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I'm studying infinite products appearing in complex analysis these days. There are lots of theorems regarding it, for example, for $0<u_n <1, \prod (1-u_n) > 0$, iff $\Sigma u_n < \infty $.

I come up with a following question which No elementary book say about it..

within above condition, $\prod u_n =1 $ is impossible, but i thought intuitively that by approaching$ u_n$ rapidly to 1, it is possible to get a value of infinite product is larger than 1-$\epsilon$ for any $\epsilon >0$ (of course our$ u_n $can depend on $\epsilon$) if anyone know about theorem regrading it or can prove about this? thanks.

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If $u_n \to 1$, using the theorem you stated, the infinite product will not converge. Another thing, why is $\prod u_n =1$ impossible ? – Teddy Dec 13 '12 at 8:05
@Teddy since infinite product is equal or less than u1<1 – Detectives Dec 13 '12 at 8:58
up vote 1 down vote accepted

Note that if $0 < u_j < 1$, $$ \prod_{j=1}^\infty u_j = \exp\left( \sum_{j=1}^\infty \ln(u_j) \right)$$ So you want $$\ln(1-\epsilon) < \sum_{j=1}^\infty \ln(u_j)$$ Can you think an infinite sum of negative terms that is greater than $\ln(1-\epsilon)$?

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I can think of some geometric series ln(1/u_j) bounded by constant ln(1/(1-e)) – Detectives Dec 13 '12 at 10:52

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