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**Laurent** Neither a proof nor a counter-example, just first thoughts...

Now I have a proof that fluctuations between points of $\displaystyle A$ can break the conjecture. I shall provide a family of examples where $\displaystyle \liminf_{x\to\infty} f(x)^{1/x}\leq 1$.

Note that $\displaystyle f(x)^{1/x}\to e$ is equivalent to $\displaystyle \frac{\log f(x)}{x}\to 1$.

Suppose $\displaystyle x$ is such that $\displaystyle m<x<M$ where $\displaystyle m,M\in A$ and $\displaystyle A\cap(m,M)=\emptyset$ (thus $\displaystyle (m,M)$ is a gap in $\displaystyle A$, thought of as a *large* gap). More precise conditions will come later.

We have the following upper bound that consists in taking all terms of the series outside the gap:

$\displaystyle f(x)\leq \sum_{a=0}^m \frac{x^a}{a!}+\sum_{a=M}^\infty \frac{x^a}{a!}$.

For the first sum, note that the largest term is the last one (each term is $\displaystyle \frac{x}{a}(>1)$ times the previous one), so that the sum is less than $\displaystyle (m+1)\frac{x^m}{m!}$.

For the second sum, I factorize by the first term and use a geometric series to bound the second factor:

$\displaystyle \sum_{a=M}^\infty \frac{x^a}{a!}=\frac{x^M}{M!}\left(1+\frac{x}{M+1} +\frac{x^2}{(M+1)(M+2)}+\cdots\right)$ $\displaystyle \leq \frac{x^M}{M!}(1+\frac{x}{M}+\frac{x^2}{M^2}+\cdot s)=\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}}$.

Summarizing, we have $\displaystyle f(x)\leq (m+1)\frac{x^m}{m!}+\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}}$.

I shall now give conditions that ensure $\displaystyle \frac{log T}{x}\to 0$ or $\displaystyle -\infty$ where $\displaystyle T$ is either of the two terms of the upper bound, and the limit is taken along some sequence $\displaystyle x_n\to\infty$ (corresponding to some intervals $\displaystyle [m_n,M_n]$).

First one: $\displaystyle \frac{1}{x}\log\frac{x^m}{m!}=\frac{m\log x-\log m!}{x}=\frac{m}{x}(\log\tfrac{x}{m}+O(1))$ $\displaystyle =\frac{\log\frac{x}{m}}{\frac{x}{m}}+O(\frac{m}{x} )$. (Using $\displaystyle \log m!=m\log m+O(m)$), hence $\displaystyle \frac{1}{x}\log\frac{x^m}{m!}\to 0$ as soon as $\displaystyle \frac{m}{x}\to 0$ and $\displaystyle m\to\infty$. (Indeed, $\displaystyle \frac{\log x}{x}\to_{x\to\infty} 0$) * I forgot $\displaystyle \frac{1}{x}\log(m+1)$, which also tends to 0 (faster) under this condition.

Second one: $\displaystyle \frac{1}{x}\log(\frac{x^M}{M!}\frac{1}{1-\frac{x}{M}})=\frac{1}{x}(M\log x-\log M!-\log(1-\frac{x}{M}))$ $\displaystyle =\frac{M\log x-M\log M+M-\frac{1}{2}\log M+O(1)}{x}+\frac{1}{M}+o(\frac{1}{M})$ if $\displaystyle \frac{x}{M}\to 0$ and $\displaystyle M\to\infty$ (using Stirling estimate: $\displaystyle \log n!=n\log n-n+\frac{1}{2}\log n+O(1)$ and $\displaystyle \log(1+x)=x+o_{x\to 0}(x)$), hence this term equals $\displaystyle \frac{M}{x}(-\log\frac{M}{x}+1-\frac{1}{2}\frac{\log M}{M}x)+O(\frac{1}{x})+\frac{1}{M}+o(\frac{1}{M})$, so that it tends to $\displaystyle -\infty$ as soon as $\displaystyle \frac{x}{M}\to 0$ and $\displaystyle M\to\infty$.

Since $\displaystyle \log(a+b)\leq \max(\log (2a),\log (2b))=\log 2+\max(\log a,\log b)$, we deduce that $\displaystyle \liminf f(x)^{1/x}\leq 1$ where the bound corresponds to a limit taken along sequences of $\displaystyle x$ that are in gaps $\displaystyle (m,M)$ of $\displaystyle A$ such that $\displaystyle m<\!\!< x<\!\!<M$ ("$\displaystyle <\!\!<$" meaning "negligible compared to"). Such sequences of $\displaystyle x$'s exist when $\displaystyle A$ grows very fast (we must have $\displaystyle m<\!\!<M$ where $\displaystyle m$ and $\displaystyle M$ are successive terms in the sequence): for instance, $\displaystyle n!<\!\!< n!\sqrt{n}<\!\!<(n+1)!$ so $\displaystyle A=\{n!|n\in\mathbb{N}\}$ and $\displaystyle x_n=n!\sqrt{n}$ is such an example.

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However, I would still be interested in knowing how you would deal with $\displaystyle A=\{n^2|n\in\mathbb{N}\}$ if you know that.