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

Monotonically increasing if $a_n \leq a_m.$

Strictly monotonically increasing if $a_n < a_m.$

Monotonically decreasing if $a_n \geq a_m.$

Strictly monotonically decreasing if $a_n > a_m.$

Monotonic or monotone if it is either monotonically increasing or monotonically decreasing.

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.