# Is $\prod_{p}{\frac{p-1}{p+1}}=0$?

Is it true that the $$\prod_{p}{\frac{p-1}{p+1}}=0$$ where the product runs over the prime numbers $p$?

• $\sum_{p}\frac{1}{p}$ is divergent hence it holds for sure. Can you prove that $\sum_{p}\frac{1}{p}$ is divergent? Commented Mar 22, 2017 at 21:41
• 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 $s=1$. Commented Mar 22, 2017 at 21:43
• Sure there is simple theorem proving this: en.wikipedia.org/wiki/… Commented Mar 22, 2017 at 21:44
• 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. Commented Mar 22, 2017 at 21:52
• A closely related product with a bit more information about the rate of divergence Commented Mar 23, 2017 at 19:09

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

Therefore, yes, it converges to zero sice $$\frac{p-1}{p+1}=1-\frac{2}{p+1}$$ and $$\sum_p \frac{1}{p} = \infty$$.
• Follows from $e^x < 1/(1-x)$ if $0 < x < 1$. Look at the power series to see a proof. Commented Mar 22, 2017 at 22:50
• 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". Commented Mar 23, 2017 at 13:43