# Inequality Mathematical Induction

1. Jul 11, 2010

1. The problem statement, all variables and given/known data
Prove the Inequality for the indicated integer values of n.

$$n!>2^n,n\geq4$$

2. Relevant equations

3. The attempt at a solution

For n=4 the formula is true because

$$4!>2^4$$

Assume the n=k

$$k!>2^k$$

Now I need to prove the equation for k+1

I can multiply both sides by 2 and have

$$2(k!)=2(k!)>(2)2^k=2^{k+1}$$

Is this what you would do next? I'm not quite sure what to do past this point.

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

2. Jul 11, 2010

### CompuChip

Why would you multiply by 2?
It's not 2k! you want to make a statement about, but (k + 1)! = (k + 1) k!

3. Jul 11, 2010

I dont know. When dealing with regular equations you would just set k=(k+1) but that's not the case here I think.

Wouldn't you be trying to prove that $$(k+1)!>2^{k+1}$$? Why would I be making a statement about '(k + 1)! = (k + 1) k!'? Where is the 2 raised to the power of (k+1) and why is there 2 factorials when there isn't in the original problem?

4. Jul 11, 2010

### CompuChip

Sorry, you are approaching the problem from a different angle than I expected.

Once you have 2 (k!) > 2k + 1, you only need to prove that (k + 1)! > 2 k! to be done with it.

The hint I gave, that (k + 1)! is equal to (k + 1) times k!, is still useful.

5. Jul 11, 2010

Ahh, so you could rewrite

(k + 1)! > 2 k!

as

k!(k+1) > 2 k!

Here you can see the when k > 1, k!(k+1) > 2 k! and since the original formula is only supposed to work for numbers greater than or equal to 4 its fine.

Thanks.

6. Jul 11, 2010

### hunt_mat

We are required to show that $$n!>2^{n}$$, for the k+1 term, multiply by k+1 to obtain:
$$(k+1)k!=(k+1)!>2^{k}(k+1)>2^{k+1}$$
Since k+1>2 for k>2 and we are done