Sums of Reciprocals of Infinite Subsets of Primes

[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.

 

[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.

 

[Office_Shredder]

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

 

[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.

 

[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.

 

[Office_Shredder]

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.

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

I got too ahead of myself

 

[Dragonfall]

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

 

[shmoe]

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

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.

 

[NateTG]

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.

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

 

[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?

 

[Dragonfall]

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'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.

 

[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 p_i by:

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 p_i 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 p_n be some prime in the sequence, with n so large that r - s > 1/p_n. Then the p_i will also include all the consecutive primes after p_n, 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/p_j for j > n is always less than 1/p_n, the first prime that puts the sum over s will not put it over r).

 

[Office_Shredder]

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

 

[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.

 

[CRGreathouse]

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

 

[Dragonfall]

Excellent. Thanks for the input.

 

[NateTG]

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...

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