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?
 A: Hint:  Recall the (elementary) proof that $\sum_{i=1}^n \frac{1}{i}$ is not an integer (for $n>1$):
Let $2^k$ be the largest power of 2 that is smaller than or equal to $n$. Consider making (smallest) common denominator of the form $2^k L$, where $L$ is odd. The numerator of every term will be even except for the term of $\frac{1}{2^k}$, which contributes an odd term $L$. Hence, the numerator is odd and the denominator is even. This cannot be an integer.
Hint: For finite sum of odd reciprocals, show that when we make a (smallest) common denominator, the numerator is not a multiple of $3$.

 Solution:  Let $3^k$ be the largest power of 3 that is smaller than or equal to $n$. Consider making common denominator of the form $3^k L$, where $L$ is not a multiple of 3. The numerator of every term will be a multiple of 3, except for terms of the form $\frac{ 1}{ a3^k}$ for some integer $a$.
 Use the fact that $a=2$ does not appear in the sequence since it is even. Also, by the definition of $3^k$, no other higher multiple can appear. Hence, there is only 1 term which contributes a non-multiple of 3. Thus, the numerator is not a multiple of 3.

A: 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.
A: You can write 
$$S=\frac{\sum_j\prod_{k\neq j}(2k+1)}{\prod (2k+1)}$$
Now the largest prime factor less or equal than $2n+1$ divides the denominator and all but one of the summands in the numerator, hence not the sum. 
