HarmonicSeries 2009-05-28 17:29:12 2009-05-28 18:07:54 Topic Taylor series preamble necessary condition of convergence Euler constant % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here The \emph{harmonic series} $$\sum_{k=1}^\infty\frac{1}{k} \;=\; 1+\frac{1}{2}+\frac{1}{3}+\ldots$$ satisfies the \emph{necessary condition of convergence} $$\lim_{k\to\infty}a_n \;=\; 0$$ for the series \,$a_1+a_2+a_3+\ldots$ of real or complex terms: $$\lim_{k\to\infty}\frac{1}{k} \;=\; 0$$ Nevertheless, the harmonic series diverges.\, It is seen if we first group the terms with parentheses: $$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right) +\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right) +\left(\frac{1}{9}+\frac{1}{10}+\ldots+\frac{1}{16}\right)+\ldots$$ Here, each parenthetic sum contains a number of terms twice as many as the preceding one.\, The sum in the first pair of parentheses is graater than\, $2\cdot\frac{1}{4} = \frac{1}{2}$,\, the sum in the second pair of parentheses is greater than\, $4\cdot\frac{1}{8} = \frac{1}{2}$;\, thus one sees that the sum in all pairs of parentheses is greater than $\frac{1}{2}$.\, Consequently, the partial sum of $n$ first terms exceeds any given real number, when $n$ is sufficiently big.\\ The \PMlinkescapetext{divergence} of the harmonic series is very slow, though.\, Its speed may be illustrated by considering the difference $$\sum_{k=1}^{n-1}\frac{1}{k}-\!\int_1^n\frac{dx}{x} \;=\; \sum_{k=1}^{n-1}\frac{1}{k}-\ln{n}$$ (see the diagram).\, We know that $\ln{n}$ increases very slowly as $n \to \infty$ (e.g. $\ln{1\,000\,000\,000} \,\approx\, 20.7$).\, The increasing of the partial sum $\sum_{k=1}^{n-1}\frac{1}{k}$ is about the same, since the limit $$\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\frac{1}{k}-\ln{n}\right)\;=\;\gamma$$ is a little positive number $$\gamma \;=\; 0.5772156649...$$ which is called the \emph{Euler constant} or \emph{Euler--Mascheroni constant}.