User Name Remember Me? Password

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

 November 29th, 2017, 08:03 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 600 Thanks: 87 Another factorial inequality Prove that $\displaystyle n!^2 \geq n^n$ for $\displaystyle n\in N$ November 29th, 2017, 01:45 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Mathematical induction can work. For n=1 or 2, equality. For n=3, 36 > 27. In general. $(n+1)!^2=(n+1)^2(n!)^2$ $(n+1)^{n+1}=(n+1)n^n(1+\frac{1}{n})^n<(n+1)n^ne<( n+1)^2(n!)^2$, for n > 2. Thanks from greg1313, Country Boy and JeffM1 December 28th, 2017, 08:39 AM   #3
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 551

Quote:
 Originally Posted by LearnerLearner Use induction and then everything can be solved
This may be the stupidest answer ever on this or any other math web site. December 28th, 2017, 08:49 AM   #4
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1588

Quote:
 Originally Posted by JeffM1 This may be the stupidest answer ever on this or any other math web site.
posted to link the study daddy pay site ... spam December 28th, 2017, 02:52 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics I would agree that this is a great exercise for using induction. In the interests of completeness, I'll add a sketch for a direct proof which doesn't require induction. 1. Note that $n!^2$ is a product of $2n$-many factors which we can group into only $n$-many factors by taking pairwise products by defining: $a_k = (n-k)(k+1) \quad k =0,1,\dots,n-1.$ 2. Now $n!^2 = a_0 a_1 \dots a_{n-1}$ and $n^n$ are each products of $n$-many factors. It suffices to prove that $a_k \geq n$ for every $k$. 3. Fix $k$ and expand $a_k - n = (n-k)(k+1)-n = nk - k^2 -k.$ 4. Since $n \geq (k+1)k$, we have $(n-k-1) \geq 0$ and since $k \geq 0$, this implies $a_k - n \geq 0$. Last edited by SDK; December 28th, 2017 at 02:54 PM. Tags factorial, inequality Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tahir.iman Algebra 9 December 26th, 2012 06:22 PM annakar Algebra 2 December 9th, 2012 01:31 AM bilano99 Calculus 2 May 1st, 2012 10:12 AM thomasthomas Number Theory 13 November 6th, 2010 06:21 AM coolaid317 Calculus 8 April 28th, 2010 06:21 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      