Matt Lee at 17:47, 21 April 2015

2015-04-21T17:47:02Z

Matt Lee: Created page with "==Useful handouts== {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Series Handout |- | %% LyX 2.0.4 created this file. For more i..."

2015-04-21T17:46:45Z

Created page with "==Useful handouts== {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Series Handout |- | <source lang = "latex"> %% LyX 2.0.4 created this file. For more i..."

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==Useful handouts==
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Series Handout
|-
|
<source lang = "latex">
%% LyX 2.0.4 created this file. For more info, see http://www.lyx.org/.
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\usepackage{amsthm}
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\begin{document}

\title{Series Handout}
\maketitle
\begin{defn*}
Given a series $\sum_{n=1}^{\infty}a_{n}=a_{1}+a_{2}+\cdots$, let
$s_{n}$ denote its $n$th partial sum:
\[
s_{n}=\sum_{i=1}^{n}a_{i}=a_{1}+a_{2}+\cdots+a_{n}
\]
If the sequence $\left\{ s_{n}\right\} $ is convergent and $\lim_{n\rightarrow\infty}s_{n}=s$
exists as a real number, then the series $\sum a_{n}$ is called \textbf{convergent
}and we write
\[
a_{1}+a_{2}+\cdots+a_{n}+\cdots=s\mbox{ \ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}or\ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}}\sum_{n=1}^{\infty}a_{n}=s
\]
The number $s$ is called the \textbf{sum }of the series. If the sequence
$\left\{ s_{n}\right\} $ is divergent, then the series is called
\textbf{divergent}.
\end{defn*}
$\mbox{ }$
\begin{fact*}
The geometric series
\[
\sum_{n=1}^{\infty}ar^{n-1}=a+ar+ar^{2}+\cdots
\]
is convergent if $\vert r\vert<1$ and its sum is
\[
\sum_{n=1}^{\infty}ar^{n-1}=\frac{a}{1-r}\ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}\ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}\vert r\vert<1
\]
If $\vert r\vert\geq1$, the geometric series is divergent.
\end{fact*}
$\mbox{ }$
\begin{thm*}
If the series $\sum_{n=1}^{\infty}a_{n}$ is convergent, then $\lim_{n\rightarrow\infty}a_{n}=0$.
\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{(Test for Divergence) }If $\lim_{n\rightarrow\infty}a_{n}$
dies not exist or if $\lim_{n\rightarrow\infty}a_{n}\neq0$, then
the series $\sum_{n=1}^{\infty}a_{n}$ is divergent.
\end{thm*}
$\mbox{ }$
\begin{thm*}
(\textbf{The Integral Test) }Suppose $f$ is a continuous, positive,
decreasing function on $[1,\infty)$ and let $a_{n}=f(n)$. Then the
series $\sum_{n=1}^{\infty}a_{n}$ is convergent if and only if the
improper integral $\int_{1}^{\infty}f(x)dx$ is convergent. In other
words:
\begin{enumerate}
\item If $\int_{1}^{\infty}f(x)dx$ is convergent, then $\sum_{n=1}^{\infty}a_{n}$
is convergent.
\item If $\int_{1}^{\infty}f(x)dx$ is divergent, then $\sum_{n=1}^{\infty}a_{n}$
is divergent.
\end{enumerate}
\end{thm*}
$\mbox{ }$
\begin{fact*}
The $p$-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent
if $p>1$ and divergent if $p=1$.

$\mbox{ }$\end{fact*}
\begin{thm*}
\textbf{(Remainder Estimate for the Integral Test) }Suppose $f(k)=a_{k}$,
where $f$ is a continuous, positive, decreasing function for $x\geq n$
and $\sum a_{n}$ is convergent. If $R_{n}=s-s_{n}$, then
\[
\int_{n+1}^{\infty}f(x)dx\leq R_{n}\leq\int_{n}^{\infty}f(x)dx
\]

\end{thm*}
$\mbox{ }$
\begin{fact*}
If we add $s_{n}$ to each side of the above inequalities:
\[
s_{n}+\int_{n+1}^{\infty}f(x)dx\leq s\leq s_{n}+\int_{n}^{\infty}f(x)dx
\]

