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January 3rd, 2017, 07:28 AM   #1
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Post Need help with (n^n)/((n!)^2)

I m struggling with this limit for quite a while.Can someone help me please?
$\displaystyle (n^n)/(n!)^2 $ as n approaches infinity
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January 3rd, 2017, 01:48 PM   #2
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Try Stirling's approximation for factorials:

$\displaystyle n!\approx n^n e^{-n}\sqrt{2\pi n}$

Then you can deal with

$\displaystyle\lim_{n\to\infty}\frac{n^n}{\left(n^ n e^{-n}\sqrt{2\pi n}\right)^2}$
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January 3rd, 2017, 03:10 PM   #3
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Another approach:

$\displaystyle \frac{n\cdot n\cdot n\cdots}{n^2\cdot(n-1)^2\cdot(n-2)^2\cdots}$

The terms in the denominator are greater than the terms in the numerator until $n-a\leq\sqrt n$ for some $a$. But what is the ratio of a root to its square? It is $\frac{1}{\sqrt n}$ and we have

$$\lim_{n\to\infty}\frac{1}{\sqrt n}=0$$

Hence we may state

$$\lim_{n\to\infty}\frac{n^n}{(n!)^2}=0$$
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January 5th, 2017, 10:16 AM   #4
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Thank you very much for your answers!
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