Does CLT fail in this case? We have mutually independent random variables $X_n$ with $P(X_n = 2^n) = P(X_n = -2^n) = \frac12$.
Of course their means $\mu_n = 0$ and variances $\sigma_n^2 = 4^k$. Let $S_n = \sum_1^n X_k$. Clearly the mean $m_n=E(S_n) = 0$ and variance $s_n^2 = E(S_n^2)=  \sum_1^n \sigma_k^2 = \frac13(4^{n+1}-1)$.
I have shown that in this case the law of large numbers does not apply, and need to show that CLT does/does not apply.
The only method I know to show that the CLT does apply would be Lindeberg's theorem, which I have tried to apply but I can't seem to get anywhere with. Since $\max_k \sigma_k^2/s_n^2$ does not go to zero, I cannot use Lindeberg's theorem to show that the CLT doesn't apply, so I am lost on how to proceed.
How might I proceeed otherwise?
 A: There is indeed convergence in distribution, but not to a normal distribution:


*

*The sequence $(2^{-n-1}S_n)_{n\geqslant1}$ converges in distribution to the uniform distribution on $(-1,1)$.


A constructive way to show this is to start from $U$ uniform on $(-1,1)$ and to define inductively $(V_n)_{n\geqslant1}$ and $(Y_n)_{n\geqslant1}$ as follows: 


*

*$V_1=U$, $Y_n=\mathrm{sgn}(V_n)$, $V_{n+1}=2V_n-Y_n$. 


Then every $V_n$ is uniform on $(-1,1)$, $U=\sum\limits_{n\geqslant1}2^{-n}Y_n$ and $(Y_n)_{n\geqslant1}$ is an i.i.d. sequence of symmetric $\pm1$ Bernoulli random variables. This shows that $2^{-n-1}S_n$ is distributed as $U_n=\sum\limits_{k=1}^n2^{-k}Y_k$. Since $U_n\to U$ almost surely, $2^{-n-1}S_n\to U$ in distribution.
Edit: A (somewhat less intuitive but perhaps) more direct way to reach the conclusion is to compute the characteristic function $\varphi_n$ of $2^{-n-1}S_n$, that is,
$$
\varphi_n(t)=\mathbb E(\exp(\mathrm it2^{-n-1}S_n))=\prod_{k=1}^n\cos(2^{-k}t)=\frac{\sin t}{2^n\sin(2^{-n}t)}.
$$
One sees that $\varphi_n\to\psi$ pointwise, where $\psi(t)=(\sin t)/t=\mathbb E(\exp(\mathrm itU))$ is indeed the characteristic function of $U$. Thus, $2^{-n-1}S_n\to U$ in distribution.
