% This is lecture no. 7 of MH3100/: Real Analysis I.

\chapter{Subsequences}%
\label{cha:subsequences}

\section{Subsequences}%
\label{sec:subsequences}

\begin{exmp}
	Alternating Harmonic Series.
	The alternating harmonic series
	\[
	1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dotsb
	\]
	can be shown to be convergent by considering two sequences:
	\begin{enumerate}
		\item The sums of an odd number of terms
			\[
			1,\quad
			1 - \frac{1}{2} + \frac{1}{3},\quad
			1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5},\quad
			\dotsc,
			\]
		\item And the sums of an even number of terms
			\[
			1 - \frac{1}{2},\quad
			1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4},\quad
			1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6},\quad
			\dotsc
			\]
	\end{enumerate}
	The first sequence is decreasing while the second is increasing, and both converge to the same limit.
	Together, they hist that the alternating harmonic series is convergent.
\end{exmp}

\begin{definition}
	Subsequence.
	A \textbf{subsequence} of a sequence $\left\{a_n\right\}^{\infty}_{n=1}$ is a sequence
	\[
	a_{n_1}, a_{n_2}, a_{n_3}, \dotsc
	,\]
	where $n_1<n_2<n_3<\dotsb$ is a strictly increasing sequence of natural numbers.
	A subsequence $\left\{a_{n_k}\right\}$ is said to be \textbf{proper} if $n_k \neq k$ for some $k\in\mathbb{N}$.
\end{definition}

\begin{exmp}
	Subsequences.
	Consider the sequence
	\[
	\left\{a_n\right\}\defas \left\{\frac{n-1}{n}\right\} = 0,\frac{1}{2},\frac{2}{3},\frac{3}{4},\dotsc
	.\]
	The sequences
	\[
	\frac{1}{2},\frac{3}{4},\frac{5}{6},\frac{7}{8},\dotsc
	\quad
	\text{and}
	\quad
	\frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \frac{9999}{10000}, \dotsc
	\]
	are subsequences, and they are indexed by the respective strictly increasing seqeunces
	\[
	2,4,6,8,\dotsc
	\quad
	\text{and}
	\quad
	10, 100, 1000, 10000,\dotsc
	.\]
	The sequences
	\[
	\frac{9}{10}, \frac{4}{5}, \frac{99}{100}, \frac{49}{50}, \frac{999}{1000}, \frac{499}{500},\dotsc
	\quad
	\text{and}
	\quad
	\frac{2}{3},\frac{2}{3}, \frac{4}{5},\frac{4}{5}, \frac{6}{7}, \frac{6}{7},\dotsc
	\]
	are not subsequences.
\end{exmp}

\section{Subsequences \& Convergence}%
\label{sec:subsequences_&_convergence}

\begin{thrm}
	Limits of Subsequences of Convergent Sequences.
	If a sequence $\left\{a_n\right\}$ \textbf{converges} to $a\in\mathbb{R}$ and $\left\{a_{n_k}\right\}$ is a subsequence, then $\left\{a_{n_{k}}\right\}$ \textbf{converges} to $a$.
\end{thrm}

\begin{proof}
	The definition of $a = \lim a_n$ is
	\[
	\forall\varepsilon>0,\ \exists N\in\mathbb{N}\suchthat\forall n\in\mathbb{N}\quad n>N\implies\abs{a_n-a}<\varepsilon
	.\]
	Therefore,
	\begin{align*}
		\forall\varepsilon>0, \exists K = N\in\mathbb{N}\suchthat\forall k\in\mathbb{N}\quad k>K &\implies n_k \geqslant k>K =N\\
													     &\implies \abs{a_{n_k}-a}<\varepsilon,
	\end{align*}
	which is the definition of $a=\lim a_{n_k}$.
\end{proof}

\begin{thrm}
	\label{divcrit}
	Divergence Criterion.
	A \textbf{bounded} sequence $\left\{a_n\right\}$ is \textbf{divergent} iff it has \textbf{two subsequences} $\left\{a_{n_k}\right\}$ and $\left\{a_{m_k}\right\}$ converging to different limits.
\end{thrm}

Note that the theorem is not true for sequences that are not bounded.
For example, $1,\frac{1}{2},3,\frac{1}{4},5,\frac{1}{6},\dotsc$ is divergent but every convergent subsequence converges to 0.
