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Prove that if $0\le p_n< 1$ and $S=\sum p_n< \infty$ then $\prod (1-p_n)>0$. Hint given: First show that if $S<1$, then $\prod (1-p_n)\ge 1-S$.

Attempt: I was able to show the hint by using recursion setting $A_n=\prod_{i=1}^{n}(1-p_i)$ re expression the $\prod (1-p_n)$ as $1-S+\sum_{n_1, n_2=1,n_1<n_2 }^{\infty} p_{n_1}p_{n_2}-..... $ and observed that every subsequent term is less than the previous term hence $\prod (1-p_n)\ge 1-S$ which is $>0$ if $S<1$ but am unsure about how to extend it to $S\ge1$.

Martin Sleziak
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Pk.yd
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    If $S\ge 1$, we can modify the $p_i$ to non-negative $q_i$ with sum $T\lt 1$. Also, $1-q_i \ge 1-p_i$. – André Nicolas Oct 08 '12 at 07:07
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    http://math.stackexchange.com/questions/158089/infinite-products-reference-needed/158099#158099 –  Oct 08 '12 at 07:09

1 Answers1

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As $\sum\limits_n p_n < \infty$, $N$ exists such that $S_{N}=\sum\limits_{n> N} p_n <1$. So $\prod\limits_{n> N} (1-p_n)> 1-S_{N}>0$.

$$\prod\limits_n(1-p_n)=\left(\prod\limits_{n \leq N}(1-p_n)\right) \times \left(\prod\limits_{n> N} (1-p_n)\right) >0 $$

Did
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francis-jamet
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