probability summation for an infinite sequence Very simple case, but don't know how to prove
say function $rand$ returns a uniformly sampled value in $(0,1)$
$x_0 = 1$
$x_1 = rand * x_0$
$x_2 = rand * x_1$
...
$x_n = rand * x_{n-1}$
Now the summation $S(n) = \sum_{i=0}^n x_i$
Is $\lim_{n\to\infty} S(n)$ finite?
It must be, but how to show ?
 A: Proof Sketch: We show that each term in the sum can be bounded by the corresponding term in a geometric sequence - well, except maybe for a finite number of them.
Proof: Let $r_1,\ldots,r_n$ be random variables denoting the outputs of the rand function. Observe that $r_1,\ldots,r_n$ and for all $i$ we have $E[r_i]=\frac 1 2$. Also, we have $x_i = \prod_{1\le j\le i} r_j$ and since $r_i$'s are independent, we can write 
$$E[x_i] = \prod_{1\le j \le i} E[r_j] = \frac 1 {2^i} $$ 
For each $i$, let $E_i$ denote the event $x_i\ge \left(\frac 3 4\right)^i$.
Then, by the Markov's inequality we have:
\begin{align}
\Pr[E_i] = \Pr\left[x_i\ge \left(\frac 3 4\right)^i\right] \le \left(\frac 4 3\right)^i \cdot \frac 1 {2^i} = \left(\frac 2 3\right)^i
\end{align}
Thus we have $\sum_{n=1}^\infty \Pr[E_n]$ is bounded by a geometric series with common ratio $2/3$, and therefore  $\sum_{n=1}^\infty \Pr[E_n] <\infty$. Now, by the Borel-Cantelli lemma, only a finite number of $E_n$'s occur. Hence, we can break the sum as follows:
$$S_n = \sum_{i: E_i\mbox{ occurs}} x_i + \sum_{i: E_i\mbox{ does not occur}} x_i,$$
where the first sum is finite, and the second sum is upperbounded by a geometric series with common ratio $3/4$. This yields $\lim_{n\rightarrow \infty} S_n <\infty$.
A: The sequence $S(n)$ increases to $S$, and by monotone convergence 
$$\mathbb{E}(S)=\lim_n\, \mathbb{E}(S(n))=\lim_n\,\sum_{i=0}^n \mathbb{E}(x_i)=\lim_n\,\sum_{i=0}^n \left({1\over 2}\right)^i=2<\infty.$$
Therefore $\mathbb{P}(S<\infty)=1$.
