Proving $\lim_{x \to \infty} \frac{\sum_{n=1}^{\infty}\frac{x^n}{n!}\sqrt{x}}{\sum_{n=1}^{\infty}\frac{x^n}{n!}\sqrt{n}} = 1$ As part of solving a problem, I am trying to prove the following identity:
$$\lim_{x \to \infty }\dfrac{\sum_{n=1}^{\infty}\dfrac{x^n}{n!}\sqrt{x}}{\sum_{n=1}^{\infty}\dfrac{x^n}{n!}\sqrt{n}} = 1$$
Intuitively, I can understand why this is true. But I don't know how to make a rigorous argument to prove it.
For small $n \ll x$, $\frac{x^n}{n!}$ does not contribute to the sums as the terms with a higher power in $x$ dominate. As well, for $n \gg x$, the denominator $n!$ dominates and the fractions become far too small.
So the main contributors to the sum are the ones with $n$ at $n\approx \text{ceil}\left[x\right]$ or $n \approx \text{floor}\left[x\right]$, at which $\dfrac{x^n}{n!}$ is maximized.
Say the width of this region is $\Delta W$. My guess at making this argument concrete, is that we have to prove that the contribution to the sum outside this interval $[x-\Delta W/2, x+\Delta W/2]$ is bounded above by an arbitrary $\epsilon/2$, and inside the interval $\sqrt{x/n}$ differs from $1$ at most by $\epsilon/\Delta W$.
 A: We have
$$\frac{\sum_{n=1}^{+\infty}\frac{x^n}{n!}\sqrt{n}}
{\sum_{n=1}^{+\infty}\frac{x^n}{n!}\sqrt{x}} 
= \frac{\sum_{n=0}^{+\infty}\frac{x^n}{n!}\sqrt{n}}{(e^x-1)\sqrt{x}} = \frac{E[\sqrt{N_x}]}{(1-e^{-x})\sqrt{x}},$$
where $N_x$ is a Poisson random variable with parameter variance $X$.
Since $\sqrt{N_x/x}$ is closer than $N_x/x$ from $1$,
$$\Big(\frac{N_x}{x}-1\Big)^2 \ge \Big(\sqrt{\frac{N_x}{x}}-1\Big)^2 = 
\frac{N_x}{x}+1-2\sqrt{\frac{N_x}{x}} \ge 0.$$
Taking expectations, since the expectation and the variance of $N_x$ equal $x$, we get
$$\frac{1}{x} \ge 2-2E\Big[ \sqrt{\frac{N_x}{x}} \Big] \ge 0.$$
As a result, $E[\sqrt{N_x/x}] \to 1$ as $x \to +\infty$, which gives the desired conclusion.
Addendum: the square root function may be replaced by any power function. Indeed, $X_x/x \to 1$ in $L^2(P)$ as $x \to +\infty$, hence in probability. Hence for every $\alpha>0$, $(X_x/x)^\alpha \to 1$ in probability, and also in $L^1(P)$, by uniform integrability. Indeed, choose any integer $p>\alpha$, and note that the family $(X_x/x)_{x \ge 1}$ is bounded in $L^p(P)$, since $E[X_x^p]$ is a polynomial with degree $p$ with regard to $x$.
A: (Don't know if this is
a comment or answer,
but it is easier as an answer.)
Here's how I would do it,
if I wanted to expend the effort.
As mentioned,
the maximum terms of each
are around $n=x$,
so a natural method of solution
is to split each sum
into three parts
(less than
$x-d(x)$,
from $x-d(x)$ to
$x+d(x)$,
and greater than
$x+d(x)$
for some $d(x)$)
and show that the first and third parts
are small compared to the
middle part
and that the ratio of
the two middle parts tends to $1$.
