# Convergence of infinite product of prime reciprocals?

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!)

• What does the few expanded product to series look like? Oct 16, 2014 at 10:38
• Since $\left(1-\frac{1}{p_n}\right) \lt 1$ we have that $$\prod_{k=1}^{n}\left(1-\frac{1}{p_k}\right) \lt 1 \;\; \forall n \in \Bbb N$$ Oct 16, 2014 at 10:41
• In case of infinite products they do not converge to zero, they diverge to zero Oct 16, 2014 at 10:42
• Arjang and @Ishfaaq: infinite products all of whose terms are between zero and one CAN converge to non-zero values. See math.stackexchange.com/questions/141705/… 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 Oct 16, 2014 at 10:44
• @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. Oct 16, 2014 at 10:53

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