Is it true that the $$\prod_{p}{\frac{p-1}{p+1}}=0$$ where the product runs over the prime numbers $p$?
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1$\begingroup$ $\sum_{p}\frac{1}{p}$ is divergent hence it holds for sure. Can you prove that $\sum_{p}\frac{1}{p}$ is divergent? $\endgroup$– Jack D'AurizioCommented Mar 22, 2017 at 21:41
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$\begingroup$ Hint: for any $s>1$, $$\sum_{p}\frac{1}{p^s} = O(1)+\sum_{p}-\log\left(1-\frac{1}{p^s}\right) = O(1)+\log\zeta(s) $$ and the Riemann $\zeta$ function has a pole (with residue $1$) at $s=1$. $\endgroup$– Jack D'AurizioCommented Mar 22, 2017 at 21:43
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$\begingroup$ Sure there is simple theorem proving this: en.wikipedia.org/wiki/… $\endgroup$– Gevorg HmayakyanCommented Mar 22, 2017 at 21:44
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2$\begingroup$ Of course. It is interesting to prove that the same sum ranging over twin primes is indeed convergent. That is usually done through Brun's sieve. $\endgroup$– Jack D'AurizioCommented Mar 22, 2017 at 21:52
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$\begingroup$ A closely related product with a bit more information about the rate of divergence $\endgroup$– Jyrki LahtonenCommented Mar 23, 2017 at 19:09
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There is a theorem stating that $$ \prod_{i=0}^{\infty} (1-a_i)=0 \iff \sum_{i=0}^{\infty}a_i=\infty \quad\quad\quad0 <a_i<1$$
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Therefore, yes, it converges to zero sice $\frac{p-1}{p+1}=1-\frac{2}{p+1}$ and $\sum_p \frac{1}{p} = \infty$.
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$\begingroup$ That's a good theorem! Does it have a name? $\endgroup$ Commented Mar 22, 2017 at 21:21
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1$\begingroup$ @StellaBiderman It has been mentioned here: en.wikipedia.org/wiki/Infinite_product $\endgroup$– MR_BDCommented Mar 22, 2017 at 21:31
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1$\begingroup$ Follows from $e^x < 1/(1-x)$ if $0 < x < 1$. Look at the power series to see a proof. $\endgroup$ Commented Mar 22, 2017 at 22:50
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1$\begingroup$ It's a discovery of Leonhard Euler that $\prod_{p\in P}\;(1-p^{-1})=0.$ His collected works, on paper, will literally fill a room. So nothing is called "Euler's Theorem". $\endgroup$ Commented Mar 23, 2017 at 13:43
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