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How can we define an infinite subset of primes such that the sum of reciprocals converges?

$S=\{p\in \mathbb{Z}^+ : p\ \text{is prime and some condition on}\ p\}$ s.t. $\sum\limits_{p\in{S}}\frac{1}{p}\neq\infty$

A few options that come to mind for the condition on $p$ are:

  • $\log_2(p+1)\in\mathbb{N}$
  • $\log_2(p-1)\in\mathbb{N}$

But it has not been proved that there are infinite many such primes for either one of these options.

Any ideas?

barak manos
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7 Answers7

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For every $n\in \mathbb{N}$ define prime $p_n$ as the smallest of the primes which are greater than $2^n$. Then the set $\{p_n \mid n\in \mathbb{N}\}$ is infinite and

$$\sum_n \frac{1}{p_n} \leq \sum_n \frac{1}{2^n} = 1\text{.}$$

Dennis Jaheruddin
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JanezNovak
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  • Nice! I was actually hoping for something a little more "concrete" (like, some sort of condition with which you could compute the exact values of the elements in the subset), but this answer is generally correct. – barak manos Aug 29 '14 at 08:08
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    To be a bit more precise: Define $p_n$ to be a (or: the smallest) prime greater than $2^n$ and greater than $p_0, p_1, \dots, p_{n-1}$. This'll make sure that you'll definitely end up with an infinite collection of primes as your sequence will be well-defined and injective. – Frunobulax Aug 29 '14 at 08:09
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    @Frunobulax My set is infinite for sure. For example, a subset $A$ of set $\mathbb{N}$ is infinite $\Leftrightarrow$ For every $n\in \mathbb{N}$ there exists $a\in A$ such that $a>n$ ... – JanezNovak Aug 29 '14 at 08:14
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    @Jaka: I wasn't meaning to say that your set isn't infinite. I just think the construction is more obvious and explicit this way. – Frunobulax Aug 29 '14 at 08:19
  • @Frunobulax Oh, I completely agree with that. – JanezNovak Aug 29 '14 at 08:26
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    @Frunobulax you can remove the condition "and greater than $p_0,p_1,\ldots,p_{n-1}$" I think because of Bertrand's postulate. There always exists a prime between $2^n$ and $2^{n+1}$ and so the least such can be chosen for each $p_n$ without worrying about them coinciding. – Dan Rust Aug 29 '14 at 13:24
  • @DanielRust: Yes, you are right. Good point! – Frunobulax Aug 29 '14 at 13:26
  • @DanielRust you can also remove it since both the inequality and the infinite cardinality are obvious if you accidentally duplicate a prime (where we don't assume Bertrand's postulate). – djechlin Aug 30 '14 at 01:47
  • Not bad for being only 2 days on this forum :) – barak manos Aug 30 '14 at 07:17
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Let $p_n$ be a prime divisor of $(n^2)! + 1$. Then we have $p_n > n^2$, which means that $$\sum_{n \geq 1} \frac{1}{p_n} < \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty.$$

user133281
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Take the smallest prime divisor of Fermat numbers $F_n=2^{2^n}+1$
It is wel-known that they are pairwise coprime and that every prime divisor of these numbers is of the form $p=k\cdot 2^{n+2}+1>2^n \Rightarrow \Sigma \frac{1}{p}<\Sigma\frac{1}{2^n}=1$ which proves that the sum of their reciprocals converges.

Konstantinos Gaitanas
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I see lots of answers giving subsets of primes yielding different upper bounds on $\sum{\frac{1}{p_n}}$, but none stating explicitly what the bound is. So I thought I'd give a subset for which the exact sum is known.

Let $x$ be any positive real number. Now define $p_n$ to be the smallest prime such that $\sum_{i=1}^{n}{\frac{1}{p_i}} < x$ and $\forall i<n: p_n \neq p_i$. Then it holds that $\sum_{i=1}^{\infty}{\frac{1}{p_i}} = x$

kasperd
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  • Interesting, thanks... You've managed to define a bound, but it doesn't really provide a concise method for generating the sequence itself. – barak manos Aug 30 '14 at 12:11
  • @barakmanos I don't believe there could be a concise method for generating such a sequence with a specific bound. The remainder of the sequence has to depend on prior numbers. Coming up with a concise method for producing an infinite sequence of primes is a difficult problem even if there are no constraints on what primes you may chose. – kasperd Aug 30 '14 at 12:23
  • Yep, I agree on that. I am currently aware of only one simple method for generating primes, which is Rowland's recurring relation (http://en.wikipedia.org/wiki/Formula_for_primes#Recurrence_relation)... And even that, as far as I know, doesn't guarantee "when the next prime is going to be found". – barak manos Aug 30 '14 at 12:32
  • @barakmanos Interesting sequence, but doesn't look like you could compute $a_n$ without first computing all predecessors. – kasperd Aug 30 '14 at 12:37
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$S=\{p\in \mathbb{Z}^+ : p\ \text{is prime and some condition on}\ p\}$ s.t. $\sum\limits_{p\in{S}}\frac{1}{p}\neq\infty$

Here's an example with a simple explicit condition on $p$.

$$ S= \{p \in \mathbb{Z}^+ | \text{$p$ is prime and there is an $n\in\mathbb{Z}$ such that $2^{n^2-1} \leq p \leq 2^{n^2}$} \}. $$

It is known that there is always a prime between $k$ and $2k$ for each $k\in\mathbb{Z}$. Thus the set is infinite. [Wikipedia]

It is also known that there is a constant $C$ such that the number of primes smaller than an integer $x$, denoted by $\pi(x)$, is less than $C\frac{x}{\ln x}$. [Wikipedia]

Now we can get a very crude upper bound: Because the number of primes between $2^{n^2-1}$ and $2^{n^2}$ is less than $\pi(2^{n^2})$ and their reciprocals are less than $\frac{1}{2^{n^2-1}}$, we get $$ \sum_{p\in{S}}\frac{1}{p} \leq \sum_{n\in\mathbb{Z}^+} \frac{1}{2^{n^2-1}} \pi(2^{n^2}) \leq \sum_{n\in\mathbb{Z}^+} \frac{1}{2^{n^2-1}} C\frac{2^{n^2}}{\ln 2^{n^2}} \leq \sum_{n\in\mathbb{Z}^+} \frac{2 C}{\ln 2} \frac{1}{n^2} = \frac{2 C}{\ln 2} \frac{\pi^2}{6}. $$

JiK
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Let $f:\Bbb N\rightarrow\Bbb N$ be such that $\sum_{n=0}^{\infty}1/\!f(n)$ converges (e.g. $f(n)=2^n$). Then the sum of the reciprocals of the $f(n)$th primes ($n=0,1,...$) converges.

John Bentin
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Define the set of primes as $\{p_n \mid n\in \mathbb{N}\}$ ,where $p_n$ is the smallest prime such that $n^2<p_n$.

Bumblebee
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