Trying to prove $\lim_{n\to\infty}\frac{n^n}{n!} = \infty$

Asked Jan 13 '22 at 19:37

Active Jan 13 '22 at 19:37

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I'm looking to prove that $\lim_{n\to\infty}{\frac{n^n}{n!}}=\infty$.

I know I can use the ratio test which gives $e > 1$, thus it diverges to $\infty$, but does the following proof also work?

$$ \frac{n^n}{n!} = \frac{n}{n} \cdot \frac{n}{n-1} \cdot ... \cdot \frac{n}{2} \cdot \frac{n}{1} $$

So the first term $\frac{n}{n} = 1$, and the other $n-1$ terms are some value $a$ such that $1 < a \leq n$.

So,

$$\lim_{n\to\infty}\prod_{i=1}^{n}a = \infty \textrm{, therefore } \lim_{n\to\infty}\frac{n^n}{n!} = \infty $$

Is this a valid proof?

asked Jan 13 '22 at 19:37

mars_plastic

1

No, because the other $n - 1$ terms are not all equal to the same fixed $a$. – balddraz Jan 13 '22 at 19:42
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Does this answer your question? Limit of $n!/n^n$ as $n$ tends to infinity. Then take the reciprocal. – Dietrich Burde Jan 13 '22 at 19:44
@0XLR They aren't equal to the same fixed $a$, but they are $>1$. Isn't the product of an infinite number of terms $>1 = \infty$? – mars_plastic Jan 13 '22 at 19:48
You may just observe $n^n/n!\ge n$. – David Mitra Jan 13 '22 at 19:48
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Note that $$\displaystyle \frac{n^n}{n!} = \frac{n\cdot n\cdots n}{n\cdot(n-1)\cdots1}=\frac{n}{n}\frac{n}{n-1}\frac{n}{n-2}\cdots\frac{n}{1}\geq 1\cdot1\cdot1\cdots n=n$$ and then take the limit as $n\to\infty$. – Axion004 Jan 13 '22 at 19:49
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@mars_plastic "Isn't the product of an infinite number of terms $> 1 = \infty$": No, not necessarily if those terms depend on $n$. If for example, you instead had $(1 + \frac{1}{n})^n$, then the limit of all those terms $> 1$ is $e$ not $\infty$. – balddraz Jan 13 '22 at 19:52
@0XLR Right, of course that's the $e$ that I get when I do the ratio test.... – mars_plastic Jan 13 '22 at 19:57
@DietrichBurde This answered my question, thank you! – mars_plastic Jan 13 '22 at 20:06

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