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Math Help - series

  1. #1
    Super Member Random Variable's Avatar
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    series

     \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \ \frac{\ln n}{n}

    The sum should be  \displaystyle \gamma \ln 2 - \frac{(\ln 2)^{2}}{2} .
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  2. #2
    MHF Contributor chisigma's Avatar
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    Remembering the definition of the 'Dirichlet eta function'...

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

    ... it is easy to see that is...

    \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n}= \eta^{'} (1) (2)

    Not so easy however is the numerical computation of (2)...

    Kind regards

    \chi \sigma
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  3. #3
    Senior Member roninpro's Avatar
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    I haven't thought about this deeply or anything, but this almost seems like it can be done using some kind of summation by parts argument.

    Summation by parts - Wikipedia, the free encyclopedia
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  4. #4
    MHF Contributor chisigma's Avatar
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    Let's start with the so called 'Dirichlet eta function'...

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

    ... where \zeta(s) is the so called 'Riemann zeta function'. The goal is to find a series expansion 'somewehre around' s=1 of its derivative...

    \displaystyle \eta^{'}(s)= \ln 2\ 2^{1-s}\ \zeta(s) + (1-2^{1-s})\ \zeta^{'}(s) (2)

    At this scope we use the [well known] Laurent expansion...

    \displaystyle \zeta(s) = -\frac{1}{1-s} + \gamma + \sum_{n=1}^{\infty} \gamma_{n}\ \frac{(1-s)^{n}}{n!} (3)

    ... where \gamma= .577215665... is the Euler's constant and the \gamma_{n} are the so called 'Stieltjes constants'. Deriving (3) we obtain...

    \displaystyle \zeta^{'}(s) = -\frac{1}{(1-s)^{2}} + \sum_{n=1}^{\infty} (-1)^{n}\ n\ \gamma_{n}\ \frac{(1-s)^{n-1}}{n!} (4)

    ... and is...

    \displaystyle 2^{1-s}= \sum_{n=0}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!} \implies 1-2^{1-s}= - \sum_{n=1}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!} (5)

    ... so that we have all the terms we need to compute (2). We are interested to the value of (2) in s=1, so that we stop the computation at the constant term obtaining...

    \displaystyle \eta^{'} (s)= -\frac{\ln 2}{1-s} + \gamma\ \ln 2 - \ln^{2} 2 +...+\frac{\ln 2}{1-s} + \frac{\ln^{2} 2}{2}+... (6)

    ... and finally...

    \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n} = \eta^{'} (1) = \gamma\ \ln 2 - \frac{\ln^{2} 2}{2} (7)

    Kind regards

    \chi \sigma
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