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.
 A: Given that André showed your proof is not correct, I have an alternate idea to give you. 
We know by the comparison test, that if the series $$\sum_{n=1}^\infty a_n$$ diverges
 and $$\lim\limits_{n \to \infty} \dfrac{b_n}{a_n}=\ell >0 $$ then 
$$\sum_{n=1}^\infty b_n$$
diverges too.
But from elementary calculus, we know that
$$\mathop {\lim }\limits_{x \to 0^+} \frac{{\log \left( {1 + x} \right)}}{x} = 1$$
This means that letting $x=1/n$
$$\mathop {\lim }\limits_{n \to +\infty } \frac{{\log \left( {1 + \frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1$$
Does $${\sum\limits_{n = 1}^\infty  {\log \left( {1 + \frac{1}{n}} \right)} }$$ converge or diverge?

Propted by Hennings comment, I add the following:
Cauchy's Condensation Test
If $a_n$ is a decreasing sequence of positive terms, then $$\sum_{n>0} a_n$$ converges if and only if  $$\sum_{n \geq 0} 2^n a_{2^n}$$ does, and
$$\sum_{n>0} a_n \leq  \sum_{n \geq 0} 2^n a_{2^n} \leq 2 \sum_{n >0} a_n $$
(This can be proven rather easily)
A: 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 $a_n=\frac{2\log n}{n+1}$. 
