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Where pn is the nth prime number, does the infinite product $$\prod_{n=1}^{\infty}\left(1-\frac{1}{p_n}\right)$$ converge to a nonzero value? (Any help would be much appreciated!)

ManRow
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  • What does the few expanded product to series look like? – jimjim Oct 16 '14 at 10:38
  • Since $ \left(1-\frac{1}{p_n}\right) \lt 1 $ we have that $$\prod_{k=1}^{n}\left(1-\frac{1}{p_k}\right) \lt 1 ;; \forall n \in \Bbb N$$ – Ishfaaq Oct 16 '14 at 10:41
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    In case of infinite products they do not converge to zero, they diverge to zero – jimjim Oct 16 '14 at 10:42
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    Arjang and @Ishfaaq: infinite products all of whose terms are between zero and one CAN converge to non-zero values. See http://math.stackexchange.com/questions/141705/result-of-the-product-0-9-times-0-99-times-0-999-times for an example...

    What I am asking here is if $$\prod_{n=1}^{\infty}\left(1-\frac{1}{p_n}\right)>0$$

    But, as of Liu Gang's answer, it appears that $$\prod_{n=1}^{\infty}\left(1-\frac{1}{p_n}\right)=0 \text{ since } \sum_{n=1}^{\infty}\frac{1}{p_n} diverges $$ so it does in fact converges to zero

     – ManRow Oct 16 '14 at 10:44
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    @ManRow : I was aware of that example, there was a comment that contained "convergence to 0" in case of infinite products that is no correct, they only diverge to 0. – jimjim Oct 16 '14 at 10:53
  • The question has already been answered here. – Dietrich Burde Feb 08 '16 at 22:42

1 Answers1

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If all $a_n \in (0,1)$, $\displaystyle\prod_{n=1}^{+\infty} (1- a_n)$ is non-zero if and only if $\sum_{n=1}^{+\infty} a_n < +\infty$

And we know that $\displaystyle\sum_{n=1}^{+\infty} \frac{1}{p_n} = +\infty$, you can find proofs here

Petite Etincelle
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