efekwulsemmay Messages 53 Reaction score 0 Sep 21, 2009 #2 We would be happy to help you if you at least try the problem yourself and show us what you have tried
We would be happy to help you if you at least try the problem yourself and show us what you have tried
Office_Shredder Staff Emeritus Science Advisor Gold Member Messages 5,706 Reaction score 1,592 Sep 21, 2009 #3 Obviously not. 31>1! Have you started trying to prove it yet? Presumably the question asks to prove that for large enough n... what are your initial thoughts?
Obviously not. 31>1! Have you started trying to prove it yet? Presumably the question asks to prove that for large enough n... what are your initial thoughts?
Gregg Messages 452 Reaction score 0 Sep 21, 2009 #4 Maybe he means n^n<n! because there is always [tex]x=n!^{1\over n}[/tex] ? If so, you need to construct your argument around n^n = n*n*n*n...*n (n times) and n! = n(n-1)(n-2)...3.2.1 I think.
Maybe he means n^n<n! because there is always [tex]x=n!^{1\over n}[/tex] ? If so, you need to construct your argument around n^n = n*n*n*n...*n (n times) and n! = n(n-1)(n-2)...3.2.1 I think.
Borek Mentor Messages 29,211 Reaction score 4,635 Sep 21, 2009 #5 Perhaps for n approaching infinity?