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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.

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    That is correct. My fault2012-06-27
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    Since this is homework, I think you have to show more precisely why $b_{n}$ is strictly increasing. Observe that it depends on $n$ both in the index of the summation and in the fraction.2012-06-27
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    You say that the sequence $(b_n)$ is strictly increasing. I don't see that, is false.2012-06-27
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    @AndréNicolas And I'm thinking it goes to $0$. It can't be all nice and rainbows to prove the harmonic series diverges!2012-06-27

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