Exercise 2.27

Proof. Let $\mathrm{\Delta }\subset {\mathbb{N}}^{\text{∗}}$ the set of square free integers.

Let $N\in {\mathbb{N}}^{\text{∗}}$. Every integer $n,1\le n\le N$ can be written as $n=a{b}^2$, where $a,b$ are integers and $a$ is square free. Then $1\le a\le N$, and $1\le b\le \sqrt{N}$, so

Therefore

As ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }\frac{1}{n}$ diverges, $\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{a\in \mathrm{\Delta },a\le N}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\frac{1}{a}=+\infty$, so the family ${\left(\frac{1}{a}\right)}_{a\in \mathrm{\Delta }}$ of the inverse of square free integers is not summable:

Let $S_N={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{p<N}(1+1∕p)$ , and $p_1,p_2,\dots ,p_l(l=l(N))$ all prime integers less than $N$. Then

We prove this last formula by induction. This is true for $l=1$ : ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{𝜀\in \{0,1\}}1∕p_1^{𝜀}=1+1∕p_1$.

If it is true for the integer $l$, then

So it is true for all $l$.

Thus $S_N=\underset{n\in {\mathrm{\Delta }}_N}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\frac{1}{n}$, where ${\mathrm{\Delta }}_N$ is the set of square free integers whose prime factors are less than $N$.

Let $A\subset \mathrm{\Delta }$ be any finite set of square free integers. There exists $N\in N$ such that $A\subset {\mathrm{\Delta }}_N$, namely $N=\max <!--\; FUNCTION\; APPLICATION\; -->(A)$. Indeed, if $n\in A$, then $n\le N$, so that every prime factor of $n$ is less than $N$.

Let $B$ be an arbitrary real. Since ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \mathrm{\Delta }}1∕n$ diverges, there is a finite set $A\subset \mathrm{\Delta }$ such that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n\in A}1∕n>B$. By the preceding argument, there is some $N\in \mathbb{N}$ such that $A\subset {\mathrm{\Delta }}_N$, thus $S_N={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n\in {\mathrm{\Delta }}_N}1∕n>B$. This proves that $\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{S}_N=+\infty$, that is

We know that $e^x\ge 1+x,x\ge \log <!--\; FUNCTION\; APPLICATION\; -->(1+x)$ for $x>0$, so

$\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\log <!--\; FUNCTION\; APPLICATION\; -->S_N=+\infty$ and $\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}l(N)=+\infty$, so

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