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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 $0 0$, 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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    @Teddy since infinite product is equal or less than u1<12012-12-13

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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))2012-12-13