For $n\geq 2$,the rational number

\begin{equation}\label{eq:343242}

1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}

\end{equation}is not an integer.

First let's look at an example:

\begin{align*}

1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}&=1+\frac{1}{2^1}+\frac{1}{3^1}+\frac{1}{2^2}+\frac{1}{5}+\frac{1}{2\times

3}=\frac{2^43^25+2^33^25+2^43\times 5+2^23^25+2^43^2+2^33\times 5}{2^43^25}

\end{align*}

In this example,we find that one term of the numerator is $2^43^2$,there is no $5$ in this term,and every other term in numerator has 5,in denominator there is also a 5.So $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}$$ is not an integer(Why?)

Now we prove the theorem formally.

 

Proof:Let $p_1<p_2<p_3<\cdots<p_k$,$p_1,p_2,\cdots,p_k$ are all the primes which are not larger than $n$.For $2\leq t\leq n$,the standard factorization of $t$ is

\begin{equation}

p_1^{\alpha_{1t}}p_2^{\alpha_{2t}}\cdots p_k^{\alpha_{kt}}

\end{equation}

where $\alpha_{1t},\alpha_{2t},\cdots,\alpha_{kt}\geq 0$.Let's turn \ref{eq:343242} into

\begin{equation}

\label{eq:824}

\sum_{i=1}^n\frac{\frac{1\times 2\times\cdots\times n}{i}}{n!}

\end{equation}

The standard factorization of $n!$ is

\begin{equation}

p_1^{l_1}p_2^{l_2}\cdots p_k^{l_k}

\end{equation}

It is easy to verify that $\forall i< p_k$,the standard factorization of

\begin{equation}

\frac{1\times 2\times \cdots \times n}{i}

\end{equation}is

\begin{equation}

\Delta_1\Delta_2\cdots p_k^{l_k}

\end{equation}(Why?)

When $n\geq i>p_k$,suppose the standard factorization of $i$ is

\begin{equation}

p_1^{h_1}p_2^{h_2}\cdots p_k^{h_k}

\end{equation}

Then it can be verified that $h_k=0$(I have to say this is not a easy thing to do ,because in order to prove this,one should use the

following lemma:

lemma:If $p$ is a prime,then there must exists a prime $q$ such that $p<q<2p$.

I believe this lemma is true,but at present I don't know how to prove it),so the standard factorization of

\begin{equation}

\frac{1\times 2\times\cdots \times n}{i}

\end{equation}is

\begin{equation}

\Delta'_1\Delta'_2\cdots p_k^{l_k}

\end{equation}

 

The standard factorization of

\begin{equation}

\frac{1\times 2\times\cdots\times n }{p_k}

\end{equation}is

\begin{equation}

\Delta_1''\Delta_2''\cdots p_k^{l_k-1}

\end{equation}

So \ref{eq:824} is not an integer(Why?)