# New member, induction problem

#### 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.

#### craig

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

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 $$\displaystyle n! > n^3$$ is 6.

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

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

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

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

It looks like you're doing fine.

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

#### pastmyprime

No, sorry, I can't prove it.

#### roninpro

Did you try using induction?

#### pastmyprime

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

Step 1 rearrange

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

Step 2 induction

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

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

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

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

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

So

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

Is that OK?