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Math Help - I need help about hypergeometric series

  1. #1
    Junior Member miss_lolitta's Avatar
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    Angry I need help about hypergeometric series

    The hyper geometric series :the infinite series

    [1+ (ab/1!c)x+[a(a+1)b(b+1)/(2!c(c+1)]x^2+{[a(a+1)(a+2)b(b+1)(b+2)]/[3!c(C+1)(c+1)(c+2)]}x^3+....]

    Where a,b and c are neither 0 nor negative integers,,

    Please can anyone help me to prove by explaining that the hyper geometric series is converges??!

    any help would be appreciated ,thank you
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  2. #2
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    There are a few ways to write this, but try looking at it like this.

    S_{n}=1+\sum_{n=0}^{\infty}\frac{(a+n)!(b+n)!x^{n+  1}}{(n+1)!(c+n)!}

    Whew. And I would definitely take the ratio test from here.
    Last edited by Jameson; April 9th 2006 at 05:10 PM.
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  3. #3
    Junior Member miss_lolitta's Avatar
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    Thanks Jameson

    but if you do'nt have any objection,can you give me the complete answer

    I'm confused and not sure if I can do this or no

    Please Jameson..

    thanks for any responses.
    Last edited by miss_lolitta; April 9th 2006 at 10:29 PM.
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  4. #4
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    Ok. I'll try to Tex this all out.

    Let b_{n}=\frac{a_{n+1}}{a_n} or a_{n+1}*\frac{1}{a_n}

    So b_n=\frac{(a+n+1)!(b+n+1)!x^{n+2}}{(n+2)!(c+n+1)!}  *\frac{(n+1)!(c+n)!}{a+n)!(b+n)!x^{n+1}}

    All I did was substitute (n+1) for in, then divide that expression by a_n. But to simplify, I just flipped a_n and multiplied. Now for some cancellation.

    b_n=\left(\frac{(a+n+1)(a+n)!(b+n+1)(b+n)!x^{n+1}*  x}{(n+2)(n+1)!(c+n+1)(c+n)!}\right) \left(\frac{(n+1)!(c+n)!}{(a+n)!(b+n)!x^{n+1}} \right)

    For some reason, that Latex won't make that last bracket! Anyway, cancel some terms out and you'll get.

    b_n=\frac{(a+n+1)(b+n+1)x}{(n+2)(c+n+1)}

    Now you have your common ratio! So take the limit as n\rightarrow\infty to see how the terms trend torwards infinity, and you should be able to show that this converges for all x.

    Hopefully that makes sense.
    Last edited by CaptainBlack; April 11th 2006 at 09:39 PM.
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  5. #5
    Junior Member miss_lolitta's Avatar
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    I try to take limit as <br />
n\rightarrow\infty<br />
I get <br />
\lim b_n=x<br />

    And conclude that the series diverges for x>1 , converges for x<1 and test failed at x=1 .

    We can use Gauss' test to see series behavior ..

    But I have one question :
    Is it necessary that we test series behavior at x=-1 or no??!

    thanks
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  6. #6
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    Yes. Sorry. The series converges for |x|<1 and you must test the endpoints of 1 and -1 for convergence.
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  7. #7
    Junior Member miss_lolitta's Avatar
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    but that's very long & hard

    anyway..

    thanks so much for Jameson
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