80 lines
4.2 KiB
TeX
80 lines
4.2 KiB
TeX
\section{Introduction to Infinite Series}
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\begin{definition}
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If $X:=(x_n)$ is a sequence in $\R$, then the \textbf{infinite series} (or simply the \textbf{series}) \textbf{generated by} $X$ is the sequence $S:= (s_k)$ defined by
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\begin{align*}
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s_1 & := x_1 \\
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s_2 & := s_1 + s_2 & (=x_1 + x_2) \\
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& \dots \\
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s_k & := s_{k-1}+x_k & (=x_1+x_2+\dots+x_k) \\
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& \dots
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\end{align*}
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\end{definition}
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\begin{theorem}[\textbf{The $n$th Term Test}]
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If the series $\sum x_n$ converges, then $\lim (x_n) = 0$.
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\end{theorem}
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\begin{theorem}[\textbf{Cauchy Criterion for Series}]
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The series $\sum x_n$ converges if and only if for every $\varepsilon >0$ there exists $M(\varepsilon) \in \N$ such that if $m>n\geq M(\varepsilon)$, then
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\[|s_m-s_n|=|x_{n+1}+x_{n+2}+\dots+x_m|<\varepsilon\]
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\end{theorem}
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\begin{theorem}
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Let $(x_n)$ be a sequence of nonnegative real numbers. Then the series $\sum x_n$ converges if and only if the sequence $S=(s_k)$ of partial sums is bounded. In this case,
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\[\sum\limits_{n=1}^{\infty}x_n = \lim (x_k) = \sup \{s_k:k \in \N\}\]
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\end{theorem}
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\begin{theorem}[\textbf{Comparison Test}]
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Let $X:=(x_n)$ and $Y:=(y_n)$ be real sequences and suppose that for some $K \in \N$ we have
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\[0 \leq x_n \leq y_n\ \ \ \text{for}\ \ \ n \geq K\]
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\begin{enumerate}
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\item Then the convergence of $\sum y_n$ implies the convergence of $\sum x_n$.
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\item The divergence of $\sum x_n$ implies the divergence of $\sum y_n$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{Limit Comparison Test}]
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Suppose that $X:=(x_n)$ and $Y:=(y_n)$ are strictly positive sequences and suppose that the following limit exists in $\R$:
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\[r:=\lim \left(\frac{x_n}{y_n}\right)\]
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\begin{enumerate}
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\item If $r \neq 0$, then $\sum x_n$ is convergent if and only if $\sum y_n$ is convergent.
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\item If $r=0$ and if $\sum y_n$ is convergent, then $\sum x_n$ is convergent.
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\item If $r=\infty$ and $\sum y_n$ diverges, then $\sum x_n$ diverges.
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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Let $(a_n):n \mapsto a(n)$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero. That is, for every $n$ in the domain of $(a_n):a_n>0,\ a_{n+1} \leq a_n$, and $a_n \to 0$ as $n \to + \infty$. The series $\displaystyle\sum_{n=1}^{\infty} 2^n a(2^n)$ is called the \textbf{condensed} form of the series $\displaystyle\sum_{n=1}^{\infty} a_n$.
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\end{definition}
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\begin{theorem}[\textbf{Cauchy Condensation Test}]
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Let $(a_n):n \mapsto a(n)$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero. That is, for every $n$ in the domain of $(a_n):a_n>0,\ a_n \geq a_{n+1}$, and $a_n \to 0$ as $n \to +\infty$. Then the series $\displaystyle\sum_{n=1}^{\infty} a_n$ converges if and only if the condensed series $\displaystyle\sum_{n=1}^{\infty} 2^na(2^n)$ converges.
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\end{theorem}
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\begin{theorem}[\textbf{Cauchy Ratio Test}]
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Let $\displaystyle\sum_{n=1}^{\infty} a_n$ be a series and $a_n>0$ for all $n \in \N$, and suppose the following limit exists in $\R$:
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\[L=\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}\]
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\begin{enumerate}
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\item If $L<1$, then the series converges absolutely.
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\item If $L>1$, then the series is divergent.
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\item If $L=1$ or the limit does not exist, then the test is inconclusive.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{Kummer's Test}]
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Let $\sum a_n$ be a positive term series.
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\begin{enumerate}
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\item If there exists a positive term sequence $b_n$, $\alpha > 0$, and $\N \in \N$ such that $\displaystyle\frac{a_n}{a_{n+1}} \cdot b_n - b_{n+1} \geq \alpha,\ \forall\ n \geq N$, then $\sum a_n$ converges.
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\item If $\displaystyle\frac{a_n}{a_{n+1}}\cdot b_n - b_{n+1} \leq 0,\ \forall\ n \geq N$, and if $\sum \frac{1}{b_n}$ diverges, then $\sum a_n$ diverges.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{Gauss' Test}]
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If $\sum a_n$ is a positive term series, and if there exists a bounded sequence $b_n$ such that $\forall\ n \geq N$, $\displaystyle\frac{a_n}{a_{n+1}} = 1 +\frac{L}{n} + \frac{b_n}{n^2}$, then
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\begin{enumerate}
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\item If $L > 1$, then $\sum a_n$ converges.
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\item If $L \leq 1$, then $\sum a_n$ diverges.
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\end{enumerate}
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\end{theorem}
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