PlanetMath: proof of divergence of harmonic series (by grouping terms)

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proof of divergence of harmonic series (by grouping terms)

(Proof)

The harmonic series can be shown to diverge by a simple argument involving grouping terms. Write

$\displaystyle \sum_{n=1}^{2^M} \frac{1}{n} = \sum_{m=1}^M \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n}.$

Since $1/n \ge 1/N$ when $n \le N$ , we have

$\displaystyle \sum_{n=2^{m-1}+1}^{2^m} \frac{1}{n} \ge \sum_{n=2^{m-1}+1}^{2^m} 2^{-m} = (2^m - 2^{m-1}) 2^{-m} = \frac{1}{2}$

Hence,

$\displaystyle \sum_{n=1}^{2^M} \frac{1}{n} \ge \frac{M}{2}$

so the series diverges in the limit $M \to \infty$ .

"proof of divergence of harmonic series (by grouping terms)" is owned by rspuzio.

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Cross-references: limit, series, terms, argument, simple, diverge, harmonic series
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This is version 6 of proof of divergence of harmonic series (by grouping terms), born on 2005-03-21, modified 2007-11-26.
Object id is 6890, canonical name is ProofOfDivergenceOfHarmonicSreies.
Accessed 3193 times total.

Classification:

AMS MSC:

40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)

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