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I am trying to take the limit of the following fraction :

$$ \lim_{N \to\infty} \frac { N !}{(N-r)!} $$

Attempts : I tried using the Stirling approximation $\ln(n!) =n \ln n - n $ but I figured it doesn't work for $n\to\infty$ and neither does the more accurate $n! = \sqrt{2 \pi n} (\frac{n}{e})^n$ for the same reason.

Either way , in both cases I get $e^{-r} N^r$ . I should note here that $r \in \mathbb{Z}^+ $.

Thank you in advance.

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    $\begingroup$ Isn't it clearly $+\infty$? (based on the edit) $\endgroup$ – Clarinetist Mar 8 '16 at 15:02
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    $\begingroup$ Based on the answer OP claims, glancing at Stirling's approximation suggests the expression whose limit is taken should have an additional factor of $n^{-r}$ (or something asymptotically equivalent anyway). $\endgroup$ – Travis Willse Mar 8 '16 at 15:05
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For all $r > 0$, $$\dfrac{N!}{(N-r)!}=N(N-1)(N-2)\cdots(N-r+1) \to \infty\text{.}$$ If $r = 0$, $$\dfrac{N!}{(N-r)!} = \dfrac{N!}{N!} = 1$$ and for all $r < 0$, let $s = -r$. Then $(N-r)! = (N+s)! > N!$ and $$\dfrac{N!}{(N-r)!} = \dfrac{N!}{(N+s)!}=\dfrac{N(N-1)\cdots 1}{(N+s)(N+s-1)\cdots (N+1)N(N-1)\cdots 1} = \dfrac{1}{(N+s)(N+s-1)\cdots (N+1)} \to 0\text{.}$$

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If $r \gt 0$, then $\frac { N !}{(N-r)!} \ge N-r-1$ and the limit is ????
If $r \lt 0$, then $\lim_{N \to\infty} \frac { N !}{(N-r)!} \lt \frac 1{N+1}$ and the limit is ????

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