Evaluate $\lim_{x\to+∞}\frac{(\sum_{n=0}^∞{(\frac{x^n}{n!})^2})^2}{(\sum_{n=0}^∞{(\frac{x^n}{n!})^1}) (\sum_{n=0}^∞{(\frac{x^n}{n!})^3})}$ Prove the following limit:
$$
\lim_{x\rightarrow +\infty} \frac{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^2} \right) ^2}{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^1} \right) \left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^3} \right)}=\frac{\sqrt{3}}{2} \tag{1}
$$
Mathematica tells me that
$$
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^2=I_0(2 x)\\
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^3 = \, _0F_2\left(;1,1;x^3\right)
$$
but they don't make sense to me for calculating the limits, since I don't have any knowledge about Special Functions.
How can I prove $(1)$ in an elementary way(not involving special functions)?
 A: We investigate a heuristic idea that lead the correct answer. Consider a Poisson random variable $N$ with rate $x$, so that
$$ \mathbb{P}(N = n) = \frac{x^n}{n!} e^{-x} $$
for $n \geq 0$. Then by the (local) central limit theorem, we know that $Z = \frac{N-x}{\sqrt{x}}$ approximates the standard normal distribution, loosely in the sense that
\begin{align*}
\mathbb{P}(N = n)
&= \mathbb{P}\left( \left| N - n \right| < \frac{1}{2}\right) \\
&\approx \mathbb{P}\biggl( \left| Z - \frac{n-x}{\sqrt{x}} \right| < \frac{1}{2\sqrt{x}} \biggr) \\
&\approx \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}}
\end{align*}
where $\phi(t) = \frac{1}{\sqrt{2\pi}} e^{-t^2/2}$ is the p.d.f. of the standard normal distribution. So we expect that, for each given $\alpha > 0$, as $x\to\infty$,
\begin{align*}
\sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
&\sim e^{\alpha x} \sum_{n=0}^{\infty} \left[ \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}} \right]^{\alpha} \\
&\sim \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\alpha t^2/2} \, \mathrm{d}t \\
&= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}}. \tag{*}
\end{align*}
Assuming that $\text{(*)}$ indeed holds, then the limit easily follows.

Indeed, some hard analysis using the idea of Laplace's method shows that, for each fixed $\alpha > 0$ and sufficiently small $\epsilon > 0$,
$$ \sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}} \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \tag{$\diamond$} $$
This can be established by splitting the sum into two parts, one over the range $\left|n - x\right| \leq x^{\frac{1}{2}+\epsilon}$ that realizes the leading term of $(\diamond)$, and the other over the range $\left|n - x\right| \geq x^{\frac{1}{2}+\epsilon}$ that only contribute to the relative error $(\diamond)$.
As the computation is a bit lengthy, let me estimate the first sum only:

*

*Fix a sufficiently small $\epsilon > 0$ and consider the case $n \in \mathcal{I} = \{ n \in \mathbb{N}_0 : \left| n - x \right| \leq x^{\frac{1}{2}+\epsilon} \}$. Also, introduce a new parameter $u$ which is related to $n$ by
$$n = x + u\sqrt{x}.$$
In particular, $\left| u \right| \leq x^{\epsilon}$ and $ \frac{n}{x} = 1 + \frac{u}{\sqrt{x}} $. Now by the Stirling's approximation,
\begin{align*}
\frac{x^n}{n!}
&= \exp\left( n \log x - \frac{1}{2}\log(2\pi n) + n - n \log n + \mathcal{O}\bigl(n^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) - \frac{1}{2}\log\frac{n}{x} + x \left( \frac{n}{x} -  \frac{n}{x} \log \frac{n}{x} \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + \mathcal{O}\bigl(x^{-\frac{1}{2}+\epsilon}\bigr) + x \left( 1  - \frac{u^2}{2x} + \mathcal{O}\bigl(x^{-\frac{3}{2}+3\epsilon}\bigr) \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + x  - \frac{u^2}{2} +  + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \right) \\
&= \bigl( 1 + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \bigr) e^{x} \phi(u) \frac{1}{\sqrt{x}}.
\end{align*}
So it follows that, for each given $\alpha > 0$,
\begin{align*}
\sum_{n \in \mathcal{I}} \left( \frac{x^n}{n!} \right)^{\alpha}
&= \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}.
\end{align*}
Now the summation part in the right-hand side can be regarded as the Riemann sum. Indeed, using the fact that
$$ \left| \sum_{a \leq n \leq b} f(n) - \int_{a}^{b} f(x) \, \mathrm{d}x \right| \leq 3\sup_{a \leq x \leq b} \left| f(x) \right| $$
holds for any continuous unimodal function $f : [a, b] \to \mathbb{R}$,
$$ \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}
= \int_{-x^{\epsilon}}^{x^{\epsilon}} \frac{1}{\sqrt{2\pi}} e^{-\alpha u^2/2} \, \mathrm{d}u + \mathcal{O}(x^{-1/2})
= \frac{1}{\sqrt{\alpha}} + \mathcal{O}(x^{-1/2}). $$
Combining all the estimates, we obtain the desired leading term in $(\diamond)$.
