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Question: How to prove that $$\lim_{n\to\infty}\sum_{k=0}^n\frac{n^{2k}}{(k!)^2}\Bigg/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}=\frac12?$$ ($0^0$ is defined to be one). In addition, can we calculate $$\lim_{n\to\infty}\sqrt n\left(\sum_{k=0}^n\frac{n^{2k}}{(k!)^2}\Bigg/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}-\frac12\right)?$$

Relating to this question, there seems to be two possible ways, one is using central limit theorem, the other one is to turn this sum into an integral and estimate it. Unfortunately, the first possible method cannot be applied because the random variable $X_n$ with $$P(X_n=x)=\frac{n^{2x}}{(x!)^2}\Bigg/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}$$ does not have good properties like Poisson distribution. I'm able to calculate $\mathrm E(X)$ and $\mathrm{Var}(X)$, which are $\frac{I_1(2n)}{I_0(2n)}$ and $n^2\left(1-\frac{I_1(2n)}{I_0(2n)}\right)$ respectively. CLT cannot be applied here. I'm not familiar with generalized CLT, so I'm hoping for an analytical method.
Analytical Attempt
Denote $\sum_{k=0}^n\frac{n^{2k}}{(k!)^2}\big/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}$ by $L_n$. $$L_n=1-\frac{n^{2n+2}{}_1F_2(1;n+2,n+2;n^2)}{((n+1)!)^2I_0(2n)}\\ =1-\left(\frac1{\sqrt{\pi n}}+O(n^{-3/2})\right){}_1F_2(1;n+2,n+2;n^2)$$ But we have $$_1F_2(\cdots)=(n+1)\int_0^1(1-t)^n{}_0F_1(2+n;n^2t)dt\\ =e^{-n}\sqrt{2\pi n}(n+O(1))\int_0^1t^{-(n+1)/2}(1-t)^nI_{n+1}(2n\sqrt t)dt\\ =e^{-n}\sqrt{8\pi n}(n+O(1))\int_0^1t^{-n}(1-t^2)^nI_{n+1}(2nt)dt$$ Where all $I$'s above denote Bessel I function.
I think the asymptotic behavior of $I_n(z)$ when $n\approx kz\gg 0$ is needed, but I don't have reference of it.

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  • $\begingroup$ Have a look at dlmf.nist.gov/10.17 . Cheers $\endgroup$ – Claude Leibovici Feb 20 '20 at 11:34
  • $\begingroup$ @ClaudeLeibovici Thank you, but in the website, it only shows the case that the argument is fixed. Actually the behavior of $I_{x}(zx)$ as $x\to+\infty$ is required. I'm aware of the behavior of $I_\alpha(x)$ when one of the variables are fixed but I don't know how to do when both varies. $\endgroup$ – Kemono Chen Feb 20 '20 at 12:07
  • $\begingroup$ Anyway, would the downvoter mind to suggest an improvement? $\endgroup$ – Kemono Chen Feb 20 '20 at 12:08
  • $\begingroup$ Sorry, since I misunderstood. I am not the downvoter. If fact I upvoted your post. $\endgroup$ – Claude Leibovici Feb 20 '20 at 12:17
  • $\begingroup$ This could help you: dlmf.nist.gov/10.41.ii $\endgroup$ – Gary Feb 20 '20 at 14:47
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Here are two possible approaches:


Method 1. Let $X_n$ be a random variable with

$$ \mathbb{P}(X_n = k) = \frac{n^{2k}}{(k!)^2} \bigg/\biggl( \sum_{l=0}^{\infty} \frac{n^{2l}}{(l!)^2} \biggr), \qquad k = 0, 1, 2, \cdots. $$

Then the characteristic function of $X_n$ is given by

$$ \varphi_{X_n}(t) = \mathbb{E}[e^{it X_n}] = \frac{I_0(2n e^{it/2})}{I_0(2n)}, $$

where $I_0$ is the modified Bessel function of the first kind and order $0$. Now we normalize $X_n$ as follows:

$$ Z_n = \frac{X_n - n}{\sqrt{n}}. $$

Then by invoking the asymptotic formula for $I_0$:

$$ I_0(z) \sim \frac{e^{z}}{\sqrt{2\pi z}} \qquad \text{as} \quad z \to \infty \quad\text{along}\quad |\arg(z)| \leq \frac{\pi}{2}-\delta, $$

for each fixed $t \in \mathbb{R}$ it follows that

$$ \varphi_{Z_n}(t) = e^{-it\sqrt{n}} \, \frac{I_0(2n\exp(it/2\sqrt{n}))}{I_0(2n)} \sim \exp\bigl( 2ne^{it/2\sqrt{n}}-2n-it\sqrt{n} \bigr) \qquad \text{as} \quad n\to\infty. $$

