$\prod\left(1-p_n\right)>0$

Question

I want to prove that if $0\le p_n<1$ and $\sum p_n<\infty$, then $\prod\left(1-p_n\right)>0$ .

There is a hint : first consider the case $\sum p_n<1$, and then show that $\prod\left(1-p_n\right)\ge1-\sum p_n$ .

How can I use this hint to show the statement above?

Answer 1

Since $1-x\geq \exp\left(-\frac{x}{1-x}\right)$ for any $x\in[0,1)$, it follows that: $$\prod(1-p_n)\geq \exp\left(-\sum \frac{p_n}{1-p_n}\right),$$ but since $\sum p_n$ is converging, it follows that for any $n$ big enough ($n\geq N$) we have $p_n\leq\frac{1}{2}$, so: $$\prod(1-p_n) \geq \exp\left(-\sum_{n=1}^{N}\frac{p_n}{1-p_n}-2\sum_{n>N}p_n\right)>0.$$

Answer 2

Why it is sufficient to prove the hint: Suppose $\sum p_n < \infty$. Then there is an integer $N$ such that $\sum_{n \geq N} p_n < 1$. Now observe that both $\prod_{n < N} (1-p_n)$ (a finite product) and $\prod_{n \geq N} (1-p_n)$ (using the hint) are positive.

Answer 3

Suppose $\sum p_n = C$. Let's notice that there exists $N: \sum_{k=1}^{N}p_n > C-1$. Let's replace N first numbers with $0$. $\prod(1-p_n) = c\prod_{n > N}(1-p_n)$ We know from hint $\prod_{n > N}(1-p_n) > 1-\sum_{n > N} p_n > 0$. Then $\prod(1-p_n) = c\prod_{n > N}(1-p_n) > 0$.