# Does this function have any exponential growth?

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!

• Have you optimized by $n$? What is the largest term of this series for $a$ and $x$ given? – Seva Nov 26 '14 at 18:27
• Thanks Seva, could you please be more precise? – Anand Nov 26 '14 at 18:30
• For $a$ and $x$ given, what is the maximum of $x^n\exp(-a^n/x)/n!$ over all $n$? – Seva Nov 26 '14 at 19:45
• Or, at least, for what $n$ isn't the second factor just nuking the first? – fedja Nov 27 '14 at 1:24

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.)