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Math Help - sum series

  1. #1
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    sum series

    Some han help to calculate the serie
    \sum _{z=1}^{\infty } \frac{(-1)^{-1+z}}{\left(-\frac{4}{3}+\sqrt{2+z}\right)^3}
    thanks
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  2. #2
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    Re: sum series

    Wolfram can't get an exact answer, but the series is convergent by the Alternating Series Test.

    http://www.wolframalpha.com/input/?i=sum+of+%28-1%29^%28z-1%29%2F%28Sqrt%282%2Bz%29-4%2F3%29^3+with+z+from+1+to+infinity
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  3. #3
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    Re: sum series

    Quote Originally Posted by Prove It View Post
    Wolfram can't get an exact answer, but the series is convergent by the Alternating Series Test.

    http://www.wolframalpha.com/input/?i=sum+of+%28-1%29^%28z-1%29%2F%28Sqrt%282%2Bz%29-4%2F3%29^3+with+z+from+1+to+infinity
    What I find amusing about the Alpha attempt to sum this series is that it reports negative results for a couple of convergence tests, and ignores the fact that it is obviously convergent (in fact it is obviously absolutly convergent).

    (Note I asked for a simplified rearrangenment of this which it failed to notice converged by the alternating series convergence test)

    CB
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  4. #4
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    Re: sum series

    \frac{13095}{121}+\frac{81}{\sqrt{2}}-\frac{243 \sqrt{3}}{121}-2 \text{PolyGamma}\left[1,\frac{1}{9}\right]+2 \text{PolyGamma}\left[1,\frac{11}{18}\right]-\frac{32}{27} \text{PolyGamma}\left[2,\frac{11}{18}\right]+\frac{32}{27} \text{PolyGamma}\left[2,\frac{10}{9}\right]
    working on it I found an aproximate sum
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  5. #5
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    Re: sum series

    Quote Originally Posted by capea View Post
    \frac{13095}{121}+\frac{81}{\sqrt{2}}-\frac{243 \sqrt{3}}{121}-2 \text{PolyGamma}\left[1,\frac{1}{9}\right]+2 \text{PolyGamma}\left[1,\frac{11}{18}\right]-\frac{32}{27} \text{PolyGamma}\left[2,\frac{11}{18}\right]+\frac{32}{27} \text{PolyGamma}\left[2,\frac{10}{9}\right]
    working on it I found an aproximate sum
    If that is an approximation how is it more useful than say: 13.3083968 +/- 0.0000359 ?

    CB
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  6. #6
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    Re: sum series

    Thanks for your comments on the previous approximation expressions lack a term 0(x^-4), but i have not clue how to calculate any idea ?
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  7. #7
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    Re: sum series

    Quote Originally Posted by capea View Post
    Thanks for your comments on the previous approximation expressions lack a term 0(x^-4), but i have not clue how to calculate any idea ?
    What is wrong with summing 100000 terms of the series?

    CB
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  8. #8
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    Re: sum series

    Quote Originally Posted by capea View Post
    Thanks for your comments on the previous approximation expressions lack a term 0(x^-4), but i have not clue how to calculate any idea ?
    What would x be? You have not told us how you obtained your approximation so we have no idea what x is here.

    CB
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