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Show $$\lim_{n\to \infty} \frac{n^2}{n!}=0$$

$$\frac{n^2}{n!}-0 = \frac{n^2}{n!} \le \frac{n^2}{n} = n < \epsilon$$

So if I let $\epsilon > 0$, then as long as I choose $1/N < \epsilon$, I have an appropriate $N$?

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  • $\begingroup$ Um. No. $n \ge 1$ so if $0< \epsilon < 1$ then $n$ will never be less then $\epsilon$. By choosing 1/N < $\epsilon$, n>N implies n > 1$/\epsilon$ or $\epsilon > 1/n$ but not $\epsilon > n$ which is obviously ridiculous. $\endgroup$
    – fleablood
    Commented Feb 15, 2016 at 18:15

4 Answers 4

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One possible way to prove it: $$ 0 \le \lim\limits_{n \to \infty} \frac{n^2}{n!} = \lim\limits_{n \to \infty} \frac{n}{(n-1)!}=\\ = \lim\limits_{n \to \infty} \frac{n-1+1}{(n-1)!} = \lim\limits_{n \to \infty}\frac{1}{(n-2)!} + \frac{1}{(n-1)!} \le \lim\limits_{n \to \infty} \frac{2}{n} = 0 $$

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No, $n$ is a way too bad estimate. You'll need to use something like

$$\frac{n^2}{n!}= \frac{n^2}{n(n-1)} \frac{1}{(n-2)!}$$

and then show the first factor on the right is bounded while the second one tends to $0$.

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The easiest way is probably using that for $n>5$, $n! > n(n-1)(n-2) \cdot 3 \cdot 2 = n(2n-2)(3n-6) > n^3$.

Then $0 \leq \frac{n^2}{n!} \leq \frac1n$.

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For $n > 2$. $\frac{n^2}{n!} = \frac{n}{(n-1)!} = \frac{n}{n-1}\frac{1}{(n-2)!}$.

$\lim_{n->\infty}\frac{n}{n-1}\frac{1}{(n-2)!}=\lim_{n->\infty}\frac{n}{n-1}\lim_{n->\infty}\frac{1}{(n-2)!} = 1*0 = 0$

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