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Math Help - divergent series

  1. #1
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    divergent series

    somebody know the value of the following divergent serie

    sum[(-1)^m e^m,{1,Infinity}]
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  2. #2
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    Re: divergent series

    A series is called "divergent" if and only if it has no value!
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  3. #3
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    Re: divergent series

    it can be divergent but summable (C,K) ect..
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  4. #4
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    Re: divergent series

    Quote Originally Posted by capea View Post
    it can be divergent but summable (C,K) ect..
    You can only sum a finite number of terms of a divergent series.
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  5. #5
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    Re: divergent series

    I believe capea is talking about "Cauchy" or "mean" summability. capea, what do your "C" and "K" refer to?
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    Re: divergent series

    Quote Originally Posted by capea View Post
    somebody know the value of the following divergent serie

    sum[(-1)^m e^m,{1,Infinity}]
    A quick look suggests that this is neither Abel nor Cesaro summable

    CB
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  7. #7
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    Re: divergent series

    Its an alternating series and by alternating series test it is divergent. Therefore it can not be calculated
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  8. #8
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    Re: divergent series

    for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),
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  9. #9
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    Re: divergent series

    Quote Originally Posted by capea View Post
    for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),
    Um, no it's not. The value partial sum value is here, and since it is divergent, it does not have a sum.
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  10. #10
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    Re: divergent series

    Quote Originally Posted by capea View Post
    for example the Sum((-1)^n Log(n),{n,1,Infinite}) is a divergent serie but is sumable and his value it is -1/2 Log(Pi/2),
    What capea wrote is 'almost true'... in the sense I try to explain now...

    The so called 'Dirichlet eta function' is defined as...

    \eta (s)= - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{s}} = (1-2^{1-s})\ \zeta(s) (1)

    Writing the (1) as...

    \eta (s)= \frac{1}{2} + \frac{1}{2}\ \sum_{n=1}^{\infty} (-1)^{n-1}\ \{n^{-s} - (n+1)^{-s}\} (2)

    ... You can, with a little of patience, demonstrate that (2) converges for \text{Re}\ s > -1 and that is...

    \eta^{'} (0)= \frac{1}{2}\ \ln \frac{\pi}{2} (3)

    Now if You compute 'directly' the derivative of (1) You obtain...

     - \eta^{'} (s) = \sum_{n=1}^{\infty} \frac{(-1)^{n}\ \ln n}{n^{s}} (4)

    ... so that setting in (4) s=0 and considering (3) You obtain...

    \sum_{n=1}^{\infty} (-1)^{n} \ln n = - \frac{1}{2}\ \ln \frac{\pi}{2} (5)

    Kind regards

    \chi \sigma
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  11. #11
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    Re: divergent series

    for example

    1^-s log(1)-2^-sLog(2).....==(2^(1-s)-1)Zetaī(s)-2^(1-s)Log(2 )Zeta(s)) so when s=0
    you get the result
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