I want to prove the equivalence of the two conditions below. If someone could help me I would appreciate very much!
$$p_1,p_2,.... \in (0,1)$$$$\sum_{i = 1}^{\infty}(1-p_i)=\infty\iff \lim_{n\to\infty} \prod_{i=1}^{n}p_i=0.$$
I've tried the $\Longleftarrow$ but got stuck:
So, by the limit definition, for every $\epsilon>0$, $\exists N$, such that for every $n>N$:$$\prod_{i=1}^{n}p_i < \epsilon.$$Let's fix an $n$ that satisfies this condition.
For each $1 \leq i \leq n$:$$p_i < \frac{\epsilon}{\prod_{j \neq i}^{n}p_j}$$so:$$1 - p_i > 1 - \frac{\epsilon}{\prod_{j \neq i}^{n}p_j} $$so, summing across all $p$'s:$$\sum_{i = 1}^{n}(1-p_i) > n - \epsilon \sum_{i = 1}^{n}\frac{1}{\prod_{j \neq i}^{n}p_j}$$Now, I could say that $\epsilon$ is "arbitrarily small" and this goes to infinity as n grows, but the denominator also goes to 0 by hypothesis, so I don't know how to proceed. The $\Longrightarrow$ I couldn't even start...