Proving an Inequality for n≥4: 2n<n! and 2n≤2((n-1)!)

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Homework Statement


Prove that if n is a natural number and is greater/equal to 4, then 2n<n!,
and show that 2n is less/equal to 2((n-1)!) follows.

The Attempt at a Solution


I'm thinking I just need to use induction on n for the first part, where I get the inequality
(n+1)! = n!*(n+1) > 2n(n+1) > 2n*(2) =2n+1.

After this conclusion, can I just say that 2n is less/equal to 2((n-1!) follows from my proof of induction or is there something else that I need to do?

Thanks in advance for your help! No answers please :)
 
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Yes, you can say that this follows from your induction proof.

But I think that the point of your exercise is to show that 2^n\leq 2(n-1)!, while using only that 2^n&lt;n!. So while saying that this follows from your induction is correct, I don't think that your exercise wants you to do that :smile: Just my interpretation, I could be wrong tho...
 
micromass said:
Yes, you can say that this follows from your induction proof.

But I think that the point of your exercise is to show that 2^n\leq 2(n-1)!, while using only that 2^n&lt;n!. So while saying that this follows from your induction is correct, I don't think that your exercise wants you to do that :smile: Just my interpretation, I could be wrong tho...

Yes, this isn't what I was too sure of, though the professor usually let's us just assume it true after our initial induction proof. If I need to, how should I start? :)
(it's late and I can't even think clearly.)
Thanks so much for your help and advice! :blushing:
 
Try this:

2^n=2.2^{n-1}\leq 2(n-1)!
 
I think I figured it out! :blushing:

From my inductive proof, I showed that 2n is less than n! or values of n greater/equal to 4.
So by my inductive proof, I can say that 2n-1 is less/equal to (n-1)! for values of n greater/equal to 5. (which is bold because I want to make sure that statement is correct)

It follows that if I multiply both sides by 2, we have that
2n is less/equal to 2(n-1)! for values of n greater/equal to 5.

That was so cool!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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