# Relal Analysis Proof

1. Sep 9, 2009

### cmajor47

1. The problem statement, all variables and given/known data
Prove if n$$\geq$$4, n!>2n.

2. Relevant equations

3. The attempt at a solution
Proof: (by induction) Prove for n=4
4!>24
24>16
true for n=4
Prove for n+1
n! (n+1) > 2n (n+1)
(n+1)! > 2n (n+1)

This is where I get stuck. I don't know where to go from here.

2. Sep 9, 2009

### snipez90

Use the fact that $n \geq 4$

3. Sep 9, 2009

### VeeEight

Expand (2n)(n+1). How does this relate to (2)(2^n)

4. Sep 9, 2009

### cmajor47

I'm guessing the last reply meant expand 2n (n+1).
I know that expanded it is 2nn + 2n but I don't see how that relates to (2)(2n) or 2n+1

5. Sep 9, 2009

### n!kofeyn

Re: Real Analysis Proof

I recommend following snipez90's hint. If $n\geq 4$, then $2^n(n+1)\geq \text{something}=2^{n+1}$. Find the something.

6. Sep 9, 2009

### lanedance

hmmmm... not sure if I'm reading it correctly, but I don't understand where the (n+1) came from on the right hand side - 2n (n+1)?

For the induction proof:

Want to show n! > 2n, for n>4

First, prove true for n = 4.... done

2nd, show (n+1)! > 2n+1 assuming n! > 2n

you know
(n+1)! = (n+1).n!
2n+1 = 2.2n

now start from n! > 2n, and see if you can manipulate it into what you want

7. Sep 9, 2009

### n!kofeyn

lanedance: The (n+1) came from multiplying both sides of n!>2n by (n+1). Also, you just repeated the work that the original poster had already completed (and posted), so I'm not for sure what the point of your post was.

8. Sep 10, 2009

### lanedance

was trying implying to start with two inequalities & multiply them together, which amounts to the same thing you're doing:
n! > 2n,
and
(n+1) > ?

9. Sep 10, 2009

### lurflurf

you were almost there
here is what has already been done
by induction
24=4!>2^4=16
let n>=4
assume
n!>2^n
it is desired to show that
(n+1)!>2^(n+1)
now moving on
let us write
(n+1)!=(n+1)n!
since
n!>2^n
we see
by usual rules for inequalities
some twelve year olds have informed me the reule we need is
Multiplication Axiom of Order
if c>0
and a>b
then ac>bc
let use take a=n!,b=2^n,c=n+1
(n+1)!=(n+1)n!>(n+1)2^n
now if it could be shown that
n+1>2 (when n>=4)
we would be able to go futher towards the solution (by applying Multiplication Axiom of Order again)

Last edited: Sep 10, 2009
10. Sep 10, 2009

### n!kofeyn

lurflurf: I think it's a little bit of bad form to come in and just solve the problem. I and others had given hints for the original poster, but now you have completely solved it. It seems you didn't even bother reading the other posts, but just read the original and then posted the answer.

11. Sep 10, 2009

### lurflurf

Thank you for your opinion. That is a mighty special talent you have there determining with what others have read, but your gift has failed you. I did have the misfortune of reading the posts between the original and my own. You raise an interesting point. The question posted was almost solved. It would be helpful if questions included a detailed explanation of what thee poster found difficult. It is often the case that when a person hass difficulties solving a problem they also have difficulties discibing their difficulties. Some people such as fans of so called Inquiry-based learning would favor erring on the side of being unhelpful. If a small hint completes the problem, it is still a small hint.

12. Sep 10, 2009

### lurflurf

a simpler take
from the step
(n+1)! > (2^n) (n+1)
it may be helpful to show
n+1>2

13. Sep 11, 2009

### n!kofeyn

Of course it is helpful to do that, and that is why it was suggested already in posts 2, 5, 6 and 8. Also, it's okay to give a full solution, but I don't think it is good when the original poster hadn't responded to the hints given. This on top of the fact that your post added nothing new, was convoluted, and was edited to add an unnecessary, downgrading remark.