TITLE: Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n^{2k}}{(k!)^2}\big/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}$
QUESTION [16 upvotes]: 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.

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