About the asymptotic behaviour of $\sum_{n\in\mathbb{N}}\frac{x^{a_n}}{a_n!}$ Let $\{a_n\}_{n\in\mathbb{N}}$ be an increasing sequence of natural numbers, and
$$ f_A(x)=\sum_{n\in\mathbb{N}}\frac{x^{a_n}}{a_n!}. $$
There are some cases in which the limit
$$ l_A=\lim_{x\to+\infty} \frac{1}{x}\,\log(f_A(x)) $$
does not exist. However, if $\{a_n\}_{n\in\mathbb{N}}$ is an arithmetic progression, we have $l_A=1$ (it follows from a straightforward application of the discrete Fourier transform). Consider now the case $a_n=n^2.$


*

*Is it true that there exists a positive constant $c$ for which
$$\forall x>0,\quad e^{-x}f_A(x)=\sum_{k\in\mathbb{N}}x^k\left(\sum_{0\leq j\leq\sqrt{k}}\frac{(-1)^{k-j^2}}{(j^2)!\,(k-j^2)!}\right)\geq c\;?$$ 

*Is it true that $l_A=1$?

There is an error in the argument. See the comments.

It is a nice exercise to show $l_A = 1$ for every polynomial $n^k$.
Let me sketch the proof for $k = 2$. Observe that for $x \geq 0$:
$$\sum_{n=0} \frac{x^{n^2}}{(n^2)!} \leq \sum_{n=0} \frac{x^n}{n!} = e^x$$
simply, because the left hand side is a subsequence of the right hand side. The next thing to do is to bound our sequence from the left. We have:
$$\sum_{n=0} (2n + 1)\frac{x^{(n+1)^2}}{(n+2)^2!} \geq \sum_{n=0}\frac{x^n}{(n+2)!}$$
because in each group of the size $(n+1)^2 - n^2 = 2n + 1$ the last nominator is the biggest and the first denominator is the smallest one. For sufficiently large $x$ (for example $x \geq 2$):
$$\sum_{n=0} \frac{x^{(n+2)^2}}{(n+2)^2!} \geq \sum_{n=0} (2n+1)\frac{x^{(n+1)^2}}{(n+2)^2!}$$
thus
$$\sum_{n=0} \frac{x^{n^2}}{(n^2)!} - x - 1= \sum_{n=2} \frac{x^{n^2}}{(n^2)!} \geq \sum_{n=2} \frac{x^{(n-2)}}{n!} = \frac{1}{x^2}\sum_{n=2} \frac{x^n}{n!} = \frac{1}{x^2}(e^x - x - 1)$$
Therefore, for sufficiently large $x$:
$$\frac{e^x}{x^2} \leq \frac{e^x}{x^2} + x + 1 - \frac{x+1}{x^2} \leq \sum \frac{x^{n^2}}{(n^2)!} \leq e^x$$