This shows that

$$ \lim_{n\to\infty} \varphi_{Z_n}(t) = e^{-t^2/4}, $$

and so, $Z_n$ converges in distribution to $Z \sim \mathcal{N}(0, \frac{1}{2})$. Then the desired limit is

$$ \mathbb{P}(X_n \leq n) = \mathbb{P}(Z_n \leq 0) \xrightarrow[]{n\to\infty} \mathbb{P}(Z \leq 0) = \frac{1}{2}. $$

The second question seems also interesting and I suspect that it may be related to the local CLT, although I have no good idea in this direction.


Method 2. Here is a sketch of the proof using the Laplace's method:

By approximating the sum by integral and invoking the Stirling's formula, for any fixed large $N_0$ and for any $N \in \{N_0+1, N_0+2, \cdots\} \cup \{+\infty\}$, we expect:

$$ \sum_{n=N_0}^{N} \frac{n^{2k}}{(k!)^2} \approx \frac{1}{2\pi} \int_{N_0}^{N} \frac{n^{2x}}{x^{2x+1} e^{-2x}} \, \mathrm{d}x. $$

Now by writing

$$ \frac{n^{2x}}{x^{2x+1} e^{-2x}} = \exp\biggl( 2n - \log n - \frac{x-n}{n} - \int_{n}^{x} (x - t)\frac{2t-1}{t^2} \, \mathrm{d}t \biggr) $$

and substituting $x = n+\sqrt{n}z$ and $t = n+\sqrt{n}u$, we get

$$ \frac{1}{2\pi} \int_{N_0}^{N} \frac{n^{2x}}{x^{2x+1} e^{-2x}} \, \mathrm{d}x = \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} \exp\biggl( -\frac{z}{\sqrt{n}} - \int_{0}^{z} (z - u) \frac{2 + \frac{2u}{\sqrt{n}}-\frac{1}{n}}{\bigl( 1 + \frac{u}{\sqrt{n}}\bigr)^2} \, \mathrm{d}u \biggr) \, \mathrm{d}z. $$

Then, as $n\to\infty$, we expect this to become close to:

$$ \approx \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} \exp\biggl( - \int_{0}^{z} 2(z - u) \, \mathrm{d}u \biggr) \, \mathrm{d}z = \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} e^{-z^2} \, \mathrm{d}z. $$

Applying this to $N = n$ and $N = +\infty$ would then show that their ratio converges to

$$ \frac{\int_{-\infty}^{0} e^{-z^2} \, \mathrm{d}z}{\int_{-\infty}^{\infty} e^{-z^2} \, \mathrm{d}z} = \frac{1}{2}. $$


Addendum. For the second question, a numerical evidence suggests that

$$ \lim_{n\to\infty} \sqrt{n}\Biggl( \frac{\sum_{k=0}^{n} n^{2k}/(k!)^2}{\sum_{k=0}^{\infty} n^{2k}/(k!)^2} - \frac{1}{2} \Biggr) = \frac{5}{6\sqrt{\pi}}. $$

However, I have no simple idea for proving this.

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    $\begingroup$ Very nice solutions, for sure ! Cheers $\endgroup$ – Claude Leibovici Feb 21 '20 at 9:02
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Preliminaries

Lemma $\bf{1}$: For $-1\lt x\lt1$, $$ e^{-\frac{x}{1-x}}\le1-x\le e^{-x}\le\tfrac1{1+x}\le e^{-\frac{x}{1+x}}\tag{1} $$ Proof: For all $x\in\mathbb{R}$, Bernoulli's Inequality gives $$ \begin{align} 1+x &\le\lim_{n\to\infty}\left(1+\frac xn\right)^n\\ &=e^x\tag{1a} \end{align} $$ Taking the reciprocal of $\text{(1a)}$, for $x\gt-1$, gives $$ e^{-x}\le\frac1{1+x}\tag{1b} $$ Substituting $x\mapsto-x$ in $\text{(1a)}$ gives $$ 1-x\le e^{-x}\tag{1c} $$ Substituting $x\mapsto\frac{x}{1+x}$ in $\text{(1c)}$ gives $$ \frac1{1+x}\le e^{-\frac{x}{1+x}}\tag{1d} $$ Substituting $x\mapsto\frac{x}{1-x}$ in $\text{(1b)}$ gives, for $x\lt1$, $$ e^{-\frac{x}{1-x}}\le1-x\tag{1e} $$ $\large\square$

