Does solution of $ x=\sum_{n=0}^\infty e^{-A_n/x}$ exist? Usually when working with indefinite sums, I want to work out the sum or whether it convergence. But now I encountered a problem they other way around and I'm clueless...
Is there even a general solution of $$ x=\sum_{n=0}^\infty e^{-A_n/x}$$
for $A_n$, where $x$ is given and real, $A_n >0\space\forall n$ and $\frac{dA_n}{dx}=0\space\forall n$?
Thank you
EDIT:
To make my question clearer for the commenters and others, I'm searching for a systematic sequence $A_n$ which, when entered in the equation above, yields $x$ and this should hold for all (real) $x$.

It cannot be done. For the proof write $x:={1\over y}$. Then we should have
$${1\over y}\ \equiv\ \sum_{n=0}^\infty e^{-A_n y}\qquad(*)\ ,$$
say for all $y\geq1$. In particular $\sum_{n=0}^\infty e^{-A_n}=1$, so necessarily $\lim_{n\to\infty} A_n=\infty$. It follows that $\alpha:=\inf_n A_n>0$ and therefore
$$\sum_{n=0}^\infty e^{-A_n y}=\sum_{n=0}^\infty e^{-A_n} \ e^{-A_n(y-1)} \leq e^{-\alpha(y-1)} \qquad (y\geq1)\ .$$
This shows that $(*)$ cannot hold for all $y\geq 1$.
