# Thread: New member, induction problem

1. ## New member, induction problem

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.

2. 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?

3. 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.

4. 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?

5. No, sorry, I can't prove it.

6. Did you try using induction?

7. 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?