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I have a math question where I am to show, using induction, that $n!> n^{3}$, $n\geq 6$

i) I have shown that $LHS > RHS$ for $n=6$

$$720 > 216$$

ii)

Set $n=p$ and assume that

$p!> p^{3}$

iii) set $n=p+1$

$$(p+1)!>(p+1)^{3}$$

by the assumption we know that

$$(p+1)p! > (p+1)p^{3}$$

this is where I am stuck and would appreciate some help.

Prime Mover
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cosine
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    $(p+1)^3 = p^3+3p^2+3p+1$ while $(p+1)p^3 = \underbrace{p^3+p^3+p^3+p^3+\dots+p^3}_{p+1\text{ times}}$. Can you compare these? – JMoravitz May 21 '21 at 17:35
  • Welcome to MSE. In order for MathJax commands to be effective, they must be enclosed in$ signs. For example, $x^2$ shows up as $x^2$. – saulspatz May 21 '21 at 17:40
  • @JMoravitz, hmm well I can tell that $(p+1)p^{3}$ is significant larger when using p=6. If I know that $(p+1)p^{3}$ is larger than $(p+1)^{3}$, how can I further prove the thesis? – cosine May 21 '21 at 17:59
  • $(p+1)p^3 = \color{red}{p^3}+\color{blue}{p^3}+\color{green}{p^3}+p^3+\dots \geq \color{red}{p^3}+\color{blue}{3p^2}+\color{green}{3p}+1$. You do see that $p^3>3p^2$ without additional explanation for $p\geq 6$ don't you? – JMoravitz May 21 '21 at 18:37

3 Answers3

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The approach I recommend you take is to write the reasoning like this:

For $n = 6$, we know $$720 = 6! > 6^3 = 216.$$ So there exists an integer $p \ge 6$ such that the claim $$n! > n^3$$ is true for $n = p$. Then we wish to show that if the claim is true for $n = p$, then it must also be true for $n = p+1$. That is to say, $$(n+1)! = (n+1)n! > (n+1)n^3$$ by the induction hypothesis. But how can get this last expression to look more like $(n+1)^3$? Well, if we can show $$n^3 > (n+1)^2$$ for $n \ge 6$ then we would be done. There are a number of ways to do this. For instance, consider $$\frac{n^3}{(n+1)^2} = \left(\frac{n \sqrt{n}}{n+1}\right)^2.$$ If $n\sqrt{n} > n+1$, then this expression is greater than $1$, which would imply the desired inequality. And since we clearly have $n(\sqrt{n} - 1) > 1$ for all $n \ge 4$ (in fact, it holds for $n \ge 3$ but this is slightly less obvious, requiring us to know that $\sqrt{3} > \frac{4}{3}$), we have established $n^3 > (n+1)^2$, hence $$(n+1)! > (n+1)n^3 > (n+1)(n+1)^2 = (n+1)^3,$$ completing the induction step.

heropup
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Hint:

$(p+1)p^3=p^4+p^3>p^3+3p^2+3p+1\iff p^4>3p^2+3p+1$.

Now observe that, since $p\ge 6$, $\enspace p^4=p^2\cdot p^2\ge 36p^2> 7p^2$.

Can you continue?

Bernard
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  • I am not sure why I am to compare $(p+1)^{3}$ and $(p+1)p^{3}$. I have been studying the whole day and I think I need to come back and tackle this problem tomorrow with full capacity, hopefully then I will understand. Thank you – cosine May 21 '21 at 18:52
  • This is because what you deduce instantly from the inductive hypothesis is that $(p+1)!=(p+1)p!>(p+1)p^3$, but what you have to show is $(p+1)!>(p+1)^3$. – Bernard May 21 '21 at 18:57
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Alternative approach, similar to heropup's answer.

For $n=6$, LHS $>$ RHS.

As $n \to (n+1)$, LHS increases by a factor of $(n+1)$,
while RHS increases by a factor of
$\left(\frac{n + 1}{n}\right)^3 = \left(1 + \frac{1}{n}\right)^3.$

Note that as $n$ increases, $(n+1)$ increases while
$\left(1 + \frac{1}{n}\right)$ decreases.

For $n \geq 6, ~(n+1) > \left(1 + \frac{1}{n}\right)^3.$

user2661923
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