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What method should I use for this limit? $$ \lim_{n\to \infty}{\frac{n^{n-1}}{n!}} $$

I tried ratio test but I ended with the ugly answer $$\lim_{n\to \infty}\frac{(n+1)^{n-1}}{n^{n-1}} $$ which would go to 1? Which means we cannot use ratio test. I do not know how else I could find this limit.

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    $\begingroup$ $\lim_{n\to \infty}\left(\frac{n+1}{n}\right)^{n-1}$ goes to $e$, not 1 $\endgroup$ Nov 25 '15 at 23:03
  • $\begingroup$ You expression has the same exponent. How about making a new base $\frac{n+1}{n}$ ? $\endgroup$
    – imranfat
    Nov 25 '15 at 23:03
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Note that for $n\ge 2$ the top is equal to $n\cdot n^{n-2}$.

The bottom is $(1)(2)\left[(3)(4)\cdots(n)\right]$. The product $[(3)(4)\cdots(n)]$ consists of $n-2$ terms, all $\le n$. So the bottom is $\le (1)(2)n^{n-2}$. It follows that for $n\ge 2$ we have $$\frac{n^{n-1}}{n!}\ge \frac{n\cdot n^{n-2}}{(1)(2)n^{n-2}}=\frac{n}{2}.$$ Since $\frac{n}{2}\to\infty$ as $n\to\infty$, it follows that our sequence diverges to $\infty$.

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You could try to use the Stirling formel which states that for large n $$n! \sim \sqrt{2\pi n} (\frac{n}{e})^n$$

So you would get : $$\frac{n^{n-1}}{n!} \sim n^{n-1}(\frac{e}{n})^n (2\pi n)^{-\frac{1}{2}}$$ $$ \sim \frac{1}{\sqrt{2\pi}n^{\frac{3}{2}}} e^n \rightarrow \infty$$ Thus the sequence diverges.

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$$\lim_{n \to\infty} \left(\frac{n+1}{n}\right)^{n-1} = e$$ so it diverges.

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