Let$\ p_n$ be the$\ n$-th prime. Can you give me a proof for$\ \prod_{i=1}^\infty \frac{p_i-1}{p_i}=P\approx \frac{1}{11.0453}$?

Question

I found$\ \prod_{i=1}^\infty \frac{p_i}{p_i-1}\approx 11.0453$ on Wolfram|Alpha. Moreover, writing a paper, should one provide a proof or it is trivial? Thanks in advance.

Answer 1

The product in the body of the question is $$ \prod_{p \text{ prime}} \frac{1}{1 - \frac{1}{p}} $$ which is $\zeta(1)$, or rather, which would be $\zeta(1)$, if the sum defining the Riemann zeta function on the half-plane $\text{Re} (s) > 1$ converged at $s = 1$.

Unfortunately, it doesn't converge; the Riemann $\zeta$ function has a pole at $s = 1$. (This is the multiplicative version of the failure of the harmonic series to converge.) Informally, the product in the body of the question is infinite (and its reciprocal, which is the product in the title of the question is zero). The value that Wolfram Alpha provides appears to be a truncation of this product.

Answer 2

It is quite clear that the result is wrong. Let me consider $$P_n=\ \prod_{i=1}^{10^n} \frac{p_i}{p_i-1}$$ The numerical values are $$P_1=6.33123$$ $$P_2=11.2676$$ $$P_3=16.0086$$ $$P_4=20.5935$$ $$P_5=25.0748$$ $$P_6=29.4866$$