Infinite sum of inverse of products of successive primes If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,  ...  \rightarrow$
$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,  ...  \rightarrow$
$1, 5, 7, 11, 13, 17, 19, 23, 25,  ...  \rightarrow$
we are consecutively removing fractions
$$\frac{1}{2}, \; \frac{1}{2·3}, \; \frac{1}{2·3·5}, \; \frac{1}{2·3·5·7},\;  ... $$
from the total amount of natural numbers. This suggests that the infinite sum
$$\frac{1}{2}+ \frac{1}{2·3}+ \frac{1}{2·3·5}+ \frac{1}{2·3·5·7}+ ...$$
is equal to 1, but I couldn't figure out a proof for this result. Some references about this particular sum would be appreciated too.
Thank you in advance

This is not the case:
$$\frac12+\frac1{2\cdot 3}+\frac1{2\cdot 3\cdot 5}+\frac1{2\cdot 3\cdot 5\cdot 7}+\frac1{2\cdot 3\cdot 5\cdot 7\cdot11}+\cdots \le  \frac12\left(1+\frac1{3}+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots \right)=\frac34$$
It seems that your series converges to
$$0.70523017179180096514743168288824851374357763910915432819226791381\ldots$$
and does so way faster than the geometric series used in the above comparison.
After all, the $n$th summand is $$\le \frac12\cdot\frac{2^n\cdot n!}{(2n)!}\approx \frac{2^nn^ne^{-n}\sqrt{2\pi n}}{2\cdot(2n)^{2n}e^{-2n}\sqrt{4\pi n} }=\frac1{2\sqrt 2}\cdot \left(\frac e{2n}\right)^n.$$
