My Math Forum Need help with (n^n)/((n!)^2)

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 January 3rd, 2017, 07:28 AM #1 Newbie   Joined: Jan 2017 From: Nowhere Posts: 5 Thanks: 1 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
 January 3rd, 2017, 01:48 PM #2 Member   Joined: Apr 2015 From: USA Posts: 44 Thanks: 27 Try Stirling's approximation for factorials: $\displaystyle n!\approx n^n e^{-n}\sqrt{2\pi n}$ Then you can deal with $\displaystyle\lim_{n\to\infty}\frac{n^n}{\left(n^ n e^{-n}\sqrt{2\pi n}\right)^2}$ Thanks from MozartInACan
 January 3rd, 2017, 03:10 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,410 Thanks: 861 Math Focus: Elementary mathematics and beyond 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 $$\lim_{n\to\infty}\frac{n^n}{(n!)^2}=0$$ Thanks from MozartInACan
 January 5th, 2017, 10:16 AM #4 Newbie   Joined: Jan 2017 From: Nowhere Posts: 5 Thanks: 1 Thank you very much for your answers! Thanks from Joppy

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