# Additive properties of sequences: trying to understand Schnirelmann density

I have started reading Gelford & Linnik's elementary methods in analytic number theory (1965).

They define a sequence $A$ of integers as:

$$0, a_1, a_2,a_3,\dots$$

where

$$0 < a_1 < a_2 < a_3 < \dots$$

Let:

$$A(n) = \sum_{0<{a_i} \le n} 1$$

So that:

$$0\le\frac{A(n)}{n}\le1$$

I am following their explanations up to this point. Then, the following definition of density $d(A)$ is offered:

$$d(A) = \inf_n \frac{A(n)}{n}$$

At this point, I am not clear on how the definition maps to the examples. I found a wikipedia article on Schnirelmann density but that didn't help. I'll reread it this evening.

Gelford & Linnik provide examples of density. I would greatly appreciate it if someone could explain me how the definition above maps to these examples.

Here are three examples from the section:

(1) If $1 \notin A$, then $d(A) = 0$

(2) $d(A) = 1$ if and only if $A$ contains all the positive integers.

(3) The densities of the sequences of squares, cubes, and prime numbers equal $0$.

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## 1 Answer

(1) Clearly if $1\notin A$ then $A(1)=0$. (we try to find the (trivial) number of natural numbers not exceeding $1$ that are not $1$)

But generally, $d(A)\geq 0$ and $\frac{A(1)}{1}=0$
So, inf $\frac{A(n)}{n}=0$.

(2) Suppose that there exists a positive integer $k$ which is the least one that is not contained in $A$. Then $\frac{A(k)}{k}<1$ and so, (clearly) inf $\frac{A(n)}{n}\leq A(k)<1$

(3)you can see that the squares,cubes etc are getting more and more rare among the positive integers so the function whose infimum you want to calculate is strictly decreasing.

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Thanks. I was not clear on the definition of infimum. With your explanations, I understand. Thanks! Infimum and supremum are new to me. – Larry Freeman Nov 21 '13 at 3:27