# What's the limit of the sequence $\lim_{n \rightarrow \infty} \frac{n!}{n^n}$?

$\displaystyle \lim_{n \rightarrow \infty} \frac{n!}{n^n}$

I have a question: Is it valid to use Stirling's Formula to prove convergence of the sequence?

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"Valid" in what sense? If this were an assignment, that would depend on what you may or may not take for granted. If you mean in the sense of a possible circular argument, I do not think that this limit is needed to derive Stirling's formula, so that would not be an issue. – Arturo Magidin Sep 4 '11 at 2:06
Stirling's formula is really overkill here. I suggest writing the numerator and denominator out as products of $n$ numbers and looking for an upper bound. – Jonas Meyer Sep 4 '11 at 2:29
FWIW, the formula is hardly needed, given that the expression is a product of $n$ positive factors, all $\le1$, the smallest of which is $1/n$, hence $n!/n^n<1/n\to0$. – anon Sep 4 '11 at 2:31
In fact, if $a_n=n!/n^n$, then $\lim\sqrt[n]{a_n}=1/e$, so for sufficiently large $n$, $\sqrt[n]{a_n}<1/2$, which implies $a_n<1/2^n$. For proofs that $\lim\sqrt[n]{a_n}=1/e$, including one that uses Stirling's formula, see here and here. – Jonas Meyer Sep 4 '11 at 8:43
– Martin Sleziak May 24 '15 at 7:30

## 2 Answers

There are two distinct questions here. The first one in the title is what the limit actually is. This is easy to see by writing out the expression as a product of $n$ positive factors: $$\frac{n!}{n^n}=\left(\frac{1}{n}\right)\left(\frac{2}{n}\right)\left(\frac{3}{n}\right)\cdots\left(\frac{n}{n}\right).$$ Every one of the factors $k/n$, $k=1,2,3,\dots,n$, is less than or equal to $1$. Hence the product is $$\le\left(\frac{1}{n}\right)\cdot1\cdot1\cdots1=1/n.$$ But $1/n$ converges to $0$ as $n\to\infty$, so by the Squeeze theorem so does the original expression.

The second question is whether or not it's allowed to use Stirling's formula to derive the limit, which I believe Arturo's comment covers: there is no apparent circularity, but in the context of classwork the answer depends on whether or not you've formally learned the formula and are allowed to use it as a given.

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Thanks everyone. – Mario De León Urbina Sep 4 '11 at 14:11

You can prove that $$n! < \left( \frac{n+1}{2} \right)^n.$$ Now observe that $$0 \leq \lim_{n\to\infty} \frac{n!}{n^n} <\lim_{n\to\infty} \frac{\left(\frac{n+1}{2}\right)^n }{n^n} = \lim_{n\to\infty} \frac{1}{2^n}\cdot\frac{(n+1)^n}{n^n}.$$ We know that $1/2^n\to 0$ as $n\to\infty$. If you know that $[(n+1)/n]^n\to e$ as $n\to\infty$, then you're done... the limit is zero!

Definitely not as nice as anon's solution, but a different approach nontheless.

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