Has anyone seen any function of the following type? 


$$
g(x):=\sum\_{n=0}^\infty \frac{x^n}{n!}\exp\left(-\frac{a^n}{x}\right),\quad a>1,x\ge 0.
$$


The question is whether for some constant $c>0$,


$$
\lim\_{x\rightarrow\infty}\frac{1}{x}\log g(x) \ge c.
$$


Thanks a lot for any hints!



Does this function have any exponential growth?
This function doesn't have exponential increase. Take $N=N(x)\sim \ln x$, with a big implied constant. Then $C^n\exp(-a^n/x)$ is small for $n\ge N$, so this part of the series cannot have the same size as an exponential function.


However, for $n\le N$, the terms of the exponential series are still increasing in $n$, so we can estimate this part as follows:
$$
\sum\_{n=0}^N \frac{x^n}{n!}\exp(-a^n/x) \lesssim N \frac{x^N}{N!} \sim \sqrt{\ln x}
\left( \frac{ex}{\ln x} \right)^{\ln x} ,
$$
by Stirling's formula for the last step, and this is way smaller than $e^{cx}$. Thus $g(x)e^{-cx}\to 0$ for all $c>0$. (These estimates can certainly be done more carefully, but it does answer your question in this form.)


