Probability of symmetric random walk being in certain interval on nth step I am considering $(X_N)_{n\ge0}$, a simple symmetric random walk on $\Bbb Z$ starting at $0$. 
I've been trying to prove that for all $x \in \Bbb R$,
$\Bbb P(\frac{X_n}{\sqrt{n}} \in [x, x+2/\sqrt{n})) \sim \sqrt{\frac{{2}}{n\pi}}\cdot e^{-x^2/2}$ as $n \rightarrow \infty $
I have already proved that $\Bbb P({X_n} =0) \sim \sqrt{\frac{{2}}{n\pi}}$, by considering the ways of choosing half the steps to be positive in order to always return to $0$ for even $n$, and simplifying by using Stirling. 
A hint in the question claims that there is exactly one value in the interval $[x, x+2/\sqrt{n})$ that $\frac{X_N}{\sqrt{n}}$ could take, but by considering the inequality $\sqrt{n}x \le X_n \lt \sqrt{n}x +2$, I think there are two possible values. For example if $\sqrt{n}x$ happened to be about $10.3$ then both $11$ and $12$ lie in the interval. Would someone clarify this? I assume I've misunderstood.
In any case, I suspect this more difficult question will again involve simplification by Stirling. 
Please could I have some help? In particular, I suspect I will be comfortable with algebraic manipulation: the most difficult part is finding the best way to tackle $\frac{X_N}{\sqrt{n}}$ reaching the interval. I did attempt to consider $\lceil\sqrt{n}x\rceil$ and adapt the idea from the first part of the question by considering $n\choose\frac{1}{2}(n+\lceil\sqrt{n}x\rceil)$ because I thought that would be the number of ways to reach the correct "height" as it were.
Thank you. 
 A: First, note that it is indeed necessary to consider intervals $[x,x+2/\sqrt n)$, because each $X_n$ has a fixed parity, namely, for every $n$, $$P(X_{2n}\in2\mathbb Z)=P(X_{2n+1}\in2\mathbb Z+1)=1$$ In particular, for every $n$ large enough, there is exactly one integer value $x_n$ in the interval $[x\sqrt n,x\sqrt n+2)$ such that $P(X_n=x_n)\ne0$, which is such that $x_n+n$ is even, say $x_n+n=2t_n$.
The rest is a standard application of Stirling's formula. More precisely, $X_n=x_n$ means that the random walk performs $t_n$ steps to the right and $n-t_n$ steps to the left, hence $$P(X_n=x_n)=\frac 1{2^n}{n\choose t_n}=\frac{n!}{2^nt_n!(n-t_n)!}$$ Since $n\to\infty$, $t_n\to\infty$ and $n-t_n\to\infty$, Stirling's formula applied thrice yields an equivalent of the RHS, which, after some obvious cancellations, reads $$P(X_n=x_n)\sim A_nB_n$$ with $$A_n=\frac{n}{\sqrt{2\pi nt_n(n-t_n)}}\qquad B_n=\left(2\frac{t_n}{n}\right)^{-t_n}\left(2-2\frac{t_n}{n}\right)^{t_n-n}$$
Now, $x_n=o(n)$ hence $t_n\sim n/2$, which implies readily that $$A_n\sim\sqrt{\frac2{\pi n}}$$
Turning to $B_n$, recall that $x_n=x\sqrt{n}+z_n$ with $0\leqslant z_n<2$, hence
$$2\frac{t_n}n=1+\frac x{\sqrt n}+\frac{z_n}n$$
and
$$B_n=\left(1+\frac x{\sqrt{n}}+\frac{z_n}n\right)^{-t_n+n/2}\left(1-\frac x{\sqrt{n}}-\frac{z_n}n\right)^{t_n-n/2}\left(1-\left(\frac x{\sqrt{n}}+\frac{z_n}n\right)^2\right)^{-n/2}$$
Recall that, for every sequences $(g_k)$ and $(h_k)$ such that $g_k\to g$ and $h_k\to h$ when $k\to\infty$, $$\left(1+\frac{g_k}k\right)^{kh_k}\to e^{gh}$$ Using this fact for $(k,g,h)$ equal to $(\sqrt n,x,-\frac12x)$, $(\sqrt n,-x,\frac12x)$ and $(n,-x^2,-\frac12)$ respectively, one gets $$B_n\sim e^{-x^2/2}e^{-x^2/2}e^{x^2/2}=e^{-x^2/2}$$ which concludes the proof of the desired equivalent of $P(X_n=x_n)$.
