inverse function of series

Question

Consider

$$f(x)=e^{0A}+e^{1A}+e^{2A}+…$$

Where $A=1/\log x.$

I’m interested in the inverse of $f.$ I think $f^{-1}(x)=e^{A(g(x))}$ for some radical expression $g$.

For a truncated sum Wolfram alpha gives closed forms for $g$ but they get very complex.

What is the rule to obtain $g$? I presume it involves finding roots of polynomials in which the degree tends to infinity. At least wolfram alpha seems to be doing that.

Answer 1

The series for $f$ is geometric. It converges if $e^A<1$, i.e. $A<0$ which occurs if $0<x<1$. Therefore,

$$0<x<1\implies f(x)=\frac{1}{1-e^{\frac{1}{\log x}}}$$

This function is continuous in this range, and monotone (decreasing) so that it is invertible.

The inverse is

$$f^{-1}(x)=\exp\left(\frac{1}{\log \left(1-\frac{1}{x}\right)}\right)$$

for $x>1$.