# 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 $1/n\geq 1/N$ when $n\leq N$, we have

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

Hence,

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

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

Title proof of divergence of harmonic series (by grouping terms) ProofOfDivergenceOfHarmonicSeriesbyGroupingTerms 2013-03-22 15:08:39 2013-03-22 15:08:39 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Proof msc 40A05