Simplify a factorial

**UDaytonFlyer** · September 20th 2008, 05:38 PM

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!

**Jhevon** · September 20th 2008, 05:41 PM

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?

**Prove It** · September 20th 2008, 05:44 PM

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

**UDaytonFlyer** · September 21st 2008, 04:18 PM

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?

**UDaytonFlyer** · September 21st 2008, 04:20 PM

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?

**Prove It** · September 22nd 2008, 06:17 AM

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

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