Harmonic Series. 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.

Unfortunately, the proof breaks down, since the sequence $(b_n)$ cannot be strictly increasing. In fact, $(b_n)$ has limit $0$. 
For if $(b_n)$ were increasing, then by your formula the sum of the first $n$ terms of the harmonic series would grow at least like a constant times $n+1$. But it only grows logarithmically. So for large $n$, $b_n$ is approximately $a_n=\frac{2\log n}{n+1}$. 