\end{fact*}
$\mbox{ }$
\begin{thm*}
\textbf{(The Comparison Test) }Suppose that $\sum a_{n}$ and $\sum b_{n}$
are series with positive terms.
\begin{enumerate}
\item If $\sum b_{n}$ is convergent and $a_{n}\leq b_{n}$ for all $n$,
then $\sum a_{n}$ is also convergent.
\item If $\sum b_{n}$ is divergent and $a_{n}\geq b_{n}$ for all $n$,
then $\sum a_{n}$ is also divergent.
\end{enumerate}
\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{(The Limit Comparison Test) }Suppose that $\sum a_{n}$ and
$\sum b_{n}$ are series with positive terms. If
\[
\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=c
\]
where $c$ is a finite number and $c>0$, then either both series
converge or both diverge.
\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{(Alternating Series Test) }If the alternating series
\[
\sum_{n=1}^{\infty}(-1)^{n-1}b_{n}=b_{1}-b_{2}+b_{3}-b_{4}+b_{5}-b_{6}+\cdots\ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}\ensuremath{\mbox{ }\mbox{ }}\ensuremath{\mbox{ }}b_{n}>0
\]
satisfies
\begin{enumerate}
\item $b_{n+1}\leq b_{n}$ for all $n$
\item $\lim_{n\rightarrow\infty}b_{n}=0$
\end{enumerate}

then the series is convergent.

\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{\textup{\emph{(Alternating Series Estimation Theorem) }}}\textup{\emph{If
$s=\sum(-1)^{n-1}b_{n}$ is the sum of an alternating series that
satisfies }}
\begin{enumerate}
\item $b_{n+1}\leq b_{n}$ for all $n$
\item $\lim_{n\rightarrow\infty}b_{n}=0$
\end{enumerate}

then
\[
\vert R_{n}\vert=\vert s-s_{n}\vert\leq b_{n+1}
\]

\end{thm*}
$\mbox{ }$
\begin{defn*}
A series $\sum a_{n}$ is called \textbf{absolutely convergent }if
the series of absolute values $\sum\vert a_{n}\vert$ is convergent.
\end{defn*}
\textbf{$\mbox{ }$}
\begin{defn*}
A series $\sum a_{n}$ is called \textbf{conditionally convergent
}if it is convergent but not absolutely convergent.
\end{defn*}
$\mbox{ }$
\begin{thm*}
If a series $\sum a_{n}$ is absolutely convergent, then it is convergent.
\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{\textup{\emph{(The Ratio Test) }}}
\begin{enumerate}
\item If $\lim_{n\rightarrow\infty}\vert\frac{a_{n+1}}{a_{n}}\vert=L<1$,
then the series $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent.
\item If \textup{$\lim_{n\rightarrow\infty}\vert\frac{a_{n+1}}{a_{n}}\vert=L>1$
}\textup{\emph{or}}\textup{ $\lim_{n\rightarrow\infty}\vert\frac{a_{n+1}}{a_{n}}\vert=\infty$,
}\textup{\emph{then the series}}\textup{ $\sum_{n=1}^{\infty}a_{n}$}\textup{\emph{
is divergent.}}
\item If $\lim_{n\rightarrow\infty}\vert\frac{a_{n+1}}{a_{n}}\vert=1$,
the Ratio Test is inconclusive; that is, no conclusion can be drawn
about the convergence or divergence of $\sum a_{n}$.
\end{enumerate}
\end{thm*}
$\mbox{ }$
\begin{thm*}
\textbf{\textup{\emph{(The Root Test)}}}
\begin{enumerate}
\item If $\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=L<1$, then
the series $\sum_{n=1}^{\infty}a_{n}$ is absolutely convergent.
\item If \textup{$\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=L>1$
}\textup{\emph{or}}\textup{ $\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=\infty$,
}\textup{\emph{then the series}}\textup{ $\sum_{n=1}^{\infty}a_{n}$}\textup{\emph{
is divergent.}}
\item If $\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=1$, the Root
Test is inconclusive.\end{enumerate}
\end{thm*}

\end{document}
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Math 9C - Revision history

Matt Lee at 17:47, 21 April 2015

Matt Lee: Created page with "==Useful handouts== {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Series Handout |- | %% LyX 2.0.4 created this file. For more i..."