# summable sequence

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 $\beta_n\in[0,1]$.

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$\beta_{17}=0$, $\beta_i=1$ for $i\ne17$. – Gerry Myerson Jun 30 '12 at 12:57
@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. – user20266 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

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.

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thanks, I have also found this link, which deals with my question and more. – h.h.543 Jul 1 '12 at 7:13