I don't know if my proof is correct. Let be 

\begin{eqnarray}
H\_n &=& 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dotsb + \dfrac{1}{n} \\
&+& \\
H\_n &=& \dfrac{1}{n} + \dfrac{1}{n-1} + \dfrac{1}{n-2} + \dotsb + 1\\
&\parallel& \\
2H\_n &=& \dfrac{n+1}{n}+ \dfrac{n+1}{2(n-1)} + \dfrac{n+1}{3(n-2)} + \dotsb + \dfrac{n+1}{k(n-k+1)} + \dotsb + \dfrac{n+1}{n} \\
&\parallel& \\
2H\_n &=& (n+1) \sum\_{k=1}^n \dfrac{1}{k(n-k+1)} \\
&\parallel& \\
H\_n &=& \dfrac{(n+1)}{2} \sum\_{k=1}^n \dfrac{1}{k(n-k+1)}\\
\end{eqnarray}

Let's say $b\_n = \sum\_{k=1}^n \dfrac{1}{k(n-k+1)}$. The sequence $b\_n$ is strictly increasing, so only two things are going to happen (this part I'm not sure) :

If $b\_n$ is convergent then $\lim\limits\_{n \rightarrow \infty} b\_n >0$ and $\lim\limits\_{n \rightarrow \infty} H\_n = \lim\limits\_{n \rightarrow + \infty} \frac{(n+1)}{2} . b\_n = + \infty$ 

If $b\_n$ diverges then $\lim\limits\_{n \rightarrow \infty} b\_n = + \infty$ and $\lim\limits\_{n \rightarrow \infty} H\_n = \lim\limits\_{n \rightarrow + \infty} \frac{(n+1)}{2} . b\_n = + \infty$

Therefore, harmonic series is divergent.