Sums of the Harmonic Series

 We shall prove the following:



Let

$$m=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

Then  $m\notin \mathbf{N}$.


We will use the following simple lemma:


Let $n$ be a positive integer and $k$ the maximum number for which $2^k\le n$. Then for all $m\le n$ then exponent of $2$ in the prime factorization is either strictly less than $k$ or else $m=2^k$.


This is because if we had $m=2^k{p}_1^{s_1}...p_t^{s_t}$  then $n\ge m>2^k{2}^{s_1}..{.2}^{s_1}=2^l$ with $l>k$ which is a contradiction.
 
Consider then

$$m=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$

and multiply both sides by $n!$ to obtain:


$$n!m=n!+1\times 3\times ...\times n+..1\times 2..(n-2)\times n+(n-1)!$$

Let $k$ be the maximum exponent such that  $2^k\le n$. Consider  $1\times 2\times ...\times n$.
In the prime factorization of each of the terms $1,2,.,k,...,n$ there will certainly be for the exponents of $2$  all the numbers from $1$ to $k$ and each at least once and possibly several times except for $k$ which can occur maximum once. There are three possibilities for $n$. Either it is $2^k$, or has a prime factorization with exponent of $2$ a $c<k$ or it has no factor $2$. In the first case it is easy to see that the exponent $p$ of $2$ of $(n-1)!$ is strictly less than the corresponding exponent of each of the other  terms of
$$n!m=n!+1\times 3\times ...\times n+..1\times 2..(n-2)\times n+(n-1)!$$

So we can divide the equation by $2^p$ and obtain an odd number equal to an even and thus arrive at a contradiction.

In the second and third cases we have that there is a $t=2^k<n$ and there will be exactly one term $1\times 2...\times (t-1)\times (t+1)\times ...n$ which will have an exponent  $p$ of $2$ strictly less than that of all the other terms (because $k$ as an exponent can occur only once by the lemma) and we can divide the equation in the same way to obtain an odd number equal to an even number and thus a contradiction.