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Partial Sums resembling sums of secant hyperbolic

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Compute the following partial sum

    [tex]\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    So far, I've tried transforming the terms into secant hyperbolic functions:

    [tex]\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}=\frac{1}{2}\sum_{k=0}^n\sech(2^k \ln(2))=\frac{1}{2}\sum_{k=0}^n\frac{1}{\cosh(2^k \ln(2))}[/tex]

    Then, what I've tried to do is to analyze whether [itex]\cosh(2^k \ln(2))[/itex] can be expressed in terms of cos(ln(2)). But so far, it involves solving the quadratic mapping

    [tex]a_k=2a_{k-1}^2-1,\forall k\geq1,\quad a_0=\cosh(\ln(2))[/tex]

    throught the identity that

    [tex]\cosh(2x)=2\cosh(x)^2-1[/tex]

    Or maybe I'm making it too complex. Can anyone shred some thought?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
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