Why $\sum_{k=1}^n \frac{1}{2k+1}$ is not an integer? Let $S=\sum_{k=1}^n \frac{1}{2k+1}$, how can we prove with elementary math reasoning that $S$ is not an integer?
Can somebody help?

Let $p=2k'+1$ be the largest odd prime number less than or equal to $2n+1$.  Now consider the sum:
$$\frac13 + \frac15 + \frac17 + \cdots + \frac{1}{2k'-1} + \frac{1}{p} + \frac{1}{2k'+3} + \cdots + \frac{1}{2n+1}.$$
We can find a common denominator for this fraction by taking the product:
$$3\cdot5\cdots (2k'-1) \cdot p \cdot (2k' +3) \cdots (2n+1) = M.$$
In the numerator we have a sum of several terms:
$$\frac{M}{3} + \frac{M}{5} + \cdots + \frac{M}{p} + \cdots + \frac{M}{(2n+1)}.$$
Each term here is an integer.
I claim that all but $\frac{M}{p}$ is divisible by $p$.  This would mean the numerator is not divisible by $p$ and is congruent to $\frac{M}{p}$ modulo $p$.  Since $p$ appears in the denominator, this sum cannot be an integer.
Suppose that $\frac{M}{p}$ was indeed divisible by $p$.  This means that we must have had another multiple of $p$ appear in one of our denominators.  The next valid term would be $3p$.  However, Bertrand's postulate tells us that for any natural number $n$ we have a prime between $n$ and $2n$.  This tells us that there would have been a new largest prime between $p$ and $3p$.  Which is a contradiction.
