# Is this a valid proof for n! >2^n for all n>3

1. Sep 25, 2016

### Battlemage!

1. The problem statement, all variables and given/known data
Show that n!>2n for all n>3.

2. Relevant equations
I will attempt to use induction.

3. The attempt at a solution
We want to show that n!>2n for all n>3.
Consider the case when n=4.
$$4! = 24 > 2^4 =16.$$

We want to show by way of induction that if the inequality is true for some k greater than 4, it is true for k+1.

Assume the inequality holds for k>4. Then,
$$2^k < k!$$
$$2*2^k < 2k!$$
$$2^{k+1} < 2k!$$
for k>4.

But 2k! < (k+1)! for all k>4. Therefore

$$2^{k+1} < (k+1)!$$

and so by induction it follows that n! > 2n for all n > 3.

2. Sep 25, 2016

### micromass

Staff Emeritus
Proof is ok, but I'm going to be annoying:

Why?

Why?

Why?

3. Sep 25, 2016

### Battlemage!

Thank you. That always helps.

The reason I assumed it held for k>4: 4 is the very first integer greater than 3. I figured if I can establish by induction that it holds for 4, then 5, 6, 7,...k, k+1,... then it must hold for all numbers greater than 3. There is no upper bound on k. I did spin my wheels a bit at first since I'm used to the base case being k=1.

As for the second part, I multiplied by 2 because that would allow me to easily get the k+1 case because the base for the power is 2, and xax = xa+1. (also I don't believe the way everything relates to the number 2 here is coincidental. I think there might be a deeper relation between the base of the power and when it is larger than the factorial).

Because I multiplied both sides by a positive integer, so the inequality should remain. The multiplication was done so I can get the k+1 case for the exponent. I then kind of lucked into the next part. I knew I had to get to some point where the left was less than the target quantity, and this worked out well.

Because:
$$\lim_{n\rightarrow ∞ } \frac{2n!}{(n+1)!} = 0$$
*EDIT I had n approaching zero for some reason

and as for k =3,2,1 and 0, I manually tested them. k>4 is overkill for this part. But I figured that in order remain consistent I'd stick with k >4, since it is necessary for comparing the exponential to the factorial.

Last edited: Sep 25, 2016
4. Sep 25, 2016

### Math_QED

This is wrong.

5. Sep 25, 2016

### PeroK

There is, however, a small but significant error in your original proof. This is not related to insufficient justification, but is a genuine flaw.

Big hint: what happened to the case $n = 5$? Where did you cover that?

6. Sep 25, 2016

### Battlemage!

I wrote it wrong. It's supposed to be as n approaches infinity.

7. Sep 25, 2016

### micromass

Staff Emeritus
That means that for all $\varepsilon>0$, there is an $N$ such that if $n>N$ we have $2n! <\varepsilon (n+1)!$.

Your claim is that for all $n>3$ we have $2n!<(n+1)!$. Those are very distinct statements.

8. Sep 25, 2016

### Battlemage!

Oh I see what you're saying.

For that part when I answered your why question by posting the limit I just meant that because I knew that limit was zero it led me to the idea of trying to set up an inequality where

2k+1 < 2k! < (k+1)! for k>4.

I did not actually give a mathematical reason haha. I will think on it and try to articulate a better one.

9. Sep 25, 2016

### Ray Vickson

If you mean $2 \times 2^k < (2k)!$ then that does not follow from the previous line. If you mean $2 \times 2^k < 2 \times k!$ then that does follow.

10. Sep 25, 2016

### SammyS

Staff Emeritus
@Battlemage!

PeroK's question has been ignored. It raises an important point!

Last edited: Sep 25, 2016
11. Sep 26, 2016

### Battlemage!

Yes that's what I meant. I figured a factorial was only grouped with other terms if there were parentheses, but then again it's not like a factorial is in PEMDAS so I would be ignorant of such rules.

I post these questions at work and unfortunately ran out of time. I was thinking about it, though. Sadly I'm at a loss. More information below...

Okay... let me see...

I have manually tested 5 and it works. 5! = 120 >25 = 32.
Likewise with my last step, 25 = 32 < 2(5!) = 240 < (5+1)! = 6! = 720.

As for the induction process, I've never done it where the base case wasn't one. Did I do it wrong with my base case of 4?

Since I included k greater than 4, doesn't that already include 5. I don't quite understand what you're getting at. Any other hints? :/

EDIT- since we're here, I did find a different way to do it working the other way.

$$(k +1)! = (k + 1)k! > (k + 1)2^k$$
and since k > 4 then
$$(k + 1)! > 2^k * 2$$
$$(k + 1)! > 2^{k+1}$$

12. Sep 26, 2016

### PeroK

What you did was:

a) Showed it was true for $n = 4$

b) Showed that if it was true for $k > 4$ it was true for $k+1$

But, $k > 4$ starts at $k =5$. So, you didn't cover the step $P(k = 4) \ \Rightarrow \ \ P(k = 5)$.

The inductive stage of your proof should have been for $k \ge 4$.

On the other issues, I think you have let the questions confuse you and over-elaborated your answers. Your proof depended only on (if $a, b$ are positive):

$a < b \ \Rightarrow \ 2a < 2b$

And

$k > 2 \ \Rightarrow \ 2a < ka$

You must see this, but recognising these simple facts and stating them simply and mathematically is something that generally does not seem to come naturally to many people.

On a final point, I would tend to write either $2(k!)$ or $(2k)!$ to avoid any ambiguity. Like you, I don't know what PEMDAS has to say about factorials.

13. Sep 26, 2016

### Battlemage!

So essentially if I replace the > with a ≥ that would fix the issue?

So I could have omitted this part?

Yeah I think I'll do that from now on.

I appreciate the help by the way.

14. Sep 26, 2016

### PeroK

Yes, but you should see that yourself.

There was nothing you could have omitted. You just needed to justify what you were doing a little more.