Math 9C
Useful handouts
| Series Handout Created by Christina Osborne |
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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 \[ \vert R_{n}\vert=\vert s-s_{n}\vert\leq b_{n+1} \]
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