Lemma $\bf{2}$: For $|x-y|\le1$, $$ \left|\,e^x-e^x\,\right|\le3|x-y|\,e^{\min(x,y)}\tag2 $$ Proof: $$ \begin{align} \left|\,e^x-e^y\,\right| &\le|x-y|\,e^{\max(x,y)}\tag{2a}\\[3pt] &=|x-y|e^{|x-y|}e^{\min(x,y)}\tag{2b}\\[3pt] &\le3|x-y|\,e^{\min(x,y)}\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: Mean Value Theorem
$\text{(2b)}$: $\max(x,y)=\min(x,y)+|x-y|$
$\text{(2c)}$: $e^{|x-y|}\lt3$ for $|x-y|\le1$

$\large\square$

Theorem $\bf{1}$: If $k\le n$ $$ e^{-\frac{k(k-1)}{2(n-k+1)}}\le\overbrace{\prod_{j=0}^{k-1}\left(1-\frac jn\right)}^{n^{\underline{k}}/n^k}\le e^{-\frac{k(k-1)}{2n}}\le\overbrace{\prod_{j=0}^{k-1}\left(1+\frac jn\right)^{-1}}^{n^k/n^{\overline{k}}}\le e^{-\frac{k(k-1)}{2(n+k-1)}}\tag3 $$ Proof: Set $x=\frac jn$ in Lemma $1$: $$ e^{-\frac{j}{n-j}}\le1-\frac{j}{n}\le e^{-\frac{j}{n}}\le\frac1{1+\frac{j}{n}}\le e^{-\frac{j}{n+j}}\tag{3a} $$ For $0\le j\le k-1$, $\text{(3a)}$ implies $$ e^{-\frac{j}{n-k+1}}\le1-\frac{j}{n}\le e^{-\frac{j}{n}}\le\frac1{1+\frac{j}{n}}\le e^{-\frac{j}{n+k-1}}\tag{3b} $$ Take the product of $\text{(3b)}$ from $j=0$ to $j=k-1$.

$\large\square$

Inequality $\bf{1}$: If $k\le n^{5/9}$, then $$ \begin{align} \frac{k(k-1)^2}{n^2-(k-1)^2} &\le\frac{n^{5/9}\left(n^{5/9}-1\right)^2}{n^2-\left(n^{5/9}-1\right)^2}\tag{4a}\\ &\le\frac{n^{10/9}\left(n^{5/9}-1\right)}{n^2-n^{13/9}}\tag{4b}\\[3pt] &=\frac1{n^{1/3}}\tag{4c} \end{align} $$ Explanation:
$\text{(4a)}$: $k\le n^{5/9}$
$\text{(4b)}$: $n^{5/9}-1\le n^{5/9}$ and $\left(n^{5/9}-1\right)^2\le n^{13/9}$
$\text{(4c)}$: cancel common factors

Inequality $\bf{2}$: If $k\gt n^{5/9}$, then $$ \begin{align} \frac{k(k-1)}{n+k-1} &\ge\frac{k(k-1)}{k^{9/5}+k-1}\tag{5a}\\ &\ge k^{1/5}-2k^{-3/5}\tag{5b}\\ &\ge k^{1/5}-\frac2{n^{1/3}}\tag{5c} \end{align} $$ Explanation:
$\text{(5a)}$: $n\lt k^{9/5}$
$\text{(5b)}$: cross multiply and compare
$\text{(5c)}$: $k\gt n^{5/9}$


Approximating the squares of $\boldsymbol{n^k/n^{\overline{k}}}$ and $\boldsymbol{n^{\underline{k}}/n^k}$

Choose $\epsilon\gt0$ and let $n\ge\max\!\left(\epsilon^{-3},8\right)$.

