Is the set of primes at even positions a Large Set?

Question

If we start numbering the prime numbers from 2, and we consider the subset of prime numbers at even positions, is it known is that set is large or not?

A subset of natural numbers is called "large" if the sum of reciprocals of its elements diverges. For example it is well known that the set of natural numbers and set of prime numbers are large sets. Surprisingly I could not find any reference to whether the set of primes at alternating positions is large or not.

I tried using Bertrands Postulate but failed. I intuitively believe the set primes at odd positions must be large as the but I cant seem to be able to prove it rigorously.

Any guidance/help would be appreciated. I am also looking for amateurish resources for studies on large subsets.

Answer 1

This follows immediately from the fact that the set of primes is large. More generally, if $A=\{a_1<a_2<a_3<...\}$ is large, then so is $\{a_1<a_3<a_5<...\}$. The same holds for even positions (just think about $A\setminus\{a_1\}$) or indeed any arithmetic sequence of positions.

This is just because we have ${1\over a_{2n-1}}>{1\over a_{2n}}$. So we get $${1\over a_1}+\color{red}{1\over a_1}+{1\over a_3}+\color{red}{1\over a_3}+...>{1\over a_1}+\color{red}{1\over a_2}+{1\over a_3}+\color{red}{1\over a_4}+...$$ or more snappily $$\sum_{n\in\mathbb{N}} {2\over a_{2n-1}}>\sum_{n\in\mathbb{N}}{1\over a_n}.$$ If $A$ is large then the right hand side diverges, so the left hand side also diverges. Now just divide by $2$.

Answer 2

You know that $\sum_{k=1}^{\infty}\frac1{p_k}=\infty$. By Cauchy condensation, we have that $\sum_{k=1}^\infty \frac{2^k}{p_{2^k}}=\infty$ as well. But we know that $$\frac{2^k}{p_{2^k}}=4\cdot 2^{k-2}\frac1{p_{2^k}}\le 4\sum\limits_{2^{k-1}< j \le 2^k\\ j\text{ even}}\frac1{p_{2j}}$$

And therefore $$\infty=\frac14\sum_{k=1}^\infty \frac{2^k}{p_{2^k}}\le\sum_{k=1}^\infty \frac1{p_{2k}}$$