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 November 3rd, 2019, 07:28 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math Range of X Find the range of X such that $\displaystyle (X+6)^{n} \cdot n!\geq n^n \; , n\in \mathbb{N} ,X\in \mathbb{R}.$
 November 3rd, 2019, 07:59 AM #2 Member   Joined: May 2013 Posts: 57 Thanks: 5 unfortunately there's no definitive answer to this question as you have two variables rather than 1. setting n=2 will give a different range for x than setting it to 9.
November 3rd, 2019, 08:32 AM   #3
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Quote:
 Originally Posted by phillip1882 unfortunately there's no definitive answer to this question as you have two variables rather than 1. setting n=2 will give a different range for x than setting it to 9.
he's looking for the infinite intersection of all ranges of X over all the natural numbers.

November 3rd, 2019, 09:48 AM   #4
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Quote:
 Originally Posted by idontknow Find the range of X such that $\displaystyle (X+6)^{n} \cdot n!\geq n^n \; , n\in \mathbb{N} ,X\in \mathbb{R}.$
$$\lim\limits_{n \rightarrow \infty} \dfrac{n}{(n!)^{1/n}} = e$$

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