The infinite series $\sum_i a_i \prod_{j=0}^{i - 1}(1 - a_j)$ has sum equal to $1$ Suppose we have an infinite sequence of constants $\{ a_i \}_{i=0}^{\infty}$, where $0 < a_i < 1$ for every $i$, and we define $b_i \equiv a_i \prod_{j=0}^{i - 1}(1 - a_j)$.  How do we prove $\sum_{i=0}^{\infty} b_i = 1$?
 A: The stated result is false in general.
Imagine that you have coins $C_k$ for $k\in\Bbb N$, and that $a_k$ is the probability that $C_k$ comes up heads when you flip it. Now you flip the coins once each in sequence. Let $E_k$ be the event that the first head flipped is on $C_k$; then the events $E_k$ are pairwise disjoint, and $b_k$ is the probability of $E_k$. The event complementary to $\bigcup_{k\in\Bbb N}E_k$ is that you flip no heads, an event whose probability is clearly $\prod_{k\in\Bbb N}(1-a_k)$. Thus, $\sum_{k\in\Bbb N}b_k=1$ if and only if $\prod_{k\in\Bbb N}(1-a_k)=0$.
Recursively construct a sequence $\langle a_k:k\in\Bbb N\rangle$ in $(0,1)$ as follows. Let $a_0=\frac12$. Given $n\in\Bbb Z^+$ and $a_k$ for $k<n$ such that $1>\prod_{k<n}(1-a_k)>\frac14$, let
$$1-a_n=\frac12\left(\frac{1/4}{\prod_{k<n}(1-a_k)}+1\right)=\frac1{8\prod_{k<n}(1-a_k)}+\frac12\;;$$
clearly $0<1-a_n<1$, so $0<a_n<1$, and $\prod_{k\le n}(1-a_k)>\frac14$, so the construction goes through. 
But $\prod_{k\le n}(1-a_k)>\frac14$ for each $n\in\Bbb N$, so $\prod_{k\in\Bbb N}(1-a_k)\ge\frac14>0$, and $\sum_{k\in\Bbb N}b_k<1$.
Alternatively, you can use the argument suggested by achille hui in the comments, but note that the partial sums of the telescoping series are of the form $b_0+c_0-c_n$, where $c_m=\prod_{k\le m}(1-a_k)$, and
$$b_0+c_0-c_n=a_0+(1-a_0)-c_n=1-c_n=1-\prod_{k\le n}(1-a_k)\;.$$
This tends to $1$ as $n\to\infty$ if and only if $\prod_{k\in\Bbb N}(1-a_k)=0$, which, as we just saw, need not be the case.
