# Sums of Reciprocals of Infinite Subsets of Primes

1. Jul 27, 2006

### Dragonfall

Can someone confirm/disprove the following:

If $$X\subset\mathbb{P}$$ is infinite, then $$\sum_{n\in X}\frac{1}{n}$$ diverges or is irrational.

2. Jul 27, 2006

### StatusX

Some of these series diverge (eg, X=P), and some converge (eg, pick the nth prime in the list to be larger than 2^n). I don't see how you could prove such a a general infinite sum to be irrational.

3. Jul 27, 2006

### Office_Shredder

Staff Emeritus
It's nowhere near rigorous, but the irrationality part probably has to do with the fact that since they're all primes, the fractions share no common factors (like 1/2 and 1/4 would... if you get what I mean). This means that as you group them, you would need increasingly larger decimals

Take
(p1+p2...)/p1p2.....

for the first, say, N primes. Now, because we know it's irreducable (the top isn't divisible by any single prime, and the bottom is only divisible by the primes), we know that the decimal is only going to get larger (the bottom keeps increasing by more than an order of ten, and the top keeps increasing.... so, for example, 3/10 is a shorter decimal than 111/1000). So as the limit tends to infinity, it needs to be irrational.

It's not rigorous, but that's probably how you would start it off

Last edited: Jul 27, 2006
4. Jul 27, 2006

### StatusX

The problem with that approach is making the transistion from arbitrarily large partial sums to the limit. For example, in the sum of a geometric series like 0.9+0.09+0.009+..., the common denominator/length of nonrepeating portion of the decimal expansion gets larger and larger as you take larger partial sums, but the limit is just 1.

5. Jul 27, 2006

### matt grime

How do you know the top 'isn't divisble by any single prime'? It's certainly divisible by some primes, and there is nothing stopping it being one of the first N.

6. Jul 27, 2006

### Office_Shredder

Staff Emeritus
Looking back on it, that really is an error on my part, sorry

I got too ahead of myself

7. Jul 28, 2006

### Dragonfall

Well, if Brun's constant were rational, you'd disprove both the twin primes conjecture and this conjecture.

8. Jul 28, 2006

### shmoe

You'd have to prove it rational first of course! It would disprove only one of twin primes and the one you have here, either one along with the others negation will work with Brun's constant being rational.

You can find a subset of the primes whose sum converges to 1 (or any other rational you like) without much difficulty given the fact that the full series diverges and the terms are tending to zero. Given a c>0 there are finite subsequences, p,...q, as far out as you'd need so that c<1/p+...+1/q<2c. Just use this to build a finite sequence whose sum is between 1-2^(-n) and 1, and repeat.

Last edited: Jul 28, 2006
9. Jul 28, 2006

### NateTG

Actually, can't you just say:
Given some positive real number $r>0[/tex] chose $p_1,p_2...p_n...$ where $p_n$ is the smallest prime such that: $$\forall i < n, p_n \neq p_i$ and [tex] \sum_{i=1}^{n} \frac{1}{p_n} < r$$

Now, there will always be a suitable prime since the prior partial sum is strictly less than $r$, and the (non-empty) set of suitable primes will always have a unique least member.

Last edited: Jul 28, 2006
10. Jul 28, 2006

### shmoe

That's the first thing I thought of too, but wasn't immediately convinced a greedy algorithm would converge to your target, so I went another direction. It seemed likely that you could construct divergent series whose terms converged to zero yet this kind of greedy approach wouldn't always give a subsequence that works so I sort of expect some kind of density of the primes would be needed here. I haven't given it much thought. Can you prove this construction will converge to your r?

11. Jul 29, 2006

### Dragonfall

You're right, I got my logic backwards.

Is there a strictly positive real number that no subseries of the harmonic series can converge to? If not then this 'conjecture' is false.

Last edited: Jul 29, 2006
12. Jul 29, 2006

### 0rthodontist

I'm not sure what you mean Nate--I don't see how you extend your series to an infinite one. This may be close to what you were saying, or it may not. For the given r, define the sequence of primes pi by:
Code (Text):

input r
r[sub]1[/sub] <-- r
i <-- 1
while true
let p[sub]i[/sub] be the smallest prime not already chosen such that r[sub]i[/sub] - 1/p[sub]i[/sub] > 0
r[sub]i+1[/sub] <-- r[sub]i[/sub] - 1/p[sub]i[/sub]
i <-- i + 1

The pi here when their reciprocals are summed have to converge to r because by definition they can't exceed it, and they can't converge to something less than it. The latter is true because if they did converge to something less than r, say s, then let pn be some prime in the sequence, with n so large that r - s > 1/pn. Then the pi will also include all the consecutive primes after pn, up until a point where their reciprocal sum is greater than s (since the sum of the reciprocals of the primes diverges) but less than r (since 1/pj for j > n is always less than 1/pn, the first prime that puts the sum over s will not put it over r).

Last edited: Jul 29, 2006
13. Jul 29, 2006

### Office_Shredder

Staff Emeritus
That probably works for rationals, but are you sure you can get to things like, say, 1/sqrt(2) using that method?

EDIT: Wait,. it doesn't matter if you can. Whoops

14. Jul 29, 2006

### 0rthodontist

Yes, you can. r can be any positive real number. Actually, the same argument works for any infinite divergent series whose terms are positive and tend to 0.

Last edited: Jul 29, 2006
15. Jul 30, 2006

### CRGreathouse

I concur. The conjecture's dead; 0rthodontist's greedy algorithm killed it.

16. Jul 31, 2006

### Dragonfall

Excellent. Thanks for the input.

17. Jul 31, 2006

### NateTG

That's pretty much exactly what I was saying would work as an algorithm - and you've written out the proof that it works.