New member, induction problem

**pastmyprime** · May 16th 2010, 02:53 PM

I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

Show that n! > n^3 if n is large enough by induction? I get stuck.

**craig** · May 17th 2010, 02:07 AM

Originally Posted by pastmyprime

I have just joined this forum. I studied maths 45+ years ago, and I have not used it since - so I have forgotten nearly everything! I decided to learn calculus again, but do not have access to tutors etc, which makes it hard when I want to check answers. Anyway ...

Show that n! > n^3 if n is large enough by induction? I get stuck.

Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?

Or is it just setting up the problem initially that you're struggling with?

**pastmyprime** · May 17th 2010, 03:21 PM

Originally Posted by craig

Have you done any induction problems before, do you know/understand the basic ideas behind the proof by induction?
Or is it just setting up the problem initially that you're struggling with?

I should have been more precise. Yes, I understand induction and have solved other problems OK.

First, the smallest n such that $n! > n^3$ is 6.

Second, assuming it is true for n = k > 6 then show that

$(k+1)! > (k+1)^3$

We know that $(k+1)! = k!(k+1) > k^3(k+1)$

At which point I get stuck!

PS: My first attempt at using LaTex and I don't know why a formulae is moved down relative to the rest of the line that it is in.

**roninpro** · May 17th 2010, 08:27 PM

It looks like you're doing fine.

I'd like to note that it suffices to show that $k^3(k+1)\geq (k+1)^3$ for $k\geq 6$ . Can you do it?

**pastmyprime** · May 18th 2010, 09:13 PM

No, sorry, I can't prove it.

**roninpro** · May 19th 2010, 03:57 AM

Did you try using induction?

**pastmyprime** · May 19th 2010, 05:40 PM

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

Step 1 rearrange

$k^3(k+1) \geq (k+1)^3 \Rightarrow k^4 + k^3 \geq k^3 + 3k^2 + 3k + 1$
$\Rightarrow k^4 - (3k^2 + 3k + 1) \geq 0$

Step 2 induction

First it is true for $k = 6$ by substitution.

Second, assume it is true for some $k > 6$ . Then

$(k+1)^4 = k^4 + 4k^3 + 6k^2 +4k + 1$
$3(k+1)^2 + 3(k+1) + 1 = 3k^2 + 9k + 7$

So show $k^4 + 4k^3 + 3k^2 -5k - 6 \geq 0$ for $k > 6$

Now for $k \geq 6 \ \ 3k^2 > 18k > 17k + 6$

So

$k^4 + 4k^3 + 3k^2 -5k - 6 > k^4 + 4k^3 + 17k +6 -5k - 6 = k^4 + 4k^3 + 12k > 0$ for all k > 6

Is that OK?

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