On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + …

It is well-known that

A: The series of the reciprocals of the primes diverges

My question is whether property A is in some sense a truth strongly tied to the nature of the prime numbers.

Property A tells us that the primes are a rather fat subset of $\mathbb{N}$. Is there a way to define a topology on $\mathbb{N}$ such that every dense subset of $\mathbb{N}$ (under this topology) corresponds to a fat subset of the natural numbers?

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btw if you take A to be the set {1} union {p+1 : p odd prime}, then sum of 1/a diverges, A contains no primes, and the kth member is greater than the kth prime for all k>1. –  Rob Harron Nov 8 '09 at 4:56
There are still fairly trivial answers to your question -- for instance, take the set 2, 4, 5, 8, 11, 14, ..., p_n, p_{n+1} + 1, and this satisfies your conditions. I'm not sure what you're trying to ask, here. –  Harrison Brown Nov 8 '09 at 5:10

Yes, it's possible. Define the closed sets to be the sets the sum of whose reciprocals converges, together with $\mathbb{N}$. This collection of subsets is closed under arbitrary intersection and finite union, so it does form the closed sets of a topology.

A subset of $\mathbb{N}$ is dense in this topology if its closure is $\mathbb{N}$, in other words, if it is not contained in any smaller closed set -- in other words, if it is not contained in any set the sum of whose reciprocals converges. This is equivalent to the sum of its reciprocals not converging.

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By Fubini's theorem, the sum of the reciprocals of the primes is equal to $\int_1^\infty \frac{\pi(x)}{x^2}\ dx$, where $\pi(x)$ is the number of primes less than x. The prime number theorem tells us that $\pi(x) \sim x/\log x$ for large x, which implies the divergence of this integral. (One does not need the full strength of the PNT here; the more elementary fact that $\pi(x)$ is bounded from below by a constant multiple of $x / \log x$ would suffice.) A variant of this argument shows that $\sum_{p \leq x} 1/p = \log \log x + O(1)$ (again, this can also be shown by more elementary means - see Mertens' theorem).

The same argument shows that slightly thinner sets than the primes would also have this property, e.g. any set for which the analogue of $\pi(x)$ is asymptotically larger than $x / \log x \log \log x$, or $x / \log x \log \log x \log \log \log x$, would still diverge. On the other hand, if the analogue of $\pi(x)$ is $O( x / \log^{1+\varepsilon} x )$ for some $\varepsilon > 0$ then one will have convergence. So the primes are close to the edge of the sparsest set with this property (as one could already guess from the double-logarithmic growth of the sum).

For instance, sieve theory tells us that the number of twin primes less than x is $O( x /\log^2 x)$, which implies Brun's theorem that the sum of reciprocals of twin primes converges.

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Thanks for taking the time to reply, Professor Tao. –  J. H. S. Nov 9 '09 at 9:27

There is also this result...

$\displaystyle\sum_{a \in A} \frac{1}{a}$ diverges if and only if the span of $\{x^a | a \in A\}$ is dense in the continuous functions on an interval. (I guess you have to include the constant $1$). There is probably no relation to the primes? Or is there?

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Probably very little relation, alas; the result (a theorem of Müntz) holds when $A$ is any set of positive real numbers. –  Noam D. Elkies Aug 14 '11 at 14:26
This is very cool, though! –  Vectornaut May 21 '12 at 6:21

Also, it is possible to prove that for any numbers a and b for which gcd(a,b)=1, the sum of all the 1/p's for p prime that satisfy p = a (mod b) diverges.

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I suppose that yes, it does give some insight into the nature of the primes; or rather, their distribution amongst the natural numbers. Realize that Zeta( a ) is finite for all a > 1, and this means that given any positive real a greater than 1 will produce a finite value of $\sum_n^\infty=\frac{1}{n^a}$ (the closer to 1 it gets, the larger it will get). However, the partial sums of the reciprocals of the primes do not converge . What does this mean? Well, consider that there exists some exponent j and coefficient k such that, for every integer n , $\frac{k}{n^j}\geq\frac{1}{p(n)}$, where p ( n ) is the n th . If we flip the reciprocals, we also flip the direction of the inequality, and arrive at $\frac{n^j}{k}\leq p(n)$, or $n^j\leq p(n)\cdot k$. This means that, given j,k such that the above conditions hold, the function p ( n ) grows at least as fast as this function with exponent j . Looking back at our $\frac{k}{n^j}$, we examine $\sum_{n=1}^{\infty}\frac{k}{n^j}=k\cdot \zeta(n)\lt\infty$, as per the finite nature of k and the Zeta function at values greater than 1. But, the n th term in the series is always greater than the n th prime number, and therefore the whole sum is therefore greater than the sum of the reciprocals of the primes, which, as we know, is infinite. This creates a contradiction, in that the sum, which we know is finite, is greater than an infinite series. Therefore, there can't exist a j and k that satisfy $\frac{k}{n^j}\geq\frac{1}{p(n)}$ for all n , which means that the prime numbers themselves grow slower than any power.
Unfortunately this isn't quite true. For instance if you take the sequence a_n = $n (log n)^2$, the reciprocals of a_n are a convergent series, but a_n grows almost as slowly as the primes. –  Harrison Brown Nov 20 '09 at 21:41