My Math Forum n! inequality

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 October 18th, 2017, 09:19 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 181 Thanks: 23 n! inequality How to prove if $\displaystyle \; {(\frac{n}{3})}^{n}\leq n!$
 October 18th, 2017, 09:32 AM #2 Member   Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 69 Thanks: 6 Salam ! Did you try to show this inequality by induction ? You may use the Binomial Expansion.
 October 18th, 2017, 09:56 AM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 223 Thanks: 120 Math Focus: Dynamical systems, analytic function theory, numerics It isn't a true statement so it would be hard to prove.
October 18th, 2017, 10:29 AM   #4
Senior Member

Joined: Dec 2015
From: Earth

Posts: 181
Thanks: 23

Quote:
 Originally Posted by SDK 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.

 October 18th, 2017, 03:00 PM #5 Global Moderator   Joined: May 2007 Posts: 6,397 Thanks: 546 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!. Thanks from Ould Youbba and idontknow Last edited by mathman; October 19th, 2017 at 02:02 PM.

 Tags inequality

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post StillAlive Calculus 5 September 3rd, 2016 12:45 AM Dacu Algebra 8 June 20th, 2014 09:20 PM dupek707 Number Theory 4 November 3rd, 2010 02:39 PM Pavhard Number Theory 4 October 18th, 2010 08:00 AM bigli Calculus 1 June 8th, 2010 07:44 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top