Suppose $ H(n)=\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}$ is natural number. There exist a natural number $ k$ which satisfy $ 3^{k}\leq n < 3^{k+1}$ Let $ P(n)= 3^{k-1}\times$ (all positive numbers which satisfy both non-multiple of 3 and smaller than $ n$) Then $ H(n) \times P(n)$ is not natural number. ($ H(n)P(n)=m+\frac{1}{3}$ where $ m \in N_{0}$ because of $ 3^{k}$) Contradiction! (sorry for bad English)

nicetry007 wrote: Let $ 3^{k}\;\leq \; 2n+1 \;<\; 3^{k+1}\;$. It is to be noted that every positive odd integer $ m \;\neq\; 3^{k}$ and $ 3\;\leq \;m \;\leq\;2n+1$ has a power of $ 3$ strictly smaller than $ k\;$. Suppose not. Let $ m \; = \; 3^{l}\cdot s$ where $ l \;\geq \;k$ and $ s$ is odd and $ s \;\geq\; 1\;$. We have $ m \; = \; 3^{l}\cdot s \;\leq \;2n+1\;< \;3^{k+1}\;\Rightarrow s \;< \;3^{k+1-l}\;\leq \; 3\;$ ( as $ l \;\geq\; k$) $ \Rightarrow s = 1$ (as $ s$ is odd ) $ \Rightarrow m \; = \; 3^{l}\;<\; 3^{k+1}\;\Rightarrow \; l \;\leq \; k$. But $ l\;\geq \;k$. Hence, $ l \; = \; k \;\Rightarrow \; m \; = \; 3^{k}$, which is a contradiction as $ m \;\neq\; 3^{k}$. Thus, we can conclude that every odd positive integer $ m \; \neq \; 3^{k}$ and $ 3\;\leq \;m \;\leq \; 2n+1$ has a power of $ 3$ strictly smaller than $ k$. The denominator of the sum of fractions is $ lcm(3\;,\;5\;,\;\cdots\;,\;2n+1)$ which has $ 3^{k}$ as a factor. Thus, when we take the sum of all the unit fractions, the numerator of every fraction other than $ \frac{1}{3^{k}}$ gets multiplied by a power of $ 3$ and the numerator of $ \frac{1}{3^{k}}$ gets multiplied by an odd number which is not divisible by $ 3$. Hence, the numerator of the sum of all fractions is a number that is not divisible by $ 3$. Therefore, the sum can never be an integer as the numerator is not divisible by $ 3$ and the denominator is a multiple of $ 3$.

Does this work? By Bertrand's postulate, there exists some $ p$ with $ n + 1 < p < 2n + 2$. Also, $ p$ is clearly odd, so we have that $ \frac{1}{p}$ is among our sum. However, it is obvious that $ \frac{1}{p}$ is the only term in our fraction with a denominator divisible by $ p$. As such, if we put all our terms over a common denominator, our number will be in the form $ \frac{pa + b}{pc}$, where $ p \not | b$. But this fraction clearly cannot be an integer, since $ p \not | pa + b$, so we are done.

The method user by a Euler- Is can be generalizaded to show that the sum of the recrocals of the two or more First terms of the arithmetic progression (1,p,2p-1,3p-1,...), where p is a prime, can never be an integer.