# How to prove $\prod_{i=1}^{\infty} (1-a_n) = 0$ iff $\sum_{i=1}^{\infty} a_n = \infty$?

Given ${a_n}$ is infinite sequence, and $0 < a_n < 1$, how to prove

$$\prod_{i=1}^{\infty} (1-a_n) = 0 \text{ if and only if } \sum_{i=1}^{\infty} a_n = \infty$$

• To study infinite products, you can consider its logarithm: if $P_n=\prod_{k=1}^n a_k$, then $S_n=\ln(P_n)=\sum_{k=1}^n\ln(a_n)$. – Taladris Oct 9 '13 at 1:17
• Do you possibly mean something else than $1 < a_n < 1$? I think you mean $-1 < a_n < 1$. – user61527 Oct 9 '13 at 1:23
• @T.Bongers Sorry, it should be $0< a_n < 1$. – Tony He Oct 9 '13 at 1:56

Use $1-a_n \leq e^{-a_n}$ for $\sum_{i=1}^{\infty} a_n = \infty \implies \prod_{i=1}^{\infty} (1-a_n) = 0$
For the other direction, define independent uniform random variables $(U_n)_{n\geq 1}$ and $A_n =\{U_n < a_n\}$, then we have $\prod_{n=1}^{+\infty}P(A_n^c) = 0$