# Induction with inequality

• Apr 30th 2008, 04:39 PM
CSkyle
Induction with inequality
How do i use induction(effectively) to prove that n! < nn for n ≥ 2 ?

let p(n): n! < nn for n ≥ 2.

we see that p(2) holds true.

now i prove it holds for p(n+1) right?

(n+1)! = (n+1)n!<(n+1)nn by the induction hypothesis.

i think this is right.. can someone show me how to finish this correctly?

thanks

Kyle
• Apr 30th 2008, 04:41 PM
CSkyle
oops
oops, n^n came out as nn...anywhere there is nn, it should read (n)^n.
• Apr 30th 2008, 05:03 PM
TheEmptySet
Quote:

Originally Posted by CSkyle
How do i use induction(effectively) to prove that n! < nn for n ≥ 2 ?

let p(n): n! < nn for n ≥ 2.

we see that p(2) holds true.

now i prove it holds for p(n+1) right?

(n+1)! = (n+1)n!<(n+1)nn by the induction hypothesis.

i think this is right.. can someone show me how to finish this correctly?

thanks

Kyle

assume \$\displaystyle k! < k^k\$

Show

\$\displaystyle (k+1)^{k+1}=(k+1)^k(k+1) > k^k(k+1)>(k!)(k+1)=(k+1)!\$

So

\$\displaystyle (k+1)^{k+1}>(k+1)!\$

QED
• Apr 30th 2008, 08:42 PM
CSkyle
thx
thanks empty set.