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Math Help - Trouble proving convergence/divergence

  1. #1
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    Trouble proving convergence/divergence

    I've got a series which i applied the ratio test to, after simplifying a bit im left with:

    (r+1)*5*(2r)!
    (2(r+1))!

    What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
    But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
    (sorry for lack of correct math symbols)

    Thanks
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by sj247 View Post
    I've got a series which i applied the ratio test to, after simplifying a bit im left with:

    (r+1)*5*(2r)!
    (2(r+1))!

    What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
    But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
    (sorry for lack of correct math symbols)

    Thanks




    \frac{(2r)!}{(2r+2)!}=\frac{1}{(2r+1)(2r+2)}
    Last edited by Also sprach Zarathustra; January 6th 2011 at 02:23 PM. Reason: Don't post the original question!
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  3. #3
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    How about this:

    <br />
\dfrac{5(r+1)(2r)!}{(2(r+1))!}=\dfrac{5(r+1)(2r)!}  {(2r+2)!}=\dfrac{5(r+1)(2r)!}{(2r+2)(2r+1)(2r)!}\d  ots<br />
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  4. #4
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    Quote Originally Posted by sj247 View Post
    I've got a series which i applied the ratio test to, after simplifying a bit im left with:

    (r+1)*5*(2r)!
    (2(r+1))!

    What should be my next step? i was thing of dividing by (2(r+1))! before tending r to inf.
    But that leaves you with exact same equation, leaving it like that it would appear the equation tends to 0.
    (sorry for lack of correct math symbols)

    Thanks
    Note that (2(r+1))! = (2r + 2)! = (2r+2)(2r+1)(2r)! so yes, the limit is zero and a conclusion can be made ....
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  5. #5
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    Thanks guys . I got similar in my working, just needed to check.
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    How about this:

    <br />
\dfrac{5(r+1)(2r)!}{(2(r+1))!}=\dfrac{5(r+1)(2r)!}  {(2r+2)!}=\dfrac{5(r+1)(2r)!}{(2r+2)(2r+1)(2r)!}\d  ots<br />
    \displaystyle = \frac{5(r+1)(2r)!}{2(r+1)(2r+1)(2r)!}...
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  7. #7
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    cheers
    Last edited by sj247; January 7th 2011 at 11:58 AM.
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