Define an infinite subset of primes such that the sum of reciprocals converges

Question

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?

Answer 1

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

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

Answer 2

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

Answer 3

Let $p_n$ be a prime divisor of $(n^2)! + 1$. Then we have $p_n > n^2$, which means that $$\sum_{n \geq 1} \frac{1}{p_n} < \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty.$$

Answer 4

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.

Answer 5

$S=\{p\in \mathbb{Z}^+ : p\ \text{is prime and some condition on}\ p\}$ 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 $$ \sum_{p\in{S}}\frac{1}{p} \leq \sum_{n\in\mathbb{Z}^+} \frac{1}{2^{n^2-1}} \pi(2^{n^2}) \leq \sum_{n\in\mathbb{Z}^+} \frac{1}{2^{n^2-1}} C\frac{2^{n^2}}{\ln 2^{n^2}} \leq \sum_{n\in\mathbb{Z}^+} \frac{2 C}{\ln 2} \frac{1}{n^2} = \frac{2 C}{\ln 2} \frac{\pi^2}{6}. $$

Answer 6

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.

Answer 7

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 $\sum_{i=1}^{\infty}{\frac{1}{p_i}} = x$