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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 $\epsilon/\Delta W$.

Sangchul Lee
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Soroush khoubyarian
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2 Answers2

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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 $x$.

Christophe Leuridan
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(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 $1$.

marty cohen
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