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Evaluating a sum by Abel's Thm

  1. May 16, 2006 #1

    benorin

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    I wish to evaluate the sum

    [tex]\sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1}[/tex]

    I think I can use Abel's theorem... just read: for |z|<1, we have

    [tex]\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = z^n\frac{1}{(1+z)^{n+1}} = (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \frac{1}{1+z}\right) [/tex]
    [tex]= (-1)^n\frac{z^n}{n!}\frac{d^n}{dz^n} \left( \sum_{k=0}^{\infty}(-1)^{k}z^{k} \right) = (-1)^n\frac{z^n}{n!}\sum_{k=n}^{\infty}(-1)^{k}\frac{k!}{(k-n)!}z^{k-n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k} [/tex]

    here obtained is the identity

    [tex]\frac{1}{1+z}\left( \frac{z}{1+z}\right) ^{n} = \sum_{k=n}^{\infty}(-1)^{n+k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k} [/tex] ​

    valid for |z|<1. Suppose that 0<x<1 and multiply the above identity by [tex](-1)^n[/tex] and then integrate over (0,x) to arrive at

    [tex]\int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \int_{z=0}^{x} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) z^{k}dz = \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1} [/tex] ​

    to the latter sum apply Abel's Theorem to determine that

    [tex]\lim_{x\rightarrow 1^-} \sum_{k=n}^{\infty}(-1)^{k}\left(\begin{array}{c}k\\n\end{array}\right) \frac{x^{k+1}}{k+1} = \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} [/tex] ​

    so that, by the previous identity, the value of the sum is given by


    [tex]\boxed{ \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} = \lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} }[/tex] ​

    Okay, so www.integrals.com (also, www.quickmath.com) gives the integral in terms of the hypergeometric function 2F1, viz.

    [tex]\lim_{x\rightarrow 1^-} \int_{z=0}^{x}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \lim_{x\rightarrow 1^-} (-1)^k\frac{x^{k+1}}{k+1} _2F_1 (k+1,k+1;k+2;-x) = \frac{(-1)^k}{k+1} _2F_1 (k+1,k+1;k+2;-1) [/tex] ​

    so here I need assistance, the sum above is to be again summed over, as in

    [tex]\zeta ^{\prime} (0) =\sum_{n=0}^{\infty} (n+1)\left[ \log(n+1) -1\right] \sum_{k=n}^{\infty}\left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^{k}}{k+1} [/tex]​

    where [tex] \zeta (s)[/tex] denotes Riemann's zeta function, and it should sum to [tex] -\log \sqrt{2\pi}[/tex]: and thoughts as to how to proceed from here?
     
    Last edited: May 16, 2006
  2. jcsd
  3. May 19, 2006 #2

    benorin

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    That integral has promise...

    That integral has promise... so put

    [tex] I_{n}=\int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} [/tex]

    my prof said to try integration by parts (thanks Mihai,) pick, say

    [tex]u=z^n\Rightarrow du=nz^{n-1}dz \mbox{ and } dv=(-1)^{n}(1+z)^{-n-1}dz\Rightarrow v=\frac{(-1)^{n+1}}{n}(1+z)^{-n}[/tex] ​

    which gives

    [tex]I_{n} = \int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = \left[ -\frac{1}{n} \left( \frac{-z}{1+z} \right) ^{n} \right] _{0}^{1} - \int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n-1}\frac{dz}{1+z} = \frac{(-1)^{n+1}}{n2^n} - I_{n-1} \, [/tex]

    a recurrence relation: I'll try using the generating function technique. To be continued...
     
    Last edited: May 19, 2006
  4. May 19, 2006 #3

    benorin

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    generating functions are sweet

    About that generating function... we will need the value [tex]I_0=\log 2[/tex] (use the series or the integral, both are easy.) From the recurrence relation

    [tex]I_{n}= \frac{(-1)^{n+1}}{n2^n} - I_{n-1} \, [/tex]​

    multiply by [tex]z^n[/tex] and sum over [tex]n\geq 1[/tex] to get

    [tex]G(z):=\sum_{n=1}^{\infty} I_{n}z^{n} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^n}z^{n} - \sum_{n=1}^{\infty}I_{n-1}z^{n} = \log \left( 1+ \frac{z}{2}\right) -zI_0-z \sum_{n=1}^{\infty}I_{n}z^{n}[/tex]​

    if we put [tex]G(z) = \sum_{n=1}^{\infty} I_{n}z^{n}[/tex] for the generating function, substitute in the value of [tex]I_0=\log 2[/tex] we have

    [tex]G(z)= \log \left( 1+ \frac{z}{2}\right) -z\log 2 -zG(z)[/tex]​

    and hence

    [tex]G(z)= \frac{\log \left( 1+ \frac{z}{2}\right) -z\log 2}{1+z} = \left( \sum_{n=0}^{\infty} (-1)^{n}z^n\right) \left( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1}2^{-n}z^{n}\right) \frac{z}{2} - z\log 2 \sum_{n=0}^{\infty} (-1)^{n}z^n [/tex]
    [tex] = \frac{z}{2}\sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} \frac{(-1)^{k}}{k+1}2^{-k}(-1)^{n-k}\right) z^{n} - \log 2 \sum_{n=1}^{\infty} (-1)^{n-1}z^{n} [/tex]
    [tex] = \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} \right) z^{n} - \log 2 \sum_{n=1}^{\infty} (-1)^{n}z^{n} = \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right) z^{n} [/tex]​

    so we can now equate coefficients of [tex]z^{n}[/tex] in

    [tex] \sum_{n=1}^{\infty} I_{n}z^{n}= \sum_{n=1}^{\infty} (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right) z^{n} [/tex]​

    to arrive at

    [tex] I_{n}=\int_{0}^{1}\left( \frac{-z}{1+z}\right) ^{n}\frac{dz}{1+z} = (-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right) [/tex]​

    which (finally) gives the value of the desired sum

    [tex]\boxed{ \sum_{k=n}^{\infty} \left(\begin{array}{c}k\\n\end{array}\right) \frac{(-1)^k}{k+1} = \left\{\begin{array}{cc}\log 2,&\mbox{ if }
    n=0\\(-1)^{n-1}\left( \sum_{k=1}^{n} \frac{1}{k2^{k}} - \log 2 \right), & \mbox{ if } n\geq 1\end{array}\right. \, }[/tex]​
     
    Last edited: May 19, 2006
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