Limit/asymptotic of a modification of the exponent's Taylor series Consider the following function on $\left[0,\infty\right)\times\left[0,1\right]$
$$f\left(x,y\right)=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}y^{n^{2}}$$
I am interested in the limit/asymptotic behavior of the following expression for $x\rightarrow\infty$
$$g\left(x,y\right)=e^{x}\dfrac{y^{2}f\left(xy^{3},y\right)}{f^{2}\left(xy,y\right)}$$
I encountered these summations in the context of the $g^{\left(2\right)}$ correlation function of some quantum state of light. On physical grounds, I suspect it might converge asymptotically to $1$ from below, but not completely sure.
Any suggestion on how to evaluate this will be appreciated. Thanks in advance!
EDIT I:
I found a mistake in my calculation, which after fixing gives a slightly different function $g\left(x,y\right)$
$$g\left(x,y\right)=\dfrac{y^{2}f\left(xy^{3},y\right)f\left(xy^{-1},y\right)}{f^{2}\left(xy,y\right)}$$
Plugging inside @SangchulLee asymptotic expansion indeed gives $g\left(x\rightarrow\infty,y\right)=1$. Since there is no dependence on $x$, I would also be interested in the next order of the asymptotic expansion. Thanks!
 A: Step 1. Let us parametrize $x = e^{s}$ and $y = e^{-\varepsilon}$ for $s \in \mathbb{R}$ and $\varepsilon > 0$. Fix the values of $k$ and $\varepsilon$, and then write
$$ f(xy^k, y) = \sum_{n=0}^{\infty} e^{-h(n)}, \qquad \text{where} \quad h(n) = \varepsilon n^2 + k\varepsilon n - sn + \log n! $$
Regarding $h(n)$ as an analytic function of $n$ and differentiating twice,
\begin{align*}
h'(n) &= 2\varepsilon n + k\varepsilon - s + \psi(n+1), \\
h''(n) &= 2\varepsilon + \psi'(n+1),
\end{align*}
where $\psi$ is the digamma function. Since $h''(n) > 0$ for $n > 0$, it follows that $h$ is strictly concave. Moreover, for each fixed $k$ and $\varepsilon$ and for large $s$, we have $h'(0) < 0$ and $h'(n) \to \infty$ as $n\to\infty$. So it follows that the equation $h'(n) = 0$ has a unique zero $n = n_*$. Then, regarding $n_*$ as a function of $s$, the Laplace's method suggests that
$$ f(xy^k, y) \sim \sqrt{\frac{2\pi}{h''(n_*)}} \, e^{-h(n_*)} \quad\text{as}\quad s\to\infty. $$
Step 2. The above observation leads us to study the asymptotic behavior of $n_*$ as $ s \to \infty $. To this end, note that $s$ and $n_*$ are related via the equation
$$ s = 2\varepsilon n_* + \psi(n_* + 1) + k\varepsilon. \tag{1} $$
Using this, we will progressively improve the asymptotic formula for $n_*$ as $s \to \infty$:

*

*Since $s$ as a function of $n_*$ is strictly increasing and unbounded, the same is true for $n_*$ as a function of $s$.


*Dividing both sides of $\text{(1)}$ by $n_*$ and letting $n_* \to \infty$, we get $s/n_* \to 2 \varepsilon$ as $n_* \to \infty$. This then implies that
$$ n_* = \frac{s}{2\varepsilon}(1 + o(1)) \qquad\text{and}\qquad \psi(n_* + 1) = \log\left(\frac{s}{2\varepsilon}\right) + o(1) \tag{2} $$
as $s \to \infty$, where we utilized the asymptotic relation $\psi(z+1) \sim \log z$ as $z \to \infty $ for the second one.


