proof of divergence of harmonic series (by grouping terms)

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

$$\sum _{n=1}^{2^M}\frac{1}{n}=\sum _{m=1}^M\sum _{n=2^{m-1}+1}^{2^m}\frac{1}{n}.$$

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

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

Hence,

$$\sum _{n=1}^{2^M}\frac{1}{n}\ge \frac{M}{2}$$

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

Title	proof of divergence of harmonic series (by grouping terms)
Canonical name	ProofOfDivergenceOfHarmonicSeriesbyGroupingTerms
Date of creation	2013-03-22 15:08:39
Last modified on	2013-03-22 15:08:39
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	9
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 40A05