As noted in the [recent answer by Yuval Peres](https://mathoverflow.net/questions/343769/is-the-sum-of-a-bunch-of-uniform-distributions-say-u-c-c-each-of-which-has-a), the sum of independent uniformly distributed random variables (r.v.'s) cannot have a normal distribution. 


The question is, what happens without the independence condition? More specifically: 



> 
> Do there exist positive real numbers $a\_1,a\_2,\dots$ and r.v.'s $U\_1,U\_2,\dots$ such that for each natural $i$ the r.v. $U\_i$ is uniformly distributed on the interval $[-a\_i,a\_i]$ and the sum $U\_1+U\_2+\cdots$ has the standard normal distribution? 
> 
> 
> 



Normality of the sum of uniformly distributed random variables
Yes. Let the random variables $V\_i$ be independent, with $V\_i$ uniform on $[-\frac 1i,\frac 1i]$. Let $N$ be an independent standard normal. Inductively define $Z\_i\in \{\pm 1\}$ by 
$$
Z\_i=
\begin{cases}
1&\text{if $\text{sgn} (N-\sum\_{j<i}Z\_jV\_j)=\text{sgn}(V\_i)$}\\
-1&\text{otherwise.}
\end{cases}
$$
Notice that $\sum Z\_iV\_i=N$ almost surely - since the $|V\_i|$ approach 0, but are almost surely not summable, the partial sums overshoot successively in one direction then the other, converging to $N$.


An observation that we need is the following: the involution that negates all of the $V$’s and $N$ preserves probabilities; and the values of the $Z$ variables are unchanged. In particular for any $a$ and $i$, we have $\mathbb P(V\_i<a,\, Z\_i=-1)=\mathbb P(V\_i>-a,\, Z\_i=-1)$.


Now we claim that $U\_i=Z\_iV\_i$ has the same distribution as $V\_i$. 
To see this, notice that $\mathbb P(U\_i<a)=\mathbb P(V\_i<a,\,Z\_i=1) +
\mathbb P(V\_i>-a,\, Z\_i=-1)$. But by the above observation, $\mathbb P(V\_i>-a,\,Z\_i=-1)=\mathbb P(V\_i<a,\,Z\_i=-1)$, so that $\mathbb P(U\_i<a)=\mathbb P(V\_i<a)$ as required. 


**added comment**: @Michael Hardy pointed out the use of conditional convergence in this answer. It turns out this is essential: there does not exist a collection of dependent uniform random variables where the sum is (non-trivially) normal and the convergence is absolute. To see this, note that if $\sum a\_i<\infty$, then the sum is compactly supported, so cannot be normal. If $\sum a\_i=\infty$, then $\mathbb E\sum\_{i=1}^M|U\_i|=\sum\_{i=1}^M a\_i/2$ by linearity of expectation. Since $\sum\_{i =1}^M |U\_i|\le \sum\_{i=1}^M a\_i$, we deduce $\mathbb P(\sum\_{i=1}^M |U\_i|\ge \sum\_{i=1}^M a\_i/4)\ge 1/4$. In particular, the limit superior of these sets has measure at least $1/4$, but on this set the sum of the absolute values diverges. 


