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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?

What do you think about this?

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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
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5 Answers 5

up vote 43 down vote accepted

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
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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.

All this does is it says the primes grow slower than any power, but it doesn't say how they grow, which the Prime number theorem does (and supports this conclusion). I suppose this could be generalized to say that any function which can be shown to, for all sufficiently large terms, grow slower than any power greater than 1, it is divergent (the primes being an example, and the evens being another).

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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
I did a test, and it turns out that n(logn)² grows faster than, say, n^1.1 fairly quickly: {{0.960906, 2.14355}, {3.62085, 3.34837}, {7.68725, 4.59479}, {12.9515, 5.87309}, {19.2624, 7.17739}, {26.506, 8.5037}, {34.5926, 9.84916}, {43.4502, 11.2116}, {53.019, 12.5893}, {63.2489, 13.9808}, {74.0971, 15.3851}, {85.5265, 16.8011}, {97.5047, 18.2281}, {110.003, 19.6653}, {122.996, 21.1121}, {136.461, 22.568}, {150.376, 24.0325}, {164.725, 25.5052}, {179.488, 26.9857}} In each pair, the first one is n(logn)², and the second is n^1.1, n>=2 –  Gabriel Benamy Nov 24 '09 at 14:53
Gabriel, if you look at larger integers you will see this turn around; e.g., consider n=10^100. But the experimental data isn't necessary because of calculus; e.g., use l'Hôpital's rule. Specifically, you can convince yourself that each positive power (no matter how large) of log(n) grows more slowly than each positive power (no matter how small) of n. –  Jonas Meyer Dec 12 '09 at 10:48
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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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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
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