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!

  • 3
    $\begingroup$ Have you optimized by $n$? What is the largest term of this series for $a$ and $x$ given? $\endgroup$ – Seva Nov 26 '14 at 18:27
  • $\begingroup$ Thanks Seva, could you please be more precise? $\endgroup$ – Anand Nov 26 '14 at 18:30
  • $\begingroup$ For $a$ and $x$ given, what is the maximum of $x^n\exp(-a^n/x)/n!$ over all $n$? $\endgroup$ – Seva Nov 26 '14 at 19:45
  • $\begingroup$ Or, at least, for what $n$ isn't the second factor just nuking the first? $\endgroup$ – 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.)

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  • $\begingroup$ Dear Christian Remling, thank you very much for your nice answer! :) $\endgroup$ – Anand Nov 27 '14 at 4:48

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