My (simpler) proof of the divergence of a harmonic series.

Asked Dec 14 '13 at 16:41

Active Dec 14 '13 at 16:48

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Let $H=1+\frac{1}{2}+\frac{1}{3}+\cdots$ . Proving that it diverges this is what I did.

I supposed that the series converges to $H$, :

$$H\geq1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{6}+\frac{1}{6}+\frac{1}{8}+\frac{1}{8}+\dots=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$

$$=\frac{1}{2}+H$$

And we have reached a contradiction with the series converging to $H$ hence concluding the proof. Do I have any mistakes?

edited Dec 14 '13 at 16:48

TonyK

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asked Dec 14 '13 at 16:41

Module

1

How do you exactly form the series to the right of the first $;\ge;$ and how then you prove the inequality? – DonAntonio Dec 14 '13 at 16:44
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I saw essentially this proof in a brief note in the Monthly in the late '90s. – Michael Hardy Dec 14 '13 at 16:47
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@DonAntonio: it seems perfectly clear to me. There is no rearrangment of terms, just term-by-term comparison. – TonyK Dec 14 '13 at 16:47
I'm happy it does to you, @TonyK . It didn't to me and that's why I asked... – DonAntonio Dec 14 '13 at 16:48
The series to the right of the inequality holds because $\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}$ and $\frac{1}{5}+\frac{1}{6}>\frac{1}{6}+\frac{1}{6}$ – Module Dec 14 '13 at 16:49
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This is the exact same proof a lecturer gave us back in calculus 1 some 28 years ago, after seeing other more usual proofs. It's nice and correct, imo. – DonAntonio Dec 14 '13 at 16:50
@TonyK Yes i had edited it before, but thank you very much. – Module Dec 14 '13 at 16:52
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possible duplicate of Why does the series $\frac 1 1 + \frac 12 + \frac 13 + \cdots$ not converge? – user91500 Dec 14 '13 at 17:00
@ALGEAN I have never even seen that post. – Module Dec 14 '13 at 17:03

My (simpler) proof of the divergence of a harmonic series.

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