Inequality inductive proof: Find the lowest integer value for n such that n! > 2^n.

[stunfiskery]

Inequality inductive proof: Find the lowest integer value for n such that n! > 2^n.

For a basic inequality proof question such as:

Find the lowest integer value for n such that n! > 2^n. Give a proof by induction.

(This isn't actually my homework question but close enough)

I am able to give a proof by induction if I am given that the lowest integer value is 4. However, if I am asked to find the lowest integer value, the only method I can plausibly think of is by guess and check. Is there a cleaner method which doesn't involve brute-force or is (educated) guess and check the only way?

 

[Steven G]

For a basic inequality proof question such as:

Find the lowest integer value for n such that n! > 2^n. Give a proof by induction.

(This isn't actually my homework question but close enough)

I am able to give a proof by induction if I am given that the lowest integer value is 4. However, if I am asked to find the lowest integer value, the only method I can plausibly think of is by guess and check. Is there a cleaner method which doesn't involve brute-force or is (educated) guess and check the only way?

Guess and check works but you need to be careful. For example, it is possible that some inequality may not work for n=1,2,3; work for n=4 and n=5; not work for 6, then work for n=7,8,9...

 

[JeffM]

For a basic inequality proof question such as:

Find the lowest integer value for n such that n! > 2^n. Give a proof by induction.

(This isn't actually my homework question but close enough)

I am able to give a proof by induction if I am given that the lowest integer value is 4. However, if I am asked to find the lowest integer value, the only method I can plausibly think of is by guess and check. Is there a cleaner method which doesn't involve brute-force or is (educated) guess and check the only way?

\(\displaystyle \dfrac{n!}{2^n} = \dfrac{\displaystyle \prod_{j=1}^nj}{\displaystyle \prod_{j=1}^n 2} = \displaystyle \prod_{j=1}^n\dfrac{j}{2}.\)

It should be obvious that the only term that will be under 1 is the first and that every term after the second is greater than 1. A bit of thought will show you that the product of the first and fourth terms is 1. All that remains is to check the case where n = 3.