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$\lim_{n \to\infty} \frac{n!}{n^n}$

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

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    Related: http://math.stackexchange.com/questions/1904113/limit-cn-n-nn-as-n-goes-to-infinity2016-12-21

2 Answers 2

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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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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.