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I would like to know to find the limit of this sequence:

$${n^n\over n!}$$

It diverges, I know, but I don't know how to come to this conclusion.

Lucas Lemos
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    I'm afraid Mathematics StackExchange was never meant to replace textbooks. –  Sep 16 '17 at 20:40
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    You can see that the sequence just increases!! Compare each term in the product of $n!$ with $n.$ – green frog Sep 16 '17 at 20:41

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Note that

$n^n = \underbrace{ n \cdot n \cdots n}_{n \text{ factors}}$

while

$n! = \underbrace{ 1 \cdot 2 \cdots n}_{n \text{ factors}}$

Therefore $$\frac{n^n}{n!} = \frac{n}{1}\cdot \frac{n}{2} \cdots \frac{n}{n} > \frac{n}{1} = n \to \infty$$

as $n\to \infty$.

flawr
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Notice that for integer $n$, $$\frac{n^n}{n!}=\frac{n\cdot n\cdot n\cdot...\cdot n}{1\cdot2\cdot3\cdot...\cdot n}$$ and $$\frac{n\cdot n\cdot n\cdot...\cdot n}{1\cdot2\cdot3\cdot...\cdot n}\ge \frac{n\cdot n\cdot n\cdot...\cdot n}{1\cdot n\cdot n\cdot...\cdot n}=n$$ so $$\frac{n^n}{n!}\ge n$$ and since the sequence $\{n\}$ diverges, the sequence $$\bigg\{\frac{n^n}{n!}\bigg \}$$ also diverges.

Franklin Pezzuti Dyer
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