The discussion focuses on proving the inequality n^3 < n! for all n >= 6 using mathematical induction. The base case is established for n = 6, where the inequality holds true. The inductive step assumes k^3 < k! for some k >= 6 and demonstrates that (k+1)^3 can be shown to be less than (k! + 3k^2 + 3k + 1) by leveraging the induction hypothesis. The proof is completed by factoring and comparing values, confirming the inequality holds for all n >= 6.
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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.