What is the following sum?
$x=\frac{1}{2}+\frac{1}{2\times 3}+\frac{1}{2\times 3\times 5}+…$
$=\Sigma_{n=0}^{\infty} \frac{1}{\prod_{i=1}^{n}p_i}$, where $p_i$ is the $i^{th}$ prime.
I'm guessing this converges to 1 using the following logic.
The sieve of Eratosthenes works as follows.
From the set $N$ of all natural numbers we first remove all multiples of 2, that is half the numbers. Then from the remaining half we remove all multiples of 3, which will amount to removing one sixth of the numbers from $N$ because $1/6=1/2\times 1/3$. Repeating this process, next we remove all multiples of 5, that is, we will remove $1/2\times 1/3\times 1/5=1/30^{th}$ more numbers from $N$. If we continue this process indefinitely, in the end we would have removed $x$ fraction of the numbers and the remaining $1-x$ fraction are the primes. Since the ratio $\pi(n)/n$ approaches 0 as $n$ approaches infinity, Hence $1-x = 0$ or $x = 1$.
Best Answer
Your sieve argument could consider:
You can see the third term of $\frac{5}{2\cdot 3 \cdot 5}$ is not your third term of $\frac{1}{2\cdot 3 \cdot 5}$, which would instead be multiples of $30$ though those have already been removed. Similarly with later terms.
The sum of those four terms is $\frac{20}{21}\approx 0.95238$, already much closer to $1$ than your infinite series which sums to just over $0.70523$.