# proof of divergence of harmonic series (by splitting odd and even terms)

Suppose that the series $\sum_{n=1}^{\infty}1/n$ converged. Since all the terms are positive, we could regroup them as we please, in particular, split the series into two series, that of even terms and that of odd terms:

 $\sum_{n=1}^{\infty}{1\over n}=\sum_{n=1}^{\infty}{1\over 2n}+\sum_{n=1}^{% \infty}{1\over 2n-1}$

Since $\sum_{n=1}^{\infty}1/n=2\sum_{n=1}^{\infty}1/(2n)$, we would conclude that

 $\sum_{n=1}^{\infty}{1\over 2n}=\sum_{n=1}^{\infty}{1\over 2n-1}.$

But $2n-1<2n$, hence $1/(2n)<1/(2n-1)$, so we would also have

 $\sum_{n=1}^{\infty}{1\over 2n}<\sum_{n=1}^{\infty}{1\over 2n-1},$

which contradicts the previous conclusion. Thus, the assumption that the series converged is untenable.

Title proof of divergence of harmonic series (by splitting odd and even terms) ProofOfDivergenceOfHarmonicSeriesbySplittingOddAndEvenTerms 2013-03-22 17:38:26 2013-03-22 17:38:26 rspuzio (6075) rspuzio (6075) 4 rspuzio (6075) Definition msc 40A05