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) $x$.
 A: 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 $y\geq 1$.
A: Certainly there are many solutions.  As $x$ grows, the right side decreases monotonically, so for any series $A_n$ that is convergent there will be an $x$ that solves the equation.  So pick any $x$ and series $A_n$ that solve the problem.  Given a different $x$, just change your favorite $A_n$(s) to make it work.
For a specific example, take $x=1, A_n=\ln 2^{n+1}$.  If you want a solution for $x=2$, just decrease any set of $A_n$'s to add enough to the RHS.
A: Divide both sides by $x$ to get:
$$1 = \Sigma\frac{e^{-A_n/x}}{x},$$
As $x\to \infty$, the left hand side is 1 while the right hand side goes to zero for each $n$.
This might give a hint on what $A_n$ won't work.
