product $\prod \limits_{i=2}^{\infty} (1+\frac{1}{(p_i -2)p_i})$ for primes $p_i$ I want to calculate the product $\alpha= \prod \limits_{i=2}^{\infty} (1+\frac{1}{(p_i -2)p_i})$ for all primes $p_i >2$.
I calculated this product first with computer and get for the first primes under ten millions $\alpha=1.5147801192603$.
Analytical I tried to expand the product as a telescope sum:
$\alpha=1+\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{2}{45}+\frac{1}{9}-\frac{1}{11}+\frac{2}{105}+\cdots)$. But this doesn't help me. I don't even know if $\alpha$ converges.

Recall that if $\{ a_n \}$ is positive and $a_n \to 0$ then $\prod (1 + a_n)$ and $\sum a_n$ converge/diverge together. Given that $p_n \sim n \log n$ convergence is clear. So at least that.
As for a closed form: this is actually the reciprocal of the twin primes constant. This appears in many, many conjectures about the prime numbers.