If $k\le n^{5/9}$, then $$ \begin{align} \left|\,\left(\frac{n^k}{n^{\overline{k}}}\right)^2-e^{-\frac{k(k-1)}{n}}\,\right| +\left|\,e^{-\frac{k(k-1)}{n}}-\left(\frac{n^{\underline{k}}}{n^k}\right)^2\,\right| &\le\left|\,e^{-\frac{k(k-1)}{n+k-1}}-e^{-\frac{k(k-1)}{n-k+1}}\,\right|\tag{6a}\\ &\le3\frac{2k(k-1)^2}{n^2-(k-1)^2}\,e^{-\frac{k(k-1)}{n}}\tag{6b}\\[6pt] &\le6\epsilon\,e^{-\frac{k(k-1)}{n}}\tag{6c} \end{align} $$ Explanation:
$\text{(6a)}$: Theorem $1$
$\text{(6b)}$: Lemma $2$
$\text{(6c)}$: Inequality $1$ implies $\frac{2k(k-1)^2}{n^2-(k-1)^2}\le\min(1,2\epsilon)$

If $k\gt n^{5/9}$, then Inequality $2$ says then $$ \frac{k(k-1)}{n-k+1}\ge\frac{k(k-1)}{n}\ge\frac{k(k-1)}{n+k-1}\ge k^{1/5}-1\tag7 $$ Thus, the squares of the remainders outside of the range where $(6)$ holds can be bounded by $$ \sum_{k\gt n^{5/9}}e^{-k^{1/5}+1}=O\!\left(n^{4/9}e^{-n^{1/9}}\right)\tag8 $$ Furthermore, using Riemann Sums, we have $$ \begin{align} \frac1{\sqrt{n}}\sum_{k=0}^n e^{-\frac{k(k-1)}{n}} &=\int_0^\infty e^{-x^2}\,\mathrm{d}x+O\!\left(\frac1{\sqrt{n}}\right)\\ &=\frac{\sqrt\pi}2+O\!\left(\frac1{\sqrt{n}}\right)\tag9 \end{align} $$ since the variation of $e^{-x^2}$ is $1$ and the step-size is $\frac1{\sqrt{n}}$.


Answer to Part $\bf{1}$

Computing the sum for $\boldsymbol{m\le n}$: $$ \begin{align} \sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2 &=\sum_{k=0}^n\left(\frac{n^{n-k}}{(n-k)!}\right)^2\tag{10a}\\ &=\left(\frac{n^n}{n!}\right)^2\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2\tag{10b}\\ &=\left(\frac{n^n}{n!}\right)^2\left[\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)\right]\tag{10c} \end{align} $$ Explanation:
$\text{(10a)}$: $m=n-k$
$\text{(10b)}$: pull out a common factor
$\text{(10c)}$: $(6)$ and $(8)$ say that $\sum\limits_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2=\sum\limits_{k=0}^ne^{-\frac{k(k-1)}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)+O\!\left(n^{4/9}e^{-n^{1/9}}\right)$
$\phantom{\text{(10c):}}$ which, by $(9)$, is $\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)$

Computing the sum for $\boldsymbol{m\gt n}$: $$ \begin{align} \sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2 &=\sum_{k=2}^\infty\left(\frac{n^{n+k-1}}{(n+k-1)!}\right)^2\tag{11a}\\ &=\left(\frac{n^n}{n!}\right)^2\sum_{k=2}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2\tag{11b}\\ &=\left(\frac{n^n}{n!}\right)^2\left[\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2-2\right]\tag{11c}\\ &=\left(\frac{n^n}{n!}\right)^2\left[\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)\right]\tag{11d} \end{align} $$ Explanation:
$\text{(11a)}$: $m=n+k-1$
$\text{(11b)}$: pull out a common factor
$\text{(11c)}$: $n^k/n^{\overline{k}}=1$ for $k=0$ and $k=1$
$\text{(11d)}$: $(6)$ and $(8)$ say that $\sum\limits_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2=\sum\limits_{k=0}^ne^{-\frac{k(k-1)}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)+O\!\left(n^{4/9}e^{-n^{1/9}}\right)$
$\phantom{\text{(11d):}}$ which, by $(9)$, is $\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)$

Thus, $(10)$ and $(11)$ imply $$ \bbox[5px,border:2px solid #C0A000]{\quad\frac{\displaystyle\sum\limits_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum\limits_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}=\frac12+O\!\left(n^{-1/3}\right)\quad}\tag{12} $$ An error term of $O\!\left(n^{-1/3}\right)$ is insufficient to get the answer to Part $2$.


