Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer? 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.

Here's a short proof: Let $H_n = \displaystyle \sum_{k=1}^n\dfrac{1}{k}.$ One can show that $\displaystyle\sum_{k=1}^{n}\dfrac{(-1)^{k-1}\binom{n}{k}}{k}= H_n.$ This can be rewritten as: $$\sum_{k=0}^{n}{(-1)^k\binom{n}{k}a_k} = b_n$$
where $a_0 =0$ and $a_i = \dfrac{1}{i}$ for $i=1,\ldots n$ and $b_n = -H_n$
This answer shows that the $b_i$ are integers if and only if the $a_i$ are integers. Clearly for $i \geq 2 $ we can see that the $a_i$ are not integers, from which it follows that neither are the $b_i, i\geq 2.$ 
