# Partial Sums resembling sums of secant hyperbolic

1. Nov 18, 2011

### canis89

1. The problem statement, all variables and given/known data

Compute the following partial sum

$$\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}$$

2. Relevant equations

3. The attempt at a solution

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

$$\sum_{k=0}^n\frac{1}{2^{2^k}+2^{-2^k}}=\frac{1}{2}\sum_{k=0}^n\sech(2^k \ln(2))=\frac{1}{2}\sum_{k=0}^n\frac{1}{\cosh(2^k \ln(2))}$$

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

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

throught the identity that

$$\cosh(2x)=2\cosh(x)^2-1$$

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