Demonstration that $\forall\; n>3,\;\;n^2

Question

How do I prove by mathematical induction that$$\forall\; n>3,\;\;n^2<n!$$

I tried, $n=4$ then $4^2<4!$ what is true, because $16<24$.$$$$Hypotesis: $n^2<n!$ $$$$Thesis: $(n+1)^2<(n+1)!$$$$$Show: $$(n+1)^2=n^2+2n+1<n!+2n+1$$ and????

You wish to prove that $(n+1)^2<(n+1)!$. This is equivalent to proving $n+1<n!$. You don't even have to use the induction hypothesis. — Git Gud, Nov 06 '13 at 17:52
Not proposing to close, just linking to the node of this network of duplicates. This post here is older, but the other one has existing duplicate links. — Lee David Chung Lin, Feb 06 '20 at 18:54

score 2 · Answer 1 · answered Nov 06 '13 at 17:52

2

Hint: $$\dfrac{(n+1)^2}{n^2} = \left(1+\dfrac1n\right)^2 < 2^2 < n+1 = \dfrac{(n+1)!}{n!}, \qquad n > 3.$$

answered Nov 06 '13 at 17:52

njguliyev

14,473
1
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score 2 · Accepted Answer · answered Nov 06 '13 at 17:54

2

$$(n+1)!=(n+1)\cdot n!>(n+1)\cdot n^2>(n+1)^2$$

Here $n^2>n+1$ since $\forall n>3, \qquad n^2-n-1=(n-\frac{1}{2})^2-\frac{5}{4}>0$

answered Nov 06 '13 at 17:54

Shaswata

5,068

Demonstration that $\forall\; n>3,\;\;n^2

2 Answers2