Is it possible to express $\displaystyle \frac{1}{e^t+e^{-t}}$ as a power series in $\displaystyle e^{-t}$?
So can we find $\displaystyle a_n$ with $\displaystyle \frac{1}{e^t+e^{-t}} = \sum_{n=0}^\infty a_n (e^{-t})^n$ ?
Setting $\displaystyle e^{-t} = x$ we obtain...
$\displaystyle \frac{1}{e^{t} + e^{-t}} = \frac{1}{x+\frac{1}{x}} = \frac{x}{1+x^{2}} = $
$\displaystyle = x\cdot (1-x^{2} + x^{4} - \dots) = x - x^{3} + x^{5} - \dots$ (1)
... so that is...
$\displaystyle \frac{1}{e^{t} + e^{-t}} = e^{-t} - e^{-3t} + e^{-5t} - \dots$ (2)
Because the series (1) converges for $\displaystyle |x|<1$, the series (2) converges for $\displaystyle t>0$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$