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Viewing Version
3
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'harmonic series'
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| Title of object: |
harmonic series |
| Canonical Name: |
HarmonicSeries |
| Type: |
Topic |
| Created on: |
2009-05-28 17:29:12 |
| Modified on: |
2009-05-29 07:39:15 |
| Classification: |
msc:02.30.-f |
| Defines: |
necessary condition of convergence, Euler constant |
Preamble:
% this is the default PlanetMath 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}
\usepackage{pstricks}
\usepackage{pst-plot}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem} |
Content:
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 parentheses is greater than\, $2\cdot\frac{1}{4} = \frac{1}{2}$,\, the sum in the second parentheses is greater than\, $4\cdot\frac{1}{8} = \frac{1}{2}$;\, thus one sees that the sum in all 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}. |
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