I have this series, and I would like to know if it converges and, if so, for what value.
$$\frac{1}{2}\prod_{n=1}^\infty\frac{\operatorname{prime}(n)-1}{\operatorname{prime}(n)}$$
Looking at the plot, it seems to converge:
Also, considering the values for big n goes close to 1, the series value would converge. What makes sense, considering the small difference that removing 1 should do proportionally in a big number.
$$\frac{\operatorname{prime}(4100000000)-1}{\operatorname{prime}(4100000000)} \approxeq 1 $$
I try to calculate by hand using this recursive function:
$$f(0) = \frac{1}{2}; \qquad f(n) = f(n-1) \cdot \frac{\operatorname{prime}(n)-1}{\operatorname{prime}(n)} $$
I am getting a stable partial result: $f(50000000) \approxeq 0.02711 4$
A similar series, $\frac{1}{2}\prod_{n=2}^\infty\frac{n!-1}{n!} \approxeq 0.1976695 $.
This is the same as $$ \prod_n \Big( 1 - \frac{1}{p_n} \Big),$$ where $p_n$ is the $n$th prime. This would typically be written as $$ \prod_p \Big(1 - \frac{1}{p} \Big).$$ This actually has a name! If you recall the Riemann zeta function, we have $$ \zeta(s) = \sum_{n \geq 1} \frac{1}{n^s} = \prod_p \Big( 1 - \frac{1}{p^s} \Big)^{-1},$$ and so your product is $1/\zeta(1) = 0$.
More generally, the convergence of infinite products is closely related to the convergence of infinite sums. See Proving a result in infinite products: $\prod (1+a_n)$ converges (to a non zero element) iff the series $\sum a_n$ converges. Since it is known that $$ \sum_{p \leq X} \frac{1}{p} \approx \log \log X, $$ the infinite sum diverges very very slowly to $\infty$, and so the product goes to $0$ very very slowly also. (See also Why do we say some infinite products "diverge" when the limit is zero, a finite value? for a confusing terminology).