How do I use induction for inequalities?

Asked Feb 23 '23 at 00:59

Active Feb 23 '23 at 01:36

Viewed 94 times

$2^n$ < $(n+1)!$

I'm not too familiar with how to do questions structured like these, as the examples I find online seem to randomly add things into the inequality to make is correct. The farthest I've gone is knowing the base case is right, substitution with k ($2^k$ < $(k+1)!$, and knowing I need to prove: $2^{k+1}$ < $(k+2)!$. Beyond this, I don't actually know what to do from here. Can someone help point me in the right direction? (it's for $n\geq 2$, and please don't give me the answer directly.)

EDIT: My idiot brain switched the inequality. Sorry for the confusion.

edited Feb 23 '23 at 01:36

WindSoul

2,178

asked Feb 23 '23 at 00:59

James

1

How do you get the left hand side of the statement you want to prove from your induction hypothesis? – GReyes Feb 23 '23 at 01:01
You replace k with k+1 don't you? – James Feb 23 '23 at 01:01
2

The inequality looks backwards – J. W. Tanner Feb 23 '23 at 01:02
1

Your $k$ is fixed, and you have to prove it for $k+1$. Do you realize that $2^{k+1}=2\cdot 2^k$? – GReyes Feb 23 '23 at 01:02
1

@J.W.Tanner Yes, it is definitely backwards. – GReyes Feb 23 '23 at 01:03
Crap you're right. Fixed – James Feb 23 '23 at 01:04
1

When you use induction, you need to connect the statement for $k$ with the statement for $k+1$, and that depends on the concrete statement. Would you agree that if you know that $2^k<\textrm{something}$, then $2^{k+1}<2\cdot\textrm{something}$? – GReyes Feb 23 '23 at 01:07
Yes. That makes sense, but when would I use this? – James Feb 23 '23 at 01:08
I'm sure this question has been asked, and answered, on this site many times. Please search for those earlier questions. – Gerry Myerson Feb 23 '23 at 01:09
https://math.stackexchange.com/questions/111146/prove-by-induction-that-n2n and https://math.stackexchange.com/questions/76946/prove-the-inequality-n-geq-2n-by-induction and https://math.stackexchange.com/questions/2668193/prove-by-induction-n-ge-2n-1-for-any-n-ge-1 and https://math.stackexchange.com/questions/3985979/show-that-n-%e2%89%a5-2n-1-is-true-by-induction-for-n-being-a-positive-integer and many, many more. – Gerry Myerson Feb 23 '23 at 01:11
1

You know that $2^k<(k+1)!$. Therefore, $2^{k+1}<2\cdot(k+1)!$. It would be enough, by transitivity, to prove that $2\cdot(k+1)!<(k+2)!$. – GReyes Feb 23 '23 at 01:12
1

Does this answer your question? factorial proof using induction - found using an Approach0 search. Note there are also many other similar questions, such as Prove the inequality $n! \geq 2^n$ by induction suggested by Henry above as a duplicate, the many ones listed in Gerry Myerson's comment, and a few other ones in that Approach0 search I mentioned. – John Omielan Feb 23 '23 at 01:15

How do I use induction for inequalities?

0 Answers0