{0.0005\log(n)}n^{0.0005} \end{align} $$ Therefore, for $n\gt1$, $$ \begin{align} \frac{n^{0.001}}{\log(n)} &\stackrel{\hphantom{n\to\infty}}{\ge}0.0005n^{0.0005}\\ &\stackrel{n\to\infty}{\to}\infty \end{align} $$


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