If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.

If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.

Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.

Hint $\ $ Since there is a unique denominator $\rm\:\color{#C00} {2^K}\:$ having maximal power of $2,\,$ upon multiplying all terms through by $\rm\:2^{K-1}$ one deduces the contradiction that $\rm\ 1/2\, =\, c/d \;$ with $\rm\: d \:$ odd, $ $ e.g.

$$\begin{eqnarray} & &\rm\ \ \ \  \color{green}{m}  &=&\ \  1 &+& \frac{1}{2} &+& \frac{1}{3} &+&\, \color{#C00}{\frac{1}{4}} &+& \frac{1}{5} &+& \frac{1}{6} &+& \frac{1}{7} \\
&\Rightarrow\ &\rm\ \ \color{green}{2m} &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &+&\, \color{#C00}{\frac{1}{2}} &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}\\
&\Rightarrow\ & -\color{#C00}{\frac{1}{2}}\ \ &=&\ \ 2 &+&\ 1 &+& \frac{2}{3} &-&\rm \color{green}{2m}  &+& \frac{2}{5} &+& \frac{1}{3} &+& \frac{2}{7}^\phantom{M^M}
\end{eqnarray}$$

The prior sum has all odd denominators so reduces to a fraction with odd denominator $\rm\,d\, |\, 3\cdot 5\cdot 7$.

Note $\ $ I purposely avoided any use of valuation theory because Anton requested an "elementary" solution. The above proof can easily be made comprehensible to a high-school student.
