1. ## Factorial Proof Help

I am trying to prove the equation for any integer (k) greater or equal to 0, that

2^(k+1) < (k+3)!

I am not sure how to expand the factorial to show that the equation will always be true?

Any Help Would Be Greatly Appreciated

2. ## Re: Factorial Proof Help

Originally Posted by doleary22
I am trying to prove the equation for any integer (k) greater or equal to 0, that

2^(k+1) < (k+3)!

I am not sure how to expand the factorial to show that the equation will always be true?

Any Help Would Be Greatly Appreciated
If we write...

$\displaystyle 2^{k+1}= \prod_{i=1}^{k+1} a_{i}$ (1)

$\displaystyle (k+3)!= \prod_{i=1}^{k+2} b_{i}$ (2)

... You observe that the product (2) has k+2 factors and the product (1) k+1 factors and for i=1,2,...,k+1 is $\displaystyle a_{i}\le b_{i}$, so that the conclusion is obvious...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Factorial Proof Help

I am still a little confused because I am not familiar with the notation. Is there any other way to solve the problem?

4. ## Re: Factorial Proof Help

Originally Posted by doleary22
I am still a little confused because I am not familiar with the notation. Is there any other way to solve the problem?
A more elementary concept probably doesn't exist...

$\displaystyle 2^{k+1}= 2 \cdot 2 \cdot ... \cdot 2\ (k+1\ \text{times})$ (1)

$\displaystyle (k+3)!= 2 \cdot 3 \cdot ... \cdot (k+2)\ \cdot (k+3)$ (2)

Do You understand now?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. ## Re: Factorial Proof Help

I understand how to solve each side of the equation, but what I am confused about is how did you determine the factors in your first response, and then how do I compare those two products to show that the one will always be larger than the other?

6. ## Re: Factorial Proof Help

Hello, doleary22!

Here is an observation.
I hope you can modify it into a proof.

$\displaystyle \text{Prove that, for any integer }k \ge 0\!:\;\;2^{k+1} \:<\: (k+3)!$

$\displaystyle \text{We have: }\;\underbrace{2\cdot2\cdot2\,\cdots\,2} \quad ^<_>\quad 1\cdot2\cdot\underbrace{3\cdot4\,\cdots\,(k+3)}$
. . . . . . . . . $\displaystyle ^{k+1\text{ factors}} \qquad\qquad\qquad \quad ^{k+1\text{ factors } >\: 2}$

7. ## Re: Factorial Proof Help

I am still a little confused. I know that the statement is correct, but I just can't seem to wrap my head around a logical way to write down a proof.