monotone

Monotone sequence

A real-valued sequence $(a_n{)}_{n\in \mathbb{N}}$ is called

increasing: $\forall n\in \mathbb{N}:a_{n+1}>a_n$
non-decreasing: $\forall n\in \mathbb{N}:a_{n+1}\ge a_n$
decreasing: $\forall n\in \mathbb{N}:a_{n+1}<a_n$
non-increasing: $\forall n\in \mathbb{N}:a_{n+1}\le a_n$

A sequence is monotone if it is non-decreasing or non-increasing, and strictly monotone if it is increasing or decreasing.

A sequence that is both non-decreasing and non-increasing must be constant.

Monotonicity principle

Let $(a_n{)}_{n\in \mathbb{N}}\subset \mathbb{R}$ be a monotone sequence. Then

$(a_n)$ is convergent $⟺(a_n)$ is bounded

If non-decreasing: $\underset{n\to \infty }{\lim }{a}_n=\sup \{a_n:n\in \mathbb{N}\}$ (climbing to the ceiling)
If non-increasing: $\underset{n\to \infty }{\lim }{a}_n=\inf \{a_n:n\in \mathbb{N}\}$ (sinking to the floor)

Every monotone unbounded sequence is definitely divergent.

Checking monotonicity

Direct: compute $a_{n+1}-a_n$ and determine the sign.

Quotient trick (for positive sequences): compute $\frac{a_{n+1}}{a_n}$ instead.

$$\frac{a_{n+1}}{a_n}\ge 1\text{ for all }n⟹\text{non-decreasing},\frac{a_{n+1}}{a_n}\le 1\text{ for all }n⟹\text{non-increasing}$$

Often cleaner because ratios of products/powers simplify better than differences.

Recursive sequences

For a recursion $a_{n+1}=f(a_n)$, the standard approach:

1