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I have the bellow limit, and I know I need to use Cauchy-d’Alembert, and the limit is 1/e but have no idea how to get to it, I get to something like $\frac{\frac{\left(n+1\right)!}{n!}}{n}$, but is not right

$$\lim _{n\to \infty }\left(\frac{\sqrt[n]{n!}}{n}\right)$$

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I would prefer using kotomord's approach. I disagree that it should be simpler; this is basically equivalent to the coarse version of Stirling's approximation.

But anyway, rewrite the limit as

$$ \lim _{n\to \infty}\sqrt[n]{\frac{n!}{n^n}} $$ the Cauchy-d'Alembert criteria says this should be equal to $$ \lim_{n\to \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \\ = \lim_{n\to \infty}\frac{(n+1)n^n}{(n+1)^{n+1}} \\ = \lim_{n\to \infty}\left(\frac{n}{n+1}\right)^n \\ = \lim_{n\to \infty}\left(\frac{1}{1+\frac{1}{n}}\right)^n \\ = \frac{1}{e} $$

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Try to use Stirling's approximation

https://en.wikipedia.org/wiki/Stirling%27s_approximation

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  • $\begingroup$ thanks, but I think must be a simpler way, didn't heard of this. $\endgroup$ – John Dec 14 '16 at 23:11
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Here's a different approach.

Let $a_n$ be the element of the sequence. We have

$$ \ln a_n = \frac{1}{n} \ln (n!) - \ln n = \frac{1}{n} \sum_{j=1}^n \ln (j/n).$$

This is Riemann sum, and converges to $\int_0^1 \ln x dx= -1$ (note it is an improper Riemann integral, so for complete rigor, we need to truncate the summation). Since $\lim_{n\to\infty} \ln a_n = -1$, it follows that $\lim_{n\to\infty} a_n = e^{-1}$.

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