Precalculus Show that n! > 2^n for all integers n >4

[PeachBlossom]

Okay so now I'm studying number theory. I don't know If I understand how to approach this question. On this question do I need to do do mathematical induction? Is my thought process correct?

Thanks in advance for help! Appreciate it allot.

 

[lev888]

You don't have to. But if it works it would be a valid proof. Give it a try.

 

[PeachBlossom]

I did like this. But in the end where I need to prove n=k+1 gets me confused. Don't know how to approach it and what to do with k>4.

 

[lev888]

In steps 2 and 3 for some reason you switched from 2^k to k².

 

[pka]

View attachment 14232View attachment 14233

I did like this. But in the end where I need to prove n=k+1 gets me confused. Don't know how to approach it and what to do with k>4.

If \(\displaystyle K!>2^K\) then look at \(\displaystyle (K+1)!=(K+1)\cdot K!\).
We know that \(\displaystyle K>2\) WHY?
So that means \(\displaystyle (K+1)\cdot K!>2\cdot 2^K=2^{K+1}\) HOW & WHY?
That proves it.

 

[PeachBlossom]

I did like this

 

[PeachBlossom]

Is this an acceptable answer? Is there an easier way to do induction or is this the easiest way?

 

[pka]

Is this an acceptable answer? Is there an easier way to do induction or is this the easiest way?

There is a much easier way: See reply #5.

 

[JeffM]

I HATE the way induction is taught. (This is purely personal; at least some other helpers here do not agree with me.) You are certainly not trying to prove that n = k + 1.

What is to be proved is

[MATH]\text {For all } n \in \mathbb N,\ (n + 3)! > 2^{(n+3)}.[/MATH]
First, we show that the set of such numbers is not the empty set.

[MATH]n = 1 \implies (n + 3)! = 4! = 24 > 16 = 2^k.[/MATH]
Therefore there is a non-empty set of natural numbers with the property desired. That set contains 1 (and we want to prove that it contains many more natural numbers, but have not proved that yet). So we choose an arbitrary member of the set. I'd write that in math language as

[MATH]\therefore \exists \ \mathbb K \text { such that } k \in \mathbb K \implies k \in \mathbb N \text { and } (k + 3)! > 2^{(k+3)}.[/MATH]
What we want to do is to show that the property that we know k has necessarily applies to k + 1 as well. But I would start by finding out how the terms involving k + 1 can be expressed using k rather than k + 1.

So the first thing I would do is to express the terms that will be used to describe the property for k + 1 in terms of what what we already know about k.

[MATH]\{(k + 1) + 3\}! = (k + 4)!.[/MATH]
That is simple enough, right? OK, that is one of the terms without k + 1.

[MATH]2^{\{(k+1)+3\}} = 2^{(k+4)}.[/MATH]
And there is the other term without k + 1. Again, just simple algebra.

(I don't guarantee that these two steps are always this easy, but very frequently they are.)

Now we said that k is in N, which means of course that k + 1 is as well. (This step is usually ignored as trivially obvious.)

But usually that is not the really important clue hidden in the fact k is in N.

[MATH]k \in \mathbb N \implies n \ge 1 \implies k + 4 \ge 5 > 2.[/MATH]
[MATH]\therefore (k + 4)! = (k + 4) * (k + 3)! \ge 5 * (k + 3)! > 2 * (k + 3)!.[/MATH]
[MATH]\text {And } (k + 3)! > 2^{(k+3)} \implies 2 * (k + 3)! > 2 * 2{(k+3)} = 2^{(k+4)}.[/MATH]
[MATH]\therefore (k + 4)! > 2 * (k + 3)! > 2^{(k+4)} \implies[/MATH]
[MATH]\{(k + 1) + 3\}! > 2^{\{(k+1)+3\}}. \text {Q.E.D.}[/MATH]