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

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

Thanks for your help.

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

Use Borel-Cantelli lemma like here enables to conclude.

For a non-pobabilistic proof, see proof and the first comment here

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