Exercise 1.3.36* ($\sum_{j=1}^n 1/j$ is not an integer)

Proof. By definition of $k$,

Assume now that $2^k\mid i$, where $1\le i\le n$. Then $i=2^{k}q$, where $q\ge 1$.

If $q\ge 2$, then $i\ge 2^{k+1}>n$. Since $i\le n$, this is a contradiction, so $q<2$, and this proves that $q=1$, so $i=2^k$.

Hence $2^k\mid 2^k\in S$, but $2^k$ is not a divisor of any other integer in $S=\{1,2,\dots ,n\}.$

Note that, with the hypothesis $n>1$, $k\ge 1$.

Write ${\nu }_p(x)$ the exposant of $p$ in the factoring of $x\in {\mathbb{N}}^{\text{∗}}$ into prime powers, i.e. the unique integer $j\ge 0$ such that $p^j\mid x$ but $p^{j+1}\nmid x$. We have proved that for all $i\in S$,

Since the fractions $1∕j,j\in S,$ have a common denominator $n!$, we can write

where $D=n!$, and

Here $\frac{n!}{i},i\in S,$ are integers.

Define $K={\nu }_2(n!)$. Then, for all $i\in S$,

Therefore,

By definition of $K={\nu }_2(n!)$, $2^K\mid D$. Since $k>1$, a fortiori $2^{K-k+1}\mid D$, so that

Then, dividing $N$ and $D$ by $2^{K-k}$, $T=\frac{N}{D}=\frac{N^{\prime }}{2D^{\prime }}$ . If $T$ was an integer, $N^{\prime }=2D^{\prime }T$ would be even : this is a contradiction, hence

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