Shape of distribution of infinite Sum of weighted Gaussians

Let $x_i$ be samples from gaussian distribution with mean $0$ and variance $\sigma^2$ and $s_n=\sum_{i=0}^n2^ix_i$. What can one say about the distribution of $s_n$ at $n\rightarrow\infty$?

Sum of weighted Gaussians is a Gaussian.

$\mu(s_n)=\sum_i2^i\mu_i=0$

$\sigma^2(s_n)=\sum_i2^{2i}\sigma_i^2=\sigma^2\sum_i2^{2i}=\sigma^2\frac{2^{2n+1}-1}{2^2-1}$

So, $\lim_{n\rightarrow\infty}\sigma^2(s_n)=\infty$

Does the distribution become an almost $\epsilon$ function in distribution since mass of the total distribution has to be $1$?

First note that $2^iX_i\sim N(0,2^{2i}\sigma^2)$. If $f_i$ is the pdf corresponding to $N(0,2^{2i}\sigma^2)$, then the distribution of $s_n$ is the convolution of $f_1,\cdots,f _n$. Now to find the asymptotic distribution of $s_n$, you need some kind of Central Limit Theorem for non-i.i.d. random variables. I think Lyapunov Condition will suffice.
If the samples are independent then $s_n$ has a gaussian distribution with mean $Es_n=0$ and variance $Es_n^2=\Sigma_{i=0}^n2^{2i}E[x_i^2]=\sigma^2\Sigma_{i=0}^n2^{2i}=\sigma^2\frac{4^{n+1}-1}{4-1}$. The result remains valid as n tends to infinity.