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

I know this question is similar to one I just posted but I am getting confused with the negatives.

Proof: $\Rightarrow$ Assume that $\prod_{n=1}^\infty (1-a_n)=0$ converges. Well, $$\ln\left(\prod_{n=1}^\infty (1-a_n)\right)=\sum_{n=1}^\infty ln(1-a_n).$$ We know that for any $0<a<1$, $$-\frac{a}{1-a}<\ln(1-a)<-a.$$ So, $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\sum_{n=1}^\infty \ln(1-a_n)<-\sum_{n=1}^\infty a_n.$$ This implies that $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\sum_{n=1}^\infty \ln(1-a_n)<\infty$$ and so $$-\sum_{n=1}^\infty \frac{a_n}{1-a_n}<\infty$$ (i.e. $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges).

We need to show that if $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges, then $\sum_{n=1}^\infty \frac{-a_n}{1-a_n}$ converges. Well if $-\sum_{n=1}^\infty \frac{a_n}{1-a_n}$ converges then $$\lim_{n\to \infty}\frac{-a_n}{1-a_n}=0$$ and so is $$\lim_{n\to \infty}\frac{a_n}{1-a_n}=0.$$ This implies that $\{a_n\}$ is bounded. So $a_n\leq M$ for all $n\geq 1$. Thus $$\frac{1}{1-M}a_n\leq \frac{a_n}{1-a_n}.$$ So $\lim_{n\to \infty} a_n=0$ which implies that $\sum_{n=1}^\infty a_n<\infty$.

$\Leftarrow$ Assume that $\sum_{n=1}^\infty a_n<\infty$.

I am not sure if I can say that that if $\sum_{n=1}^\infty a_n<\infty$ then $-\sum_{n=1}^\infty a_n<\infty$.

Answer 1

This statement is incorrect. Pick $a_i=1/2$ for all $i$. $\sum_{i=1}^\infty a_i=\infty$. On the other hand $\prod_{i=1}^n(1-a_i)=0.5^n$ which converges to $0$ as $n\rightarrow\infty$.

Answer 2

Whenever $\sum_{i=1}^\infty a_i$ converges $\prod_{i=1}^\infty (1-a_i)$ converges as well. It follows from the fact that $s_n=\prod_{i=1}^n (1-a_i)$ satisfies $s_{n-1}-s_{n}<a_n$ followed by the fact that increasing bounded sequences converge. I believe this is the correct statement (and intended question).

Note that $\prod_{i=1}^\infty (1-a_i)$ converging doesn't imply the other way around. Pick $a_i=1/2$ for all $i$ as a counterexample.