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Explicitly summing a series involving hyperbolic tangent

  1. Nov 19, 2008 #1

    gabbagabbahey

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    I recently came across a problem where I was able to show that [tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \tanh \left( \frac{n \pi}{2} \right)=\frac{\ln 2- \pi}{4}[/tex] through numerical approximation....However, I don't have much practice evaluating such summations analytically, and I was wondering if anyone had any ideas on how to evaluate this one analytically?
     
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  3. Dec 17, 2008 #2
    Express tanh as a sum of complex exponentials.
     
  4. Dec 17, 2008 #3

    gabbagabbahey

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    Ermm...was the 'complex' a typo, or does expressing [itex]\tanh\left(\frac{n\pi}{2}\right)[/itex] as [tex]\frac{e^{i\left(\frac{-in\pi}{2}\right)}-e^{i\left(\frac{in\pi}{2}\right)}}{e^{i\left(\frac{-in\pi}{2}\right)}+e^{i\left(\frac{in\pi}{2}\right)}}[/tex] actually help in evaluating this sum? If so, I don't see how.
     
  5. Dec 17, 2008 #4

    gabbagabbahey

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    Okay, so I've made some headway by expressing tanh in terms of 'real' exponentials:

    Using [itex]\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}=1-\frac{2}{e^{2x}+1}}[/itex] , the sum becomes:

    [tex]\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \tanh \left( \frac{n \pi}{2} \right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{n} -2\sum_{n=1}^{\infty} \frac{(-1)^n}{n} \left( \frac{1}{e^{n\pi}+1} \right)[/tex]

    The first term is just the alternating harmonic series and sums to [itex]-\ln(2)[/itex], and the second one converges very quickly, but now I need to review my old calc text and see if there is something in there that will help me analytically compute the second summation.
     
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