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!
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.)
