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Math Help - Series Problem With Factorials

  1. #1
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    Series Problem With Factorials

    I have redone this problem many times, and can't seem to figure it out!

    The problem is to use the ratio test to find the limit and convergence of this series:


    \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}


    Now, using the ratio test, I come up with this:

    \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}



    How do I expand this to cancel the series out? This is what I tried:

    \lim_{n \rightarrow \infty} \frac{(1*2...n)(1*2...(n+1)(n+2)(n+3)) (24*720...(2(n+1)) )}{(2*24*720...(2n)) (1*2...(n+1)) (1*2...(n+1)(n+2)(n+3)(n+4))}


    However, when I cancel everything out, I get \lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)} which results in the limit being 0.

    But 0 is not the correct limit.

    Any help would be very appreciated!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lysserloo View Post
    I have redone this problem many times, and can't seem to figure it out!

    The problem is to use the ratio test to find the limit and convergence of this series:


    \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}


    Now, using the ratio test, I come up with this:

    \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}
    Try cancelling!
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  3. #3
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    Haha yes I know. I didn't mean to post when I did, I wasn't quite finished typing out my problem. Thank you for your quick reply though!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lysserloo View Post
    I have redone this problem many times, and can't seem to figure it out!

    The problem is to use the ratio test to find the limit and convergence of this series:


    \sum_{i=1}^\infty \frac{(2n)!}{n!n+3)!}


    Now, using the ratio test, I come up with this:

    \lim_{n \rightarrow \infty} \frac{n!(n+3)!(2(n+1))!}{(2n)!(n+1)!(n+4)!}
    \frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!}  {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3)  !}\cdot\frac{n!(n+3)!}{(2n)!} =\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4



    How do I expand this to cancel the series out? This is what I tried:

    \lim_{n \rightarrow \infty} \frac{(1*2...n)(1*2...(n+1)(n+2)(n+3)) (24*720...(2(n+1)) )}{(2*24*720...(2n)) (1*2...(n+1)) (1*2...(n+1)(n+2)(n+3)(n+4))}


    However, when I cancel everything out, I get \lim_{n \rightarrow \infty} \frac{2n}{4n(n+4)} which results in the limit being 0.

    But 0 is not the correct limit.

    Any help would be very appreciated!
    Quote Originally Posted by lysserloo View Post
    Haha yes I know. I didn't mean to post when I did, I wasn't quite finished typing out my problem. Thank you for your quick reply though!
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    \frac{(2(n+1))!}{(n+1)!(n+4)!}\cdot\frac{n!(n+3)!}  {(2n)!}=\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n+4)n!(n+3)  !}\cdot\frac{n!(n+3)!}{(2n)!} =\frac{(2n+2)(2n+1)}{(n+1)(n+4)}\to 4
    I do not follow what you did here at all : (

    I think the problem is that I really don't understand factorials very well...I'm not sure.

    Thank you for your reply though.

    Edit: After looking at it, the only part I don't understand is after the first equals sign. Where did that n!(n+3)! on the bottom come from? The rest I understand.
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