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If, $0<a_n< 1, \forall n$, show that $\sum_{n=1}^{+\infty} a_n$ converges iff $\prod_{n=1}^{+\infty} (1 - a_n)$ converges to a non-zero number.

As $1-x \leq e^{-x}$, we have that $\prod_{n=1}^N (1 - a_n) \leq \prod_{n=1}^N e^{-a_n} = e^{-\sum_{n=1}^N a_n}$. Then, by the Monotone Convergence Theorem, if $\sum_{n=1}^{+\infty}a_n$ converges, then $\prod_{n=1}^{+\infty} (1 - a_n)$ converges.

I'm having some trouble to conclude the other direction.

Thank you very much!

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    $\begingroup$ In your application of the monotone convergence theorem, how can you conclude, given only an upper bound, that the non-increasing sequence converges to a non-zero number? What you can conclude is the other way: if the series diverges (then to $+\infty$ since all terms are positive), then the infinite product must converge to $0$. $\endgroup$ – Clement C. May 19 '17 at 18:47
  • $\begingroup$ Because I know that it converges, by the Monotone Convergence Theorem. As $a_n < 1$, $\forall n$, we have that the convergence must be to a non-zero number (because $(1- a_n) > 0$). Am I correct? $\endgroup$ – Luísa Borsato May 19 '17 at 18:49
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    $\begingroup$ No, you are not. You know that it converges anyway, since the sequence of partial products is positive and decreasing (so it converges to some $\ell\in[0,1)$). By giving an upper bound, you cannot conclude that $\ell>0$ -- it still could very well be zero. @LuísaBorsato $\endgroup$ – Clement C. May 19 '17 at 18:50
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    $\begingroup$ Well, I am not being mean-- if I were, I'll just ignore what you did and not try to provide some help. I am merely trying to make sure I convey my point clearly (since, after your edit, you still have, for instance "if the series converges the product converges", which while true is vacuous: the product converges even if the series doesn't) $\endgroup$ – Clement C. May 19 '17 at 19:03
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    $\begingroup$ @LuísaBorsato no, what about $1-a_n = 2/3$? Then the product is $0$ even though the factors are not. Remember, limits aren't the same as finite products. If you have a finite product of non-zero things the result is always non-zero. $\endgroup$ – Adam Hughes May 19 '17 at 19:07
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Use that $\displaystyle e^x< {1\over 1-x}$ (Bernoulli's inequality) when $|x|<1$. Then WLOG we may assume $|a_n|<1$. As $\prod_{i=1}^\infty (1-a_n)$ converges to a nonzero number iff $\prod_{i=1}^\infty (1-a_n)^{-1}$ does, we see that

$$e^{\sum a_n}<\prod_{i=1}^\infty {1\over 1-a_n}$$

proving the other direction.

As Clement C. has noted, your conclusion does not show the product converges to something non-zero. For that, you can use that as the sum $\sum a_n<\infty$ converges, since then we just note that

$$\lim_{n\to\infty} {\log (1-a_n)\over a_n}=-1$$

so $\sum \log(1-a_n)$ converges, which is only possible if the product does not converge to $0$.

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hint

use the equivalence test $$-\ln (1-a_n)\sim a_n \;\;(n\to+\infty) $$ so the series $$\sum a_n \;\;\text {and}\;\; \sum -\ln (1-a_n) $$ are both convergent or both divergent.

if $\prod (1-a_n)$ converges to zero, it means that $\sum -\ln (1-a_n) $ is divergent and so is $\sum a_n $.

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  • $\begingroup$ I'm not going to downvote, but this answer misses a very important point: what the OP did in their question is incorrect -- any good answer should point that out. $\endgroup$ – Clement C. May 19 '17 at 18:55

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