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I am trying to solve this by mathematical induction but I cannot do it.

$$n! \geq n^3$$

I made this:

  • First element: $$n=6$$ $$6! ≥ 6^3$$ $$720 ≥ 216$$
  • Hypothesis: $$n=k$$ $$k!≥k^3$$
  • Thesis: $$n=k+1$$ $$(k+1)!≥(k+1)^3$$

I would appreciate your help. Many thanks!

Ezequiel Pace
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    at which step of mathematical induction are you stuck with? – Siong Thye Goh Aug 10 '17 at 23:56
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    what steps have been done ? maybe showing them in the post will help narrow down where you are having trouble. –  Aug 10 '17 at 23:57
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    People here like to see more than just a problem statement and a plea for help. If you don't show some work or give motivation for the problem then it is likely your question will be closed. See the meta threads https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question and https://math.meta.stackexchange.com/questions/1803/how-to-ask-a-homework-question for advice. – Trevor Gunn Aug 10 '17 at 23:59
  • @TrevorGunn I think most mathematical places on the web, would care for a showing of effort. it helps in mistake finding, helping a less studied audience understand. it also show specifics of the problem that may not come through the rest of the wording. –  Aug 11 '17 at 00:04
  • @SiongThyeGoh I added more information! Thank U 4 your help. – Ezequiel Pace Aug 11 '17 at 00:07
  • @RoddyMacPhee I've updated to give more details. Thanks! – Ezequiel Pace Aug 11 '17 at 00:08
  • okay so not even started to prove the last part ? –  Aug 11 '17 at 00:09
  • @RoddyMacPhee I cannot even figure it out what to sum or multiply to get to the thesis... I don't even know how to start. – Ezequiel Pace Aug 11 '17 at 00:11
  • what is $(k+1)!\over k!$ ? what is $(k+1)^3\over k^3$ ? –  Aug 11 '17 at 00:12
  • Hypothesis and Thesis. – Ezequiel Pace Aug 11 '17 at 00:14
  • Not proposing to close, just linking to the node of this network of duplicates. – Lee David Chung Lin Feb 06 '20 at 20:16

4 Answers4

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It's not true for $n=2$, $n=3$, $n=4$, or $n=5$.

For $n=6$, we do have

$$6!=720>216=6^3$$

After that, consider that $7>\left(\frac76\right)^3$, which gives you two inequalities to multiply together and obtain the next case. Can you generalize that step to see why $n!\geq n^3\implies(n+1)!\geq(n+1)^3$, when $n\geq 6$?

G Tony Jacobs
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It's not true for $n < 6$. But if $n \ge 6$:

Notice: $(n+1)! = n!*(n+1)$

And: $(n+1)^3 = n^3 + 3n^2 + 3n +1 < n^3 + n*n^2 + n^2*n +n^3= 4n^3$

And: if $n! \ge n^3$ then $n!*(n+1) \ge n^3*(n+1)$.

Put those three together.

fleablood
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Suppose $k! \geq k^3$

$$(k+1)! = k! (k+1) \geq k^3(k+1)$$

Hence it suffices to show that $k^3 (k+1) \geq (k+1)^3$

or $k^3 \geq (k+1)^2$ for $k \geq 6$

$$(k\sqrt{k}-k-1)(k\sqrt{k}+k+1)\geq 0$$

That is we are interested in showing that if $k \geq 6$, then

$$(k(\sqrt{k}-1)-1)(k(\sqrt{k}+1)+1)\geq 0$$

Hence it suffices to show that

if $k\geq 6$, then $$k(\sqrt{k}-1)-1 \geq 0$$

can you complete the task?

Siong Thye Goh
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$$n!>n^3\Longleftrightarrow \log n!>\log n^3\Longleftrightarrow \sum_{k=2}^{n}\log k>3\log n\Longleftrightarrow \int_{1}^{n}\log x\space dx>3\log n\\\Longleftrightarrow n\log n -(n-1)>3\log n\Longleftrightarrow n\log n>3\log n +(n-1)\\\Longleftrightarrow n>3+\frac{n-1}{\log n}$$

Now, as $n\geq 6$ $$\frac{n-1}{\log n}\leq \frac{n-1}{\log 6}<n-3\Rightarrow n>3+\frac{n-1}{\log n}$$

MAN-MADE
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