1
$\begingroup$

This question already has an answer here:

I have a question about the limit of the factorial function. How can I evaluate

$$ (1) \lim_{n \rightarrow \infty} \frac{10^n}{n!} $$

$$ (2) \lim_{n \rightarrow \infty} \frac{n^n}{n!} $$

Is there any alternative expression for the factorials to solve this problem?

$\endgroup$

marked as duplicate by Xander Henderson, Arnaud D., Vladhagen, Xam, A. Goodier Mar 23 '18 at 22:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

8
$\begingroup$

Hint: Use Stirling's approximation:

$$n! \sim \sqrt{2\pi n}\left(\frac ne\right)^n$$

$\endgroup$
3
$\begingroup$

I thought it might be instructive to present approaches that rely on crude and simple to obtain estimates for the factorial, rather than develop (or quote) something tantamount to Stirlin'g Formula. To that end, we now proceed.


Note that $n!\ge \left(\frac{n}{2}\right)^{n/2}$. Hence, we have

$$\frac{10^n}{n!}\le \frac{10^n}{\left(\frac{n}{2}\right)^{n/2}}=\left(\frac{200}{n}\right)^{n/2}$$

from which it is clear that

$$\lim_{n\to \infty}\frac{10^n}{n!}=0$$


It is easy to show that $n^n\ge nn!$ by using, for example, induction. Hence, we have

$$\frac{n^n}{n!}\ge n$$

whence the coveted limit is $\infty$.

$\endgroup$
1
$\begingroup$

Well, you can write each of $n!,10^n,$ and $n^n$ as products of $n$ integers. This allows you to write $\frac{10^n}{n!}$ and $\frac{n^n}{n!}$ as a product of $n$ numbers, and you just have to determine how large those $n$ numbers are to find the limits.

$\endgroup$
1
$\begingroup$

Andrew Li's answer is concise. But if you don't know Stirling's approximation, there is a elementary proof:

(1) We can see that n can be much larger than 10 as $n\to \infty$, so let $$a_n=\frac{10^{n}}{n!}\ \ \ and\ \ \ a_{20}=\frac{10^{20}}{20!}$$

Then $$\lim_{n \rightarrow \infty} \frac{10^n}{n!}=\lim_{n \rightarrow \infty} a_n=\lim_{n \rightarrow \infty} a_{20}\frac{10^{n-20}}{\frac{n!}{20!}}=a_{20}\lim_{n \rightarrow \infty} \prod\limits_{i=1}^{n-20}\left (\frac{10}{20+i}\right)\\\leq a_{20}\lim_{n \rightarrow \infty} \left(\frac{1}{2}\right)^{n-20}\to0$$

Hence$$\lim_{n \rightarrow \infty} \frac{10^n}{n!}=0$$

(2) We have

$$\lim_{n \rightarrow \infty} \frac{n^n}{n!}=\lim_{n \rightarrow \infty} \left(\prod\limits_{i=2}^{n}\left(\frac{n}{i}\right)\right)n\ge\lim_{n \rightarrow \infty}n\to \infty$$

Hence$$\lim_{n \rightarrow \infty} \frac{n^n}{n!}=\infty$$

$\endgroup$
1
$\begingroup$

Trick useful in this case: for (a) consider the series $$ \sum_{n=1}^\infty\frac{{10}^n}{n!} $$ and apply the quotient test: $$ \lim_{n\to\infty}\frac{\frac{{10}^{n+1}}{(n+1)!}}{\frac{{10}^n}{n!}} = \lim_{n\to\infty}\frac{10}{n+1} = 0 < 1. $$ As the series converge, the general term (your sequence) $\to 0$. For (b), the series $$ \sum_{n=1}^\infty\frac{n!}{n^n} $$ is also convergent: $$ \lim_{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n = \frac1e < 1. $$ Then, $$ \lim_{n\to\infty}\frac{n!}{n^n} = 0 $$ and $$ \lim_{n\to\infty}\frac{n^n}{n!} = \infty $$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.