Proof using induction: $n! > n^2$, for $n\geq4$

Question

So, $(k+1)! > k^3 + k^2 > k^2 + 2k + 1$
$(k+1)! > (k+1)^2$ ---> What we want to proof

Does this serve as a proof for my sentence?

I would make one small adjustment in your proof. In you line:$$k^3 + k^2 > k^2 + 2k + 1\text{ for }n\ge4$$you should change the $n$ to a $k$. — Mufasa, Mar 12 '16 at 12:48
Not proposing to close, just linking to the node of this network of duplicates. — Lee David Chung Lin, Feb 06 '20 at 18:44

Archis Welankar · Answer 1 · 2016-03-12T12:43:51.077

3

Alternatively lesser steps. We want to prove $(k+1)!>(k+1)^2$ ie proving $k!>(k+1)$ which is by observation or gamma function true thats all.

edited Mar 12 '16 at 12:43

answered Mar 12 '16 at 12:38

Archis Welankar

The Gamma function certainly seems too advanced for this level of problem. More important, the problem statement says to prove "using induction," which your answer does not do. But your idea is certainly good. – Rory Daulton Mar 12 '16 at 13:00
I have just written last induction step and reason for the same nothing more and thanks! – Archis Welankar Mar 12 '16 at 13:10

score 1 · Accepted Answer · edited Apr 27 '16 at 12:16

1

@Vinicius No need to make that so complicated.

Assuming $k! > k^2$. $$(k+1)!>(k+1)^2$$ $$(k+1)k! > (k+1)(k+1)$$ $$k!>k+1$$

The last statement requires little proof for $k \in \{4, 5, 6, \cdots\}$.

edited Apr 27 '16 at 12:16

N.S.JOHN

answered Mar 12 '16 at 14:43

TheRandomGuy

2 Answers2