TITLE: Why $\sum_{k=1}^n \frac{1}{2k+1}$ is not an integer?
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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.