Exercise 1.3.38* ($1 + \frac{1}{3}+\cdots + \frac{1}{2n-1}$ is not an integer.)

Proof. The integer $3^r$ is odd, so $3^r=2k-1$ for some $k\in [[1,n]]$. Since $3^r$ = 2k-1 is the highest power of $3$ in $𝒯$,

We will show that, for $2j-1\in 𝒯$ such that $j\ne k$, $3^r\nmid 2j-1$.

Indeed, assume for contradiction that $2j-1=3^{r}q\in 𝒯$, then $q>1$, because $3^r=2k-1\ne 2j-1$), and $q\ne 2$, because so $2j-1$ is odd. Thus $q\ge 3$, so

This is a contradiction, thus $3^r\nmid 2j-1$ if $j\ne k$.

To conclude this part, there is some $k\in [[1,n]]$ such that

We will prove that, for $n>1$,

is not an integer.

Since $n>1$, we know that $r\ge 1$ since $3\in 𝒯=\{1,3,\dots ,2n-1\}$.

The reduction of $S$ to the same denominator gives $S=N∕D$, where

so $N={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^n{n}_j$, where $n_j=\frac{D}{2j-1}$ is an integer.

Define $K={\nu }_3(D)$. Then

Therefore $3^{K-r}\mid N$, but $3^{K-r+1}\nmid N$, so

Since $K={\nu }_3(D)$, $3^K\mid D$. Knowing that $r\ge 1$, a fortiori $3^{K-r+1}\mid D$, so

Dividing $N$ and $D$ by $3^{K-r}$, we obtain

If $S$ is an integer, then $N^{\prime }=3D^{\prime }S$ is a multiple of $3$ . This is a contradiction, because $3\nmid N^{\prime }$. Therefore $S$ is not an integer if $n>1$. □