Induction with inequality

CSkyle · #1 $Old$ April 30th, 2008, 05:39 PM

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

CSkyle · #2 $Old$ April 30th, 2008, 05:41 PM

oops, n^n came out as nn...anywhere there is nn, it should read (n)^n.

TheEmptySet · #3 $Old$ April 30th, 2008, 06:03 PM

Quote:

Originally Posted by CSkyle $View Post$

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 $k! < k^k$

Show

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

So

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

QED

CSkyle · #4 $Old$ April 30th, 2008, 09:42 PM

thanks empty set.