Real Analysis: Sequences
Monotonic Sequences
Monotonic Sequences
Consider a sequence $\{a_n\}$ and the indices $n, m \in \mathbb{N}$ where $n < m$. Then $\{a_n\}$ is
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Monotonically increasing if $a_n \leq a_m.$
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Strictly monotonically increasing if $a_n < a_m.$
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Monotonically decreasing if $a_n \geq a_m.$
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Strictly monotonically decreasing if $a_n > a_m.$
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Monotonic or monotone if it is either monotonically increasing or monotonically decreasing.
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Strictly monotonic or strictly monotone if it is either strictly monotonically increasing or strictly monotonically decreasing.
Monotone Convergence Theorem
The monotone convergence theorem states that a monotonic sequence converges if and only if it is bounded. This theorem breaks down into two parts, each of which should be intuitive. If a sequence is monotonic and bounded, then surely it must stop at some point, namely its limit. Likewise, if a sequence is monotonic and converges to $p$, then it is clearly bounded by $a_0$ on one end and $p$ on the other.
Problems
Show that $\{n^2\}$ is monotonically increasing.
Proof by induction.
Base case: $0^2 = 0$ and $1^2 = 1$, and $0 < 1.$
Inductive step: Assume $n^2 \leq (n+1)^2$. Then
$ n^2 \leq n^2 + 2n + 1 \\ n^2 + (2n + 1) \leq (n^2 + 2n + 1) + (2n + 1) \\ n^2 + 2n + 1 \leq n^2 + 4n + 2 \\ n^2 + 2n + 1 \leq n^2 + 4n + 4 \\ (n+1)^2 \leq (n+2)^2 $
The result follows by induction.
Show that $\{a_n\}$ is both monotonically increasing and monotonically decreasing if and only if $\{a_n\}$ is constant.
Assume $a_n = c$ for all $n \in \mathbb{N}$. Then for $n, m \in \mathbb{N}$ where $n < m$, it follows that $a_n = a_m$, and therefore $a_n \leq a_m$, so $\{a_n\}$ is monotonically decreasing, and $a_n \geq a_m$, so $\{a_n\}$ is monotonically decreasing.
Conversely, assume $\{a_n\}$ is both monotonically increasing and monotonically decreasing. Then for all $n, m \in \mathbb{N}$ where $n < m$, it follows that $a_n \leq a_m$ and $a_n \geq a_m$. Therefore $a_n = a_m$.
Show that if $\{a_n\}$ is monotonically increasing, then it is bounded below by $a_0$ and that $a_0$ is its greatest lower bound.
Since $\{a_n\}$ is monotonically increasing, for all $n, m \in \mathbb{N}$ where $n < m$, it follows that $a_n \leq a_m$. Pick $n = 0$. Then $0 < m$ for all $m \neq 0$, and so $a_0 \leq a_m$. Therefore $a_0$ is a lower bound of $\{a_n\}$. Since $a_0 \in \{a_n\}$, it follows that $a_0$ is the greatest lower bound of $\{a_n\}$.
Monotone Convergence Theorem: Show a monotonic sequence converges if and only if it is bounded.
Let $\{a_n\}$ be a monotonically increasing sequence in $\mathbb{R}$. Then for all $n, m \in \mathbb{N}$, $n < m$ implies that $a_n \leq a_m$. Additionally, $\{a_n\}$ is bounded below by $a_0$.
Assume $\{a_n\}$ converges to $p$. Then $\{a_n\}$ is bounded.
Conversely, assume $\{a_n\}$ is bounded. Consider the set $B$ of upper bounds of $\{a_n\}$. By the Least Upper Bound property, $B$ contains a least upper bound, call it $p$. For any $\epsilon > 0$, there is an $N \in \mathbb{N}$ such that $p - a_N < \epsilon$, as otherwise $p - \epsilon$ is an upper bound for $\{a_n\}$, which contradicts the leastness of $p$. Since $\{a_n\}$ is monotonically increasing, for all $n > N$ it follows that $|a_n - p| = p - a_n \leq p - a_N < \varepsilon$. Therefore $\{a_n\}$ converges to $p$.
A similar argument shows that monotonically decreasing sequences converge if and only if they are bounded.
Bolzano Weierstrass Theorem (alternate proof): Use the monotone subsequence theorem to prove the that every bounded sequence has a convergent subsequence.
Let $\{a_n\}$ be a bounded sequence. By the monotone subsequence theorem, there exists a monotone subsequence $\{a_{n_k}\}$ of $\{a_n\}.$ Since $\{a_n\}$ is bounded, so is $\{a_{n_k}\}.$ By the monotone convergence theorem, it follows that $\{a_{n_k}\}$ converges.