Prove $n!>n^2$ for $n>3$

Question

I'm aware that induction is necessary. I have been stuck on this problem for a few days now. I'm having a hard time understanding how to apply the inductive hypothesis to the inequality to arrive at the $P_{n+1}$ step.

Base case clearly holds as $24 > 16$.

Assume $P_n: n!>n^2, n\geq 4$ holds.

$(n+1)n! > (n+1) n^2$

$(n+1)! > (n+1) n^2 $ <--- needs to be in form $(n+1)! > (n+1)^2$.

Thanks for any advice.

Induction isn't necessary, it's one of many ways. What's left in your proof is to show that $n^2 \geqslant n+1$. — Daniel Fischer, Sep 03 '13 at 19:16
The exercise requires that induction is used. It is prefaced with, "The principle of mathematical can be extended as follows. A list $P_m,P_{m+1},\cdots$ of propositions is true provided (i) $P_m$ is true, (ii) $P_{n+1}$ is true whenever $P_n$ is true and $n \geq m$. — Neurax, Sep 03 '13 at 19:18
It would seem that this user outsources his third-year math homework to stackexchange. The community might want to think about whether encouraging this is prudent. — Jeff Hammond, Sep 03 '13 at 21:40
Not proposing to close, just linking to the node of this network of duplicates. This post here is older, the node has existing duplicate links. — Lee David Chung Lin, Feb 06 '20 at 18:43

score 8 · Accepted Answer · answered Sep 03 '13 at 19:19

8

HINT: Show that $n^2\ge n+1$, then multiply that inequality by $n+1$.

answered Sep 03 '13 at 19:19

Brian M. Scott

616,228

Wow, thank you so much. I knew subconciously that this is what I was missing. I just wasn't aware that it was the next step in the proof. – Neurax Sep 03 '13 at 19:24
@Neurax: You’re very welcome. – Brian M. Scott Sep 03 '13 at 19:25

JLA · Answer 2 · 2013-09-03T22:52:23.090

4

Induction isn't necessary. For $n>3$, $\displaystyle n!\ge n(n-1)(n-2)\ge 2n(n-1)>2n\frac{n}{2}=n^2$.

edited Sep 03 '13 at 22:52

answered Sep 03 '13 at 19:33

JLA

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score 3 · Answer 3 · answered Sep 03 '13 at 19:23

Start with:

$$n(n! - 2) > 1$$

Note that this inequality holds, because $n \geq 2$ and $n! \geq 6$

$$n\times n! > 2n + 1$$ $$n \times n! + n! > n^2 + 2n + 1$$ $$n!(n+1) > n^2 + 2n + 1$$ $$(n+1)! > (n+1)^2$$

Using little algebra we finally get what we want.

score 1 · Answer 4 · answered Sep 03 '13 at 19:34

1

Note that $(n-1)(n-2)-n=n^2-4n+2=(n-2)^2-2$

If $n\gt 3$ then $(n-2)^2\ge 4\gt 2$ so that $(n-2)^2-2\gt 0$ whence $(n-1)(n-2)\gt n$

So that for $n\gt 3$ we have $$n!\ge n\cdot (n-1)(n-2)\gt n\cdot n=n^2$$

answered Sep 03 '13 at 19:34

Mark Bennet

100,194

score 0 · Answer 5 · answered Sep 03 '13 at 20:12

0

We need to prove that $$n!> n^2$$ for $n>3$

or,

$$(n-1)!>n$$ for $n>3$

But what is $(n-1)!$ ? It is the number of ways ordering $n$ people around a circle, and the inequality is follows.

answered Sep 03 '13 at 20:12

Salech Alhasov

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Prove $n!>n^2$ for $n>3$

5 Answers5