1. ## Simplify a factorial

Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

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

And another:

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

My goal for both of the above is to simplify the ratio of factorials. I know that a factorial is the product of all the numbers (i.e. 6! is 1*2*3*4*5*6= 720) but I can't grasp how that works when the n's get thrown in there. Any help would be greatly appreciated!

2. Originally Posted by UDaytonFlyer
Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

[(2n-1)!]/[(2n+1)!]
hint: $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2) \cdots 2 \cdot 1$

(2n+2)!]/[(2n)!]
hint: $(2n + 2)! = (2n + 2)(2n + 1)(2n) \cdots 2 \cdot 1$

look at the tail of both factorials that i gave you. what are they?

3. Originally Posted by UDaytonFlyer
Hello! Thanks for taking the time to look at my post. I am studying sequences and series and I don't understand the part about factorials. An example problem is below:

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

And another:

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

My goal for both of the above is to simplify the ratio of factorials. I know that a factorial is the product of all the numbers (i.e. 6! is 1*2*3*4*5*6= 720) but I can't grasp how that works when the n's get thrown in there. Any help would be greatly appreciated!
$(2n+1)! = (2n+1)\times 2n \times (2n-1)!$

So $\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\times 2n \times (2n-1)!} = \frac{1}{2n(2n+1)}$

4. Originally Posted by Jhevon
hint: $(2n + 1)! = (2n + 1)(2n)(2n - 1)(2n - 2) \cdots 2 \cdot 1$

hint: $(2n + 2)! = (2n + 2)(2n + 1)(2n) \cdots 2 \cdot 1$

look at the tail of both factorials that i gave you. what are they?
Hm, I think it would be the last two "terms" of the factorial? and they would "cancel" when divided by a similar factorial because they share the same terms?

5. Originally Posted by Prove It
$(2n+1)! = (2n+1)\times 2n \times (2n-1)!$

So $\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n+1)\times 2n \times (2n-1)!} = \frac{1}{2n(2n+1)}$
So for the (2n-1) part you stopped expanding the factorial there because from that point forward the two factorials would have the same terms and thus cancel, right?

6. Originally Posted by UDaytonFlyer
So for the (2n-1) part you stopped expanding the factorial there because from that point forward the two factorials would have the same terms and thus cancel, right?
That's right