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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 $-\sum_{n=1}^\infty a_n<\infty$.

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  • $\begingroup$ Can't you just use your previous question with $b_n = \frac{1}{1-a_n}-1 = \frac{a_n}{1-a_n} > 0$? You have $(1+b_n) = \frac{1}{1-a_n}$, so $\prod (1+b_n)$ converges to some $\ell$ (note that if it converges, it it to a positive value $\geq 1$) iff $\prod (1-a_n)$ converges to $1/\ell$. And $\ln(1-a_n) = -\ln(1+b_n)$. $\endgroup$ – Clement C. Feb 11 '16 at 2:23
  • $\begingroup$ If I do that then will I need to multiply the other inequality I have given by a negative? $\endgroup$ – Username Unknown Feb 11 '16 at 2:33
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    $\begingroup$ You would not really need to do anything with the inequalities: you can just apply the result of your previous question (no need to even dive into its proof), to the new sequence $(b_n)$. $\endgroup$ – Clement C. Feb 11 '16 at 3:15
  • $\begingroup$ The only reason I was tempted to do it again is because the next question is 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 $\sum_{n=1}^\infty a_n=\infty$. $\endgroup$ – Username Unknown Feb 11 '16 at 3:17
  • $\begingroup$ But that is also another case -- with the above reduction, $\prod_1^\infty (1-a_n) = 0$ is equivalent to $\prod_1^\infty (1+b_n) = \infty$, which (by the previous question applied to the positive sequence $(b_n)_n$) is equivalent to $\sum_n b_n = \infty$. You should then be able to use that to show $\sum_n a_n = \infty$. $\endgroup$ – Clement C. Feb 11 '16 at 3:22
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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.

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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 $n\rightarrow\infty$.

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  • $\begingroup$ OP meant that the product converges to zero iff the sum diverges. $\endgroup$ – marty cohen Feb 11 '16 at 3:03
  • $\begingroup$ It is still incorrect. Suppose $a_i=0$ for all $i$. In this case, both of them converge. $\endgroup$ – ecstasyofgold Feb 11 '16 at 3:10
  • $\begingroup$ But $a_i > 0$ by assumption. $\endgroup$ – Clement C. Feb 11 '16 at 3:13
  • $\begingroup$ So is there a typo with the question. $\endgroup$ – Username Unknown Feb 11 '16 at 3:16
  • $\begingroup$ Yes, it should be "converges to some positive value iff." $\endgroup$ – Clement C. Feb 11 '16 at 3:17

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