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If we define the morphism from the positive integers to the interval $(0,1]$ as follows:

$f(x)=\left(\frac{1}{x+1},\frac{1}{x}\right]$

Then we can measure the density of any set of positive integers as a proportion of the whole, e.g. the odd numbers as follows:

$\displaystyle\sum_{x = 1}^{\infty} \left(\frac{1}{2x-1}-\frac{1}{2x}\right)=\ln{2}$

Obviously the even numbers have density $1-\ln{2}$ and every set of multiples of some integer has measure less than $1$, with the set only being fully closed if we include the point at infinity in our sum.

I presume this is a known measure?

What is the density of the prime numbers by this measure?

$\displaystyle\sum_{p\in\text{prime}} \left(\frac{1}{p}-\frac{1}{p+1}\right)\approx0.3302$

It looks to simplify to $\displaystyle\sum_{p\in\text{prime}}\frac{1}{p(p+1)}$

It looks like this has something to do with the relationship of the Zeta function to the distribution of primes. I imagine this sum over primes can be translated into an Euler product, a Dirichlet series and ultimately into a value of the Zeta function but I'm struggling to see how to proceed further.

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    $\begingroup$ That's not consistent: if the second sum is the density of the primes, the first one should be that of the odd numbers, I'm not sure it's really a density, it's more a weight, and puts more weight on small numbers. The sum is not immediately related to the Zeta function, but it can be expressed (and numerically calculated) as an alternating series of values of the Prime zeta function. It's approximately $0.33022992626420383$. $\endgroup$ – Professor Vector Nov 26 '17 at 11:13
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    $\begingroup$ Well, in that case, would you mind the write down the corresponding sum for the odd numbers, and point out where the odd number $1$ enters it? $\endgroup$ – Professor Vector Nov 26 '17 at 13:33
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    $\begingroup$ I'm sorry, I don't see any change in your post. BTW, it would also be better to write an interval as $\left(\frac{1}{x+1},\frac{1}{x}\right]$, so that the left limit is smaller than the right one. $\endgroup$ – Professor Vector Nov 26 '17 at 15:15
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    $\begingroup$ oeis.org/A179119 $\endgroup$ – Charles Nov 27 '17 at 6:16
  • $\begingroup$ @ProfessorVector i fixed those bits now. I wanted to take time to not change it in such a way as to make reuns's answer (and other comments) incorrect. $\endgroup$ – samerivertwice Nov 27 '17 at 11:13
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Use the geometric series $$\frac{1}{p+1} = \frac{1}{p} \frac{1}{1+\frac1p} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{p^{k}}$$ thus $$\sum_p \frac{1}{p}-\frac{1}{p+1} = \sum_{k=2}^\infty (-1)^{k} P(k) \\= \sum_{k=2}^\infty (-1)^{k} \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(nk)=\sum_{m=1}^\infty (\frac{\mu(m)}{m}-a_m) \log \zeta(m)$$ where $a_m = \sum_{d | m} (-1)^{m/d+1} \frac{\mu(d)}{d}=\prod_{p^k \| m} a_{p^k}$ and $P(s) = \sum_p p^{-s} = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$

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  • $\begingroup$ Can you compare the rate of convergence of the original and last series ? @RobertFrost $\endgroup$ – reuns Nov 27 '17 at 11:58
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    $\begingroup$ p.s. although I've already accepted this answer, it would I think be improved by including the oeis reference given by Charles above which is also really useful. $\endgroup$ – samerivertwice Nov 27 '17 at 12:10
  • $\begingroup$ @RobertFrost It seems you don't understand Dirichlet series is the answer. The exact rate of convergence of $\sum_p p^{-1}-(p+1)^{-1}$ depends on the prime number theorem, I'm asking for a crude estimate, and the same for the last series. $\endgroup$ – reuns Nov 27 '17 at 13:06

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