# Thread: Series involving factorials and alternating terms

1. ## Series involving factorials and alternating terms

Hello,

I'm not very familiar with how to post the expression here as it is, but this is a link to an entry for it on WolframAlpha: sum from i&#61;1 to n of &#40;-1&#41;&#94;&#40;i&#43;1&#41;&#42;&#40;i&#41;&#42;& #40;n&#43;1&#41;&#33;&#42;1&#47;&#40;&#40;i&#43;1& #41;&#33;&#42;&#40;n-i&#41;&#33;&#41; - Wolfram|Alpha

I have also attached an image that shows what the series is. It also shows that the answer is equal to 1. However, Wolfram doesn't show any steps and despite my attempts, I haven't been able to prove that. I would rather not have the entire answer spoiled, but any help would be appreciated.

2. ## Re: Series involving factorials and alternating terms

Hey Erdoss.

Note that n*(n-1)! = n! so if you simplify the denominator (in terms of it being a factorial) then see if you get a cancellation.

3. ## Re: Series involving factorials and alternating terms

I'm not sure if I misunderstood you, but I know why the right hand side of the equation is equal to 1. My question is about how to get to that expression from the series on the left hand side.

4. ## Re: Series involving factorials and alternating terms

I think I misunderstood you then, and I'm not sure what identity was used.

5. ## Re: Series involving factorials and alternating terms

Exactly, I don't know how the series was simplified.

6. ## Re: Series involving factorials and alternating terms

Do you know where to get (or do you have) a table of combinatorial identities that you can comb through?

7. ## Re: Series involving factorials and alternating terms

Binomial coefficient - Wikipedia, the free encyclopedia

There are some in that page, but I haven't been able to put any of them to good use. :S