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October 18th, 2017, 09:19 AM   #1
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n! inequality

How to prove if $\displaystyle \; {(\frac{n}{3})}^{n}\leq n! $
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October 18th, 2017, 09:32 AM   #2
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Salam !

Did you try to show this inequality by induction ? You may use the Binomial Expansion.
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October 18th, 2017, 09:56 AM   #3
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It isn't a true statement so it would be hard to prove.
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October 18th, 2017, 10:29 AM   #4
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Quote:
Originally Posted by SDK View Post
It isn't a true statement so it would be hard to prove.
In my book it says true, but I want to see how to get there?
And I forgot to write $\displaystyle n \in N$.

Last edited by skipjack; October 18th, 2017 at 12:14 PM.
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October 18th, 2017, 03:00 PM   #5
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An approximate proof would start using Stirling's formula:

$\displaystyle n!\simeq \sqrt {2\pi n}(\frac{n}{e})^n>(\frac{n}{3})^n$

Checking on the derivation of Stirling's formula. $\displaystyle \sqrt {2\pi n}(\frac{n}{e})^n$ is a lower bound for n!.
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Last edited by mathman; October 19th, 2017 at 02:02 PM.
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