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

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
up vote 26 down vote accepted

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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