My question is regarding the expression below, where $\varepsilon\ll1$. $$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}}$$ Is it possible to express this in the form $$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} = a_0+a_1\varepsilon+a_2\varepsilon^2...$$
Where I am guessing $a_0=\frac{1}{\varepsilon}$ for example. I'm a bit rusty on my asymptotics so i'm sorry if this is a silly question, just need my mind jogged with some input.
Many thanks in advance.
EDIT: My idea, could we say
$$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} = \varepsilon^{\cfrac{1}{\varepsilon-1}} = \left[ 1+(\varepsilon-1) \right]^{\cfrac{1}{\varepsilon-1}} $$ Therefore as $|\varepsilon-1|=|1-\varepsilon|<1$ we may expand
$$\left(\frac{1}{\varepsilon}\right)^{\cfrac{1}{1-\varepsilon}} \simeq 1+\cfrac{1}{\varepsilon-1}(\varepsilon-1)+ \cfrac{1}{\varepsilon-1}\frac{(\varepsilon-1)^2}{2} $$ This doesn't look that promising.
Consider $$y=\left(\frac{1}{x}\right)^{\frac{1}{1-x}}$$ Take logarithms $$\log(y)=-{\frac{1}{1-x}}\log(x)$$ Now expand $\frac{1}{1-x}$. This gives $$\log(y)=-\log(x)\sum_{k=0}^\infty x^k$$ Using now $$y=e^{\log(y)}=\frac{1}{x}-\log (x)+x \left(\frac{\log ^2(x)}{2}-\log (x)\right)+x^2 \left(-\frac{1}{6} \log ^3(x)+\log ^2(x)-\log (x)\right)+O\left(x^3\right)$$
Edit
Just for illustration puroposes, let us try using $x=10^{-6}$. The above formula gives $$y=1000013.815619808284995098$$ for an "exact" value $$y=1000013.815619808284998235$$
Update
If, as according to your comments, the first order is sufficient you can solve for $x$ equation $$y=\frac{1}{x}-\log (x)$$ using $$x=\frac{1}{W\left(e^y\right)}$$ where appears again the beautiful Lambert function