Sum of Reciprocals of Prime Numbers Equals 1

[Borek]

reciprocals of primes summed to 1

As the other thread about sum of primes started it reminded me about the idea I had long ago.

My starting point was Erathostenes sieve. It occurred to me that multiples of 2 make half of all natural numbers, multiples of 3 make 1/3 of all natural numbers and so on. And as multiplies of prime numbers have to cover all natural numbers, correctly constructed sum of their reciprocals must equal 1. This is not as obvious as it seemed to me at first, as some multiplies of 2 and 3 will be calculated twice, so it has to be 1/2+1/3-1/(2*3) - but it still can be done.

Let P be set of all prime numbers. Let's define some more sets:

$$\forall {a, b} \in P, a < b \rightarrow ab \in P_2$$

$$\forall {a, b, c} \in P, a < b < c \rightarrow abc \in P_3$$

and so on P₄, P₅...

each of these sets has elements

$$p_{1i} \in P, p_{2i} \in P_2, p_{3i} \in P_3 ...$$

when combined

$$\sum {\frac 1 p_{1i} } - \sum {\frac 1 p_{2i} } + \sum {\frac 1 p_{3i} } - \sum {\frac 1 p_{4i} } ... = 1$$

(note: could be I am misusing notation, what I mean is "if a < b < c then abc is a member of set P₃ - that's a way of using each combination of three primes only once, could be it can be done much simpler; please remember I am a chemist )

 

[gel]

What you've called P_k is simply the set of integers which is a product of k distinct primes. Then the identity you have written is (almost) the same as expanding the following expression
$$\prod_{p \rm{\ prime}}\left(1-\frac{1}{p}\right) = 0,$$
which is true.

You have to be careful when you expand this though. That's because the series you get is not absolutely convergent, so the order in which you sum it is important.
In fact, in your expression, each of the summations is infinite, so it doesn't make sense as it is written.

 

[Borek]

So if I understand you correctly I should write it as

$$\lim_{n\rightarrow \infty} (\sum^n \frac{1}{p_{1i}} - \sum^n \frac{1}{p_{2i}} + \sum^n \frac{1}{p_{3i}} - \sum^n \frac{1}{p_{4i}} + ...) = 1$$

But product version is much more elegant :grumpy:

 

[gel]

I'm not sure about how you have written it. If you let P_k,N be the products of distinct primes less than N, then write your sum using these, then let N go to infinity, then it will converge 1.

You might want to look at the http://en.wikipedia.org/wiki/Riemann_zeta_function" and, in particular the Euler product formula,
$$\sum_{n\geq 1}\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}.$$
or, the reciprocals
$$\left(\sum_{n\geq 1}\frac{1}{n^s}\right)^{-1} = \prod_{p \text{ prime}} (1-p^{-s}).$$

If s>1 you can expand the right hand side, and it should vanish as s->1.