- #1
canis89
- 30
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Homework Statement
Compute the following partial sum
[tex]\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}[/tex]
Homework Equations
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?