*Rearranging $\text{(1)}$, we get
$$ n_* = \frac{s}{2\varepsilon} - \frac{k}{2} - \frac{1}{2\varepsilon}\psi(n_* + 1). $$
In light of $\text{(2)}$, we define the quantity $s_*$ as
$$ s_* := s - \log\left(\frac{s}{2\varepsilon}\right) - k\varepsilon $$
and use this to recast $ n_* $ as
$$ n_* = \frac{1}{2\varepsilon}\left[s_* - \psi(n_* + 1) + \log\left(\frac{s}{2\varepsilon}\right) \right]. \tag{3} $$


*Using $\text{(3)}$ and the asymptotic expansion $\psi(z+1) = \log z + (2z)^{-1} + \mathcal{O}(z^{-2})$ as $z \to \infty$,
\begin{align*}
&\psi(n_* + 1) - \log\left(\frac{s}{2\varepsilon}\right) \\
&= \biggl(\log n_* + \frac{1}{2n_*} + \mathcal{O}(s^{-2}) \biggr) - \log\left(\frac{s}{2\varepsilon}\right) \\
&= \log\left(\frac{2\varepsilon n_*}{s}\right) + \frac{1}{2n_*} + \mathcal{O}(s^{-2}) \\
&= \log \left(\frac{s_*}{s}\right) + \log \biggl( 1 - \frac{\psi(n_* + 1) - \log\left(\frac{s}{2\varepsilon}\right)}{s_*} \biggr) + \frac{1}{2n_*} + \mathcal{O}(s^{-2})
\end{align*}
In light of the asymptotic formulas in $\text{(2)}$, this further reduces to
\begin{align*}
&= -\log\left(1 + \frac{s - s_*}{s_*}\right) + \log \biggl( 1 - \frac{o(1)}{s_*} \biggr) + \frac{\varepsilon}{s_*(1 + o(1))} + \mathcal{O}(s^{-2}) \\
&= -\frac{1}{s_*}\log\left(\frac{s_*}{2\varepsilon}\right) - \frac{(k-1)\varepsilon}{s_*} + o(s^{-1})
\end{align*}
Therefore we obtain
\begin{align*}
n_*
&= \frac{1}{2\varepsilon}\left[ s_* + \frac{1}{s_*}\log\left(\frac{s_*}{2\varepsilon}\right) + \frac{(k-1)\varepsilon}{s_*} + o(s^{-1}) \right], \tag{5}
\end{align*}
Step 3. Now we are ready to estimate $h(n_*)$. Using $\text{(1)}$ and the Stirling's approximation together,
\begin{align*}
h(n_*)
&= \varepsilon n_*^2 - (s - k\varepsilon)n_* + \log (n_*!) \\
&= \varepsilon n_*^2 - (2\varepsilon n_* + \psi(n_* + 1))n_* + \log (n_*!) \\
&= -\varepsilon n_*^2 - n_* [\psi(n_* + 1) - \log n_*] - n_* + \frac{1}{2}\log n_* + \log\sqrt{2\pi} + o(1) \\
&= -\varepsilon n_*^2 - n_* + \frac{1}{2}\log n_* - \frac{1}{2} + \log\sqrt{2\pi} + o(1)
\end{align*}
Then plugging $\text{(4)}$ into the last line,
\begin{align*}
&= -\frac{1}{4\varepsilon}\left[ s_*^2 + 2\log\left(\frac{s_*}{2\varepsilon}\right) + 2(k-1)\varepsilon \right] - \frac{s_*}{2\varepsilon} + \frac{1}{2}\log \left(\frac{s_*}{2\varepsilon}\right) - \frac{1}{2} + \log\sqrt{2\pi} + o(1) \\
&= -\frac{1}{4\varepsilon}s_*^2 - \frac{1}{2\varepsilon}\left[ s_* + \log\left(\frac{s_*}{2\varepsilon}\right) + k\varepsilon \right] + \frac{1}{2}\log \left(\frac{s_*}{2\varepsilon}\right) + \log\sqrt{2\pi} + o(1) \\
&= -\frac{1}{4\varepsilon}s_*^2 - \frac{s}{2\varepsilon} + \frac{1}{2}\log \left(\frac{\pi s}{\varepsilon}\right) + o(1)
\end{align*}
On the other hand, using $\psi'(z+1) \sim \frac{1}{z}$, we get $h''(n_*) \sim 2\varepsilon$. Therefore,
\begin{align*}
f(e^{s-k\varepsilon}, e^{-\varepsilon})
\sim \sqrt{\frac{\pi}{\varepsilon}} e^{-h(n_*)}
\sim \exp\left\{ \frac{1}{4\varepsilon}s_*^2 + \frac{s}{2\varepsilon} - \frac{1}{2}\log s + o(1) \right\}
\end{align*}
