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 $l_A=1$?
 A: This is an answer to the first question only, for now.
There does not exist such a constant $c>0$. Assume it does, i.e., that $e^{-x} f_{A}(x) \ge c > 0$ for all $x>0$. Define $f_{A,0} = f_A$, and inductively
$$
f_{A,{k}}(x) = c+ \int_0^x f_{A,k-1}(t) \, dt = c \sum_{j=0}^{k-1} \frac{x^j}{j!} + \sum_{n \in \mathbb{N}} \frac{x^{a_n+k}}{(a_n+k)!}
$$
for $k \ge 1$. (The power series representation is easily checked by induction.) By assumption we have $f_{A,0}(x) \ge c e^x$, and then the induction step
$$
f_{A,k}(x) \ge c + \int_0^x ce^t \, dt = c+ ce^x-c = ce^x
$$
shows that $f_{A,k}(x) \ge c e^x$ for all $k \ge 0$ and $x>0$.
Now if $(a_n)$ is any sequence with $\lim\limits_{n\to\infty} (a_{n+1}-a_n) = \infty$, such as your example $a_n = n^2$, the power series representation for $f_{A,k}$ shows that
$$
F_{A,m}(x) = \sum_{k=0}^{m-1} f_{A,k}(x) \le P_{A,m}(x) + e^x
$$
where $P_{A,m}$ is a polynomial. Then
$$
P_{A,m}(x) + e^x \ge \sum_{k=0}^{m-1} ce^x = mc e^x
$$
for all $x>0$. This implies $mc \le 1$, so $c \le \frac1m$. However, as $m$ was arbitrary, this contradicts $c>0$.
A: It is true that $l_A=1$. The logic is similar to my answer to $\lim\limits_{x\to\infty}f(x)^{1/x}$ where $f(x)=\sum\limits_{k=0}^{\infty}\cfrac{x^{a_k}}{a_k!}$. Firstly, the terms after $n=3x$ don't matter:
$$\sum_{n=3x}^\infty x^n/n! <\sum_{n=3x}^\infty x^n /(3x/e)^n=C$$ by Stirling approximation. 
But for $n<3x$ there is going to be a perfect square $s$ between $n$ and $n+2\sqrt{3x}$ (this is just $(k+1)^2-k^2=2k+1$ and $k< \sqrt{3x}$). Then the values $x^s/s!$ and $x^n/n!$ differ by a factor of at most $(3x)^{2\sqrt{3x}}$, 
So if I multiply each term $x^s/s!$ by that ratio and take at least $2\sqrt{3x}$ copies I will have for each $x^n/n!$ (with $n<3x$) at least one term at least as big.  This means that $$f_A(x) (3x)^{2\sqrt{3x}}{2\sqrt{3x}} +C > e^x$$ and so $l_A=1$.
(I think the ratio of terms can actually be made $3^{2\sqrt{3x}}$, but it works as is too.)
A: 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 $x$:
$$\frac{e^x}{x^2} \leq \frac{e^x}{x^2} + x + 1 - \frac{x+1}{x^2} \leq \sum \frac{x^{n^2}}{(n^2)!} \leq e^x$$
