Variance of sum of n distinct Uniform Random variables

Question

I came across a problem in Quantguide which read:
Suppose you continually randomly sample nested intervals from [0,1], halving the size each time. That is, the next interval is [x,x+0.5], where x∼U(0,0.5), and so on. What is the variance of the point this converges to?
My Approach:
We model the problem as a sum of distinct uniform random variables such that the random variables are $U_1 \sim U(0,1/2)$ , $U_2 \sim U(0,1/4)$, $U_3 \sim U(0,1/8)$ ,...., $U_n \sim U(0,1/(2^n))$. Now the point of convergence will be $X=U_1 + U_2 + U_3 + \dots +U_n$.
Hence the Variance(X) is $\Sigma_{i=1}^{n}Variance(U_{i})$. Now we know the variance of U(a,b) to be $\frac{(b-a)^2}{12}$. Hence the final answer we get is 1/36.

Answer 1

Not sure what the question is here, but, your general approach is correct. I'll just add to your answer by computing the limit:

$$ \begin{align*} \lim_{n\to\infty} \sum_{i=1}^n\text{Var}(U_i) &= \lim_{n\to\infty}\sum_{i=1}^n \dfrac{2^{-2i}}{12}\\ &= \dfrac{1}{12}\times \dfrac{1}{3}\qquad\text{Result Below}\\ &= \dfrac{1}{36} \end{align*} $$

How do I know that $\sum_{i}^{\infty} 2^{-2i}=1/3$?

$$ \begin{align*} S&=\dfrac14+\dfrac{1}{16}+\dfrac{1}{64}+\cdots\\ 4S &= 1 + \dfrac{1}{4}+\dfrac{1}{16}+\cdots\\ 4S-S &= 1 + \dfrac{1}{4}+\dfrac{1}{16}+\cdots-(\dfrac14+\dfrac{1}{16}+\dfrac{1}{64}+\cdots)\\ 3S &= 1 \end{align*} $$

So, that's how you get $1/36$.