What's the limit of the sequence $\lim\limits_{n \to\infty} \frac{n!}{n^n}$?

Solution 1:

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.

Solution 2:

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.