Inequality Mathematical Induction

[themadhatter1]

Homework Statement

Prove the Inequality for the indicated integer values of n.

n!>2^n,n\geq4

Homework Equations

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

 

[CompuChip]

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

 

[themadhatter1]

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

I don't 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?

 

[CompuChip]

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

Once you have 2 (k!) > 2^{k + 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.

 

[themadhatter1]

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

Once you have 2 (k!) > 2^{k + 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.

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.

 

[hunt_mat]

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