|
|
|
Main Menu
|
|
Sections
Meta
Talkback
Downloads
Information
|
|
|
|
|
|
Viewing Version
5
of
'harmonic series'
|
[ view 'harmonic series'
|
back to history
]
| Title of object: |
harmonic series |
| Canonical Name: |
HarmonicSeries |
| Type: |
Topic |
| Created on: |
2009-05-28 17:29:12 |
| Modified on: |
2009-05-29 08:35:31 |
| Classification: |
msc:02.30.-f |
| Defines: |
necessary condition of convergence, Euler constant |
Revision comment (for changes between this and next version):
Preamble:
% 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
\usepackage{pstricks}
\usepackage{pst-plot}
% define commands here |
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 \PMlinkescapetext{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 \PMlinkescapetext{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}.
\begin{center}
\begin{pspicture}(-0.5,-0.5)(6,6)
\psaxes[Dx=1,Dy=1]{->}(0,0)(-0.5,-0.5)(5.5,5.5)
\rput(5.5,-0.2){$x$}
\rput(-0.2,5.5){$y$}
\rput(-0.3,-0.3){$0$}
\psplot[linecolor=red]{0.22}{5.2}{1 x div}
\psline(1,0)(1,1)
\psline[linecolor=blue](1,1)(2,1)(2,0.5)(3,0.5)(3,0.333)(4,0.333)(4,0.25)(5,0.25)(5,0.2)
\psline(2,0)(2,0.5)
\psline(3,0)(3,0.333)
\psline(4,0)(4,0.25)
\rput(1,3.5){$y = \frac{1}{x}$}
\end{pspicture}
\end{center} |
|
|
|
|
|
