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 $a_n=n^2.$


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*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 $l_A=1$?

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.)