More Preliminaries

Squaring the two leftmost inequalities from $(3)$: $$ e^{-\frac{k^2-k}{n-k+1}}\le\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)^2\le e^{-\frac{k^2-k}{n}}\tag{13} $$ Similar to Theorem $1$, but setting $x=\frac{j^2}{n^2}$, $$ \begin{align} e^{\frac{2k^3-3k^2+k}{3n^2}}-1&\le\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\le e^{\frac{2k^3-3k^2+k}{3n^2-3(k-1)^2}}-1\tag{14a}\\ \frac{2k^3-3k^2+k}{3n^2}&\le\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\le\frac{2k^3-3k^2+k}{3n^2-2k^3+5k-1}\tag{14b} \end{align} $$ Explanation:
$\text{(14a})$: $\sum\limits_{j=0}^{k-1}j^2=\frac{2k^3-3k^2+1}6$
$\text{(14b})$: $x\le e^x-1$ and $e^x-1\le\frac{x}{1-x}$

For $k\le n^{5/9}$, $(13)$ is $e^{-\frac{k^2}n}\left(1+O\!\left(n^{-1/3}\right)\right)$ and $(14)$ is $\frac{2k^3}{3n^2}\left(1+O\!\left(n^{-1/3}\right)\right)$.

For $k\gt n^{5/9}$, the bounds of $(8)$ still hold.

Using Riemann Sums we have $$ \begin{align} \sum_{k=0}^\infty\frac{2k^3}{3n^2}e^{-\frac{k^2}{n}} &=\frac23\int_0^\infty x^3e^{-x^2}\,\mathrm{d}x\tag{15a}+O\!\left(\frac1{\sqrt{n}}\right)\\ &=\frac13+O\!\left(\frac1{\sqrt{n}}\right)\tag{15b} \end{align} $$ because the variation of $x^3e^{-x^2}$ is $\sqrt{\frac{27}2}e^{-3/2}$ and the step size is $\frac1{\sqrt{n}}$.


Approximating the difference of the squares of $\boldsymbol{n^k/n^{\overline{k}}}$ and $\boldsymbol{n^{\underline{k}}/n^k}$ $$ \begin{align} \left(\frac{n^k}{n^{\overline{k}}}\right)^2-\left(\frac{n^{\underline{k}}}{n^k}\right)^2 &=\prod_{j=0}^{k-1}\left(1+\frac jn\right)^{-2}-\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16a}\\ &=\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\left(1-\frac{j^2}{n^2}\right)^{-2}-\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16b}\\ &=\left[\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\right]\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16c}\\[3pt] &=\frac{2k^3}{3n^2}\,e^{-\frac{k^2}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)\tag{16d} \end{align} $$ Explanation:
$\text{(16a)}$: write the fractions as products
$\text{(16b)}$: $(1+x)^{-1}=(1-x)\left(1-x^2\right)^{-1}$
$\text{(16c)}$: redistribute a common factor
$\text{(16d)}$: $(13)$ and $(14)$


Answer to Part $\bf{2}$

$$ \begin{align} \frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}-\frac12 &=\frac12\,\frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2-\sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2+\sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2}\tag{17a}\\ &=\frac12\,\frac{\displaystyle\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2-\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2+2}{\displaystyle\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2+\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2-2}\tag{17b}\\ &=\frac12\,\frac{\displaystyle2-\sum\limits_{k=0}^\infty\frac{2k^3}{3n^2}\,e^{-\frac{k^2}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)}{\sqrt{\pi n}+O\!\left(n^{1/6}\right)}\tag{17c}\\ &=\frac12\,\frac{\displaystyle\frac53+O\!\left(n^{-1/3}\right)}{\sqrt{\pi n}+O\!\left(n^{1/6}\right)}\tag{17d}\\[9pt] &=\frac5{6\sqrt{\pi n}}+O\!\left(n^{-5/6}\right)\tag{17e} \end{align} $$ Explanation:
$\text{(17a)}$: split the sum in the denominator into two parts
$\text{(17b)}$: apply equations $\text{(10b)}$ and $\text{(11c)}$ and cancel the factors of $\left(\frac{n^n}{n!}\right)^2$
$\text{(17c)}$: apply $(16)$ and $\text{(10c)}$ and $\text{(11d)}$
$\text{(17d)}$: apply $(15)$
$\text{(17e)}$: simplify

Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\quad\sqrt{n}\left[\frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}-\frac12\right] =\frac5{6\sqrt\pi}+O\!\left(n^{-1/3}\right)\quad}\tag{18} $$

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  • $\begingroup$ Great approximation for the summation! $\endgroup$ – Kemono Chen Mar 5 '20 at 9:29
  • $\begingroup$ @KemonoChen: Thanks! I tried to keep things as elementary as possible, with just a couple of integrals (when $\sqrt\pi$ shows up, it's hard to avoid calculus altogether). Equation $(16)$ is the key to the second answer. My guess is that the error term in $(18)$ is actually $O\!\left(\frac1n\right)$, but I need better estimates to show that. $\endgroup$ – robjohn Mar 5 '20 at 17:57

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