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<record version="15" id="2588">
 <title>alternating series test</title>
 <name>AlternatingSeriesTest</name>
 <created>2002-02-24 07:44:00</created>
 <modified>2009-10-05 17:13:55</modified>
 <type>Theorem</type>
 <creator id="127" name="Koro"/>
 <author id="127" name="Koro"/>
 <author id="1182" name="Larry Hammick"/>
 <author id="1858" name="matte"/>
 <author id="225" name="saforres"/>
 <author id="148" name="vitriol"/>
 <classification>
	<category scheme="msc" code="40-00"/>
	<category scheme="msc" code="40A05"/>
 </classification>
 <synonyms>
	<synonym concept="alternating series test" alias="Leibniz's theorem"/>
	<synonym concept="alternating series test" alias="Leibniz test"/>
 </synonyms>
 <related>
	<object name="AlternatingSeries"/>
 </related>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}</preamble>
 <content>\PMlinkescapeword{states}
\PMlinkescapeword{simple}
The {\bf alternating series test}, or
the {\bf Leibniz's Theorem}, states the following:

{\bf Theorem} \cite{rudin, kreyszig93}
Let $(a_n)_{n=1}^\infty$ be a non-negative, non-increasing sequence 
or real numbers such that $\lim_{n \rightarrow \infty} a_n = 0$.  
Then the infinite series $\sum_{n=1}^\infty (-1)^{(n+1)} a_n$ converges.

This test provides a necessary and sufficient condition for the convergence of an alternating series, since if $\sum_{n=1}^\infty a_n$ converges then $a_n\to 0$. 

{\bf Example:}  The series
$\sum_{k = 1}^{\infty}\frac{1}{k}$
does not converge, but the alternating series
$\sum_{k = 1}^{\infty}(-1)^{k+1}\frac{1}{k}$
converges to $\ln(2)$.

\begin{thebibliography}{9}
\bibitem{rudin}
 W. Rudin, \emph{Principles of Mathematical Analysis}, McGraw-Hill Inc., 1976.
 \bibitem {kreyszig93} E. Kreyszig,
 \emph{Advanced Engineering Mathematics},
 John Wiley \&amp; Sons, 1993, 7th ed.
 \end{thebibliography}</content>
</record>
