# How do I get this?

1. Jul 19, 2014

### shreddinglicks

1. The problem statement, all variables and given/known data

[(2n-1)!]/[(2n+1)!]

lim→∞

2. Relevant equations

3. The attempt at a solution

I have the answer from my solutions book, I just don't understand how

(2n+1)! = (2n+1)(2n)(2n-1)!

Can someone please explain this to me?

2. Jul 19, 2014

### SteamKing

Staff Emeritus
Think about what (2n+1)! means. If you were counting to (2n+1), what would be the other numbers you encountered if you counted all the way to (2n-1), stopped for a bit, and then started up again.

3. Jul 19, 2014

### Fredrik

Staff Emeritus
Is that equality true when n=1? Can you prove that for all integers k≥2, if it's true for n=k then it's also true for n=k+1?

4. Jul 19, 2014

### shreddinglicks

I see how it works. I was just wondering how you would come up with that without knowing it in the first place.

5. Jul 19, 2014

### Fredrik

Staff Emeritus
It's actually pretty easy. You want to evaluate $(2n-1)!/(2n+1)!$, so you should immediately think that it would be pretty nice if one of these numbers (the numerator or the denominator) is equal to the other times an integer. Then you remember the definition of !, and see that
$$(2n+1)!=(2n+1)(2n)(2n-1)(2n-2)\cdots 2\cdot 1 =(2n+1)(2n)(2n-1)!$$ If you want to avoid the ... notation, you can do it like this:
$$(2n+1)!=(2n+1)(2n)! =(2n+1)(2n)(2n-1)!$$

6. Jul 19, 2014

### shreddinglicks

I see, so basically you want to come up with some kind of combination that would include the numerator that also applies the factorial (!)

I assume I can apply this to similar problems in the future?

7. Jul 19, 2014

### SteamKing

Staff Emeritus
Why not? The stuff you learn is not supposed to have an expiration date.

8. Jul 19, 2014