The discussion revolves around proving the inequality n^3 < n! for all n >= 6 using mathematical induction. Participants are exploring the steps involved in the proof and the reasoning behind them.
Some participants have offered guidance on how to approach the final steps of the proof, while others are confirming the reasoning and exploring the implications of their calculations. Multiple interpretations of the factoring and comparison are being discussed.
Participants note that they are working under the constraints of the problem statement and are considering the validity of their assumptions regarding the values of k.
1MileCrash said:Homework Statement
Show that n^3 < n! for all n >= 6.
Homework Equations
The Attempt at a Solution
We see that for the base case of n = 6, the claim holds.
Suppose that k^3 < k! for some natural number k >= 6.
Consider that:
(k+1)^3
= k^3 + 3k^2 + 3k + 1
< k! + 3k^2 + 3k + 1 [By induction hypothesis]
What's a neat way to finish this? I'm a bit rusty, apparently.[/B]
1MileCrash said:Alright. And then, what remains after being factored has its largest value at k=6, and its value is smaller than any (k+1), and so I may write < k!(k+1), completing the induction.