An unusual setup where sampling without replacement from an infinite sequence might lead to a central limit theorem Let $a_1, ..., a_N$ denote the alternating sequence $-1, 1, -1, 1, ..., -1, 1$ (assume $N$ is even for simplicity). Let $x_1, ..., x_N$ denote a sequence obtained by random sampling without replacement from the entire alternating sequence. Consider the random variable:
\begin{equation}
s_N = N^{-1/2} \sum_{n=1}^N x_n a_n .
\end{equation}
For finite $N$, $s_N$ is a discrete random variable and when $N$ is small it is straightforward to just write the distribution down. For example, if $N=2$, then $\mathbb{P}(s_N = -2/\sqrt{2}) = 1/2$ and $\mathbb{P}(s_N = 2/\sqrt{2}) = 1/2$. As $N$ gets large, the number of possible outcomes for $s_N$ gets very large.
Interestingly, under simulation, when $N$ is large, $s_N$ starts to look an awful lot like a like a Standard Normal random variable.
Question: Does $s_N$ have a CLT as $N \rightarrow \infty$? If so, how would one go about proving this?
UPDATE: md5 method is very nice. However I've also realized that once you write down the finite sample distribution as md5 has, this is pretty much all you need to do, since that probability distribution just so happens to be a special case of the Hypergeometric distribution, which further just so happens to satisfy the assumptions in equation 7.4 in Chapter 7 of Feller's An Introduction to Probability Theory and its Applications: Volume 1 and so the Normal limit is known. See also this question here

Here it is not too hard to explicitly compute the distribution of $s_{2n}$. By counting the number of $x_k$ that match $a_k$, we see that for all $x\in\{0,\ldots,n\}$,
$$
P(\sqrt{2n}s_{2n}=2n-4x)=\frac{\binom{n}{x}\binom{n}{n-x}}{\binom{2n}{n}}.
$$
Now it suffices to apply some local limit theorem. What we need to prove is a statement of the form: for all
$$
x_n=\frac{n}{2}+\frac{\sqrt{n}t}{4}+o(\sqrt{n}),
$$
it holds that $P(s_{2n}=x_n)/\sqrt{n}\to e^{-t^2/2\sigma^2}/\sqrt{2\pi\sigma^2}.$ This should not be too hard, using standard estimates such as
$$
\binom{n}{\frac{n}{2}+\sqrt{n}t/4+o(\sqrt{n})}2^{-n}\sim\sqrt{\frac{2}{\pi n}}e^{-t^2/2}.
$$
