Borel-Cantelli-related exercise: Show that $\sum_{n=1}^{\infty} p_n < 1 \implies \prod_{n=1}^{\infty} (1-p_n) \geq 1- S$.

This is supposed to be related to the 2nd Borel-Cantelli Lemma (my justification for the independence tag). In Williams' Probability with Martingales, 2BCL is proven and then the following is given as an exercise:

Prove $S \doteq \sum_{n=1}^\infty p_n < \infty \implies \prod_{n=1}^\infty (1-p_n) > 0$ assuming $0 \leq p_n < 1$.

Hint: Show that $S < 1 \implies \prod_{n=1}^\infty (1-p_n) \geq 1 - S$.

Proving the hint:

I tried to find $\{a_n\}$ s.t.

$$\prod_{n=1}^\infty (1-p_n) \geq \prod_{n=1}^\infty e^{a_n} = e^{\sum_{n=1}^\infty a_n} \geq 1 - \sum_{n=1}^\infty a_n \geq 1 - S > 0$$

That is, find $\{a_n\}$ s.t.:

1a. $1 - p_n \geq e^{a_n}$

1b. $a_n \leq p_n$

I thought of $a_n = \ln(1-p_n)$ (a reason why $p_n < 1$, I guess)

Is that right?

Edit: Actually, assuming my proof is right, is $S < 1$ used in the proof? If so, where?

From Williams book:

Solution 1:

Hint for the case $S<1$: Prove this by induction for a finite number of probabilities using that $(1-a)(1-b)\ge 1-a -b$ and then proceed to the limit.

Hint for the rest of proof: If $\sum_{n=1}^\infty p_n<\infty$, then there exists $N$ such that $\sum_{n=N}^\infty p_n<1$.