I'm not math major, but I am interested in maths, and was at home thinking about primes, and thought about the quistion in the title. It might seem complex or kind of vague (I'm not a native english speaker), but I simply want to know if, and if so, what the equation below converges to:
$$\sum_{n=1}^{\infty}\big(\prod_{k=1}^n\frac{1}{p_k}\big)$$ Where $p_k$ is the k'th prime. (For example: $p_2=3$)
I have tried looking all over the internet and using different kinds of calculators to estimate it, and even trying to compute it manual. My best guess is that converges to 1.
For if we take all natural numbers, and took away all multiples of 2, we would have taken half of all numbers. if we now did 3, we would have taken another sixth of all numbers away;
Since 3 is a factor of a third of numbers, but every second multiple of 3 is also a multiple of 2, we would have only take a sixth of all numbers away.
Continuing should correspond to the equation above, and we should have taking all natural numbers away in the end, since all natural numbers are a multiple of primes (except 1).
But it doesn't seem to converge to 1 when I try to compute it manually. I would appreciate some help from some real mathematicians
This answer summarizes my comments. Let $\{p_1, p_2, p_3, ...\}=\{2, 3, 5, ...\}$ be the sequence of primes. We are interested in the number $$ x = \sum_{n=1}^{\infty} \prod_{k=1}^n\frac{1}{p_k}$$ Fix $m$ as a positive integer. We get $$ \sum_{n=1}^m \prod_{k=1}^n \frac{1}{p_k} \leq x \leq \sum_{n=1}^m \prod_{k=1}^n \frac{1}{p_k} + \underbrace{\sum_{n=m+1}^{\infty} (1/2)^n}_{1/2^m}$$ which uses the fact that $\prod_{k=1}^n(1/p_k)\leq 1/2^n$ for all $n\in \{1, 2, 3, ...\}$. So we can get an estimate to within an error $1/2^m$ just by evaluating the first $m$ terms of the sum. For $m=5$ we get $$ \underbrace{\frac{543}{770}}_{0.705} \leq x \leq \underbrace{\frac{543}{770} + \frac{1}{2^5}}_{0.737}$$