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: 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. 
 A: 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.
