Suppose $1>a_n>0$ for $n\in \mathbb{N}$. Prove that $\prod_{n=1}^\infty (1-a_n)=0$ converges if and only if $\sum_{n=1}^\infty a_n=\infty$.

Question

Suppose $1>a_n>0$ for $n\in \mathbb{N}$. Prove that $$\prod_{n=1}^\infty (1-a_n)=0$$ converges if and only if $\sum_{n=1}^\infty a_n=\infty$.

Proof of $\Rightarrow$: Assume that $\prod_{n=1}^\infty (1-a_n)=0$ converges. Well then at some $i$, $1-a_i=0$, which implies $a_i=1$. Knowing this if we were to $$\sum_{n=1}^\infty a_n=\underbrace{1+\dots+1}_{\infty \text{ times}}=1\cdot \infty=\infty.$$

Proof of $\Leftarrow$: Assume $\sum_{n=1}^\infty a_n=\infty$. How would I do this direction?

Answer 1

The convention for infinite products is to say that the infinite product diverges to $0$ when the limit of partial products is $0$.

To get you on the correct track, here is one side of a correct proof. Start with the well-known inequality $1-x \leqslant e^{-x}$ for $0 \leqslant x < 1$.

With $ 0 < a_n < 1$, it follows that for any $m \in \mathbb{N}$,

$$0 < P_m = \prod_{n=1}^m(1-a_n) \leqslant \prod_{n=1}^m e^{-a_n} = \exp\left(-\sum_{n=1}^ma_n\right).$$

Hence if $\sum_{n=1}^{\infty}a_n = \infty$ then $\lim_{m \to \infty} P_m = 0.$

Answer 2

For the other direction: There is an easy generalization of Bernoulli's inequality that reads $$ \prod_{n=N}^\infty (1-a_n)\ge 1-\sum_{n=N}^\infty a_n $$ so if the series converges one can find an $N$ such that the remainder is smaller than $\sum_{n=N}^\infty a_n<\frac12$ and thus $$ \prod_{n=1}^\infty (1-a_n)\ge\frac12\prod_{n=1}^{N-1} (1-a_n)>0. $$