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I am looking for a sequence $\beta_n\in(0,1)$ such that

(i) $~ \prod_{n\in\mathbb N} \beta_n =0$,

(ii) $~\sum_{n\in\mathbb N} (1-\beta_n)< +\infty$.

Does such a sequence exist?

edit: i have changed to $\beta_n\in(0,1)$ instead of $\beta_n\in[0,1]$.

h.h.543
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    $\beta_{17}=0$, $\beta_i=1$ for $i\ne17$. – Gerry Myerson Jun 30 '12 at 12:57
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    @GerryMyerson I was in the process of writing something more elaborate, but since yours is likely to be one of the $17$ simplest solutions one can think of I deleted mine. –  Jun 30 '12 at 13:01
  • Is there such a sequence when $0<\beta_n<1$ for each $n$ – pritam Jun 30 '12 at 13:03
  • With the appropriate hypotheses, convergence of $\prod(1+a_n)$ is equivalent to convergence of $\sum a_n$. – Gerry Myerson Jun 30 '12 at 13:10
  • No such example exists. Any text that discusses infinite products surely includes this. Perhaps written differently, see the topic "absolute convergence" for infinite series. – GEdgar Jun 30 '12 at 13:19
  • A convergent product in which all terms are nonzero can't take the value zero (by convention). – Cocopuffs Jun 30 '12 at 13:22
  • @Cocopuffs $\prod_{i=1}^n 1/2=1/2^n\rightarrow 0$ as $n\rightarrow \infty$ – pritam Jun 30 '12 at 13:44
  • @pritam that is correct. As soon as $a_n\le \theta<1$ for all $n>N_0$ you get convergence to zero. For this reason this case is, by definition, excluded if one talks about convergent infinite products. This is why Cocopuffs wrote 'by convetion'. –  Jun 30 '12 at 14:18

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Assume without loss of generality when (ii) holds that $\beta_n\geqslant1-\frac12\log2$ for every $n\in\mathbb N$. Then, for every $n\in\mathbb N$, $\beta_n\geqslant\mathrm e^{-2(1-\beta_n)}$, hence $$ \prod_{n\in\mathbb N}\beta_n\geqslant\exp\left(-2\sum_{n\in\mathbb N}(1-\beta_n)\right). $$ Then (ii) implies that the RHS is positive hence the LHS is positive and (i) cannot hold.

Did
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