Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 recurssion 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 $S\ge1$.

share|improve this question
1  
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 8 '12 at 7:07
1  
math.stackexchange.com/questions/158089/… –  user17762 Oct 8 '12 at 7:09
add comment

1 Answer 1

up vote 2 down vote accepted

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

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

share|improve this answer
    
+1. Please use \prod. –  Did Oct 8 '12 at 7:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.