On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ... - MathOverflow

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

2009-11-08T04:48:03Z

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?

Answer by Terry Tao for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-08T05:11:25Z

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.

Answer by Gerald Edgar for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-08T13:54:22Z

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?

Answer by Hugh Thomas for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-08T16:22:49Z

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.

Answer by Marko Amnell for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-08T18:04:47Z

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.

Answer by Gabriel Benamy for On the series 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ...

2009-11-20T21:36:29Z

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).