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This question already has an answer here:

I want to prove that the harmonic series $H_n=1+\frac{1}{2}+\frac{1} {3}+\dots+\frac{1}{n}$ does not converge.

My attempt: to see it does not converge, I try to prove that it is unbounded and it is also an increasing sequence.

$H_{2n}-H_{n}= \frac{1}{n+1}+\frac{1}{n+2}+\dots +\frac{1}{2n}\geq \frac{1}{2n}+\frac{1}{2n}+\dots+\frac{1}{2n}=\frac{1}{2}$

And I am stucked here.

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marked as duplicate by Watson, Lucian, user91500, Namaste, hardmath Feb 1 '17 at 15:20

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This shows that $H_{2n} \ge H_{n} + \frac 12$. Assume by contradiction that $\{H_n\}$ converges to a finite number. Then passing to the limit you would get $H \ge H + \frac{1}{2}$, a contradiction.

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  • $\begingroup$ I am not sure if I understand this answer. What you are trying to say here is that if I choose $M$ so that $H_n \leq M \forall n$ then when $n$ tends to infinity, $H_{2n}=H_{n}$ and therefore $H \geq H +1/2$ which is false and therefore $M$ does not exist and $H_n$ is not bounded, right? $\endgroup$ – J doeoeo Jan 30 '17 at 0:47
  • $\begingroup$ The ingredients are all there, but let me rephrase what you've just said. Since $H_n$ is an increasing sequence, it is either convergent to a finite number $H$ or it is unbounded. Since we want to exclude the first one, let's assume by contradiction that it is true. Then taking the limit, $H_{2n} \to H$ and $H_n \to H$, giving the contradiction as mentioned above. I hope this clarifies! $\endgroup$ – Giovanni Jan 30 '17 at 1:04
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If it converged, $H_n$ would be a Cauchy sequence, and in particular, there would exist in integer $N$ such that $\lvert H_m-H_n\rvert<\frac12$ for any $m,n\ge N$. In particular $$H_{2N}-H_N<\frac12.$$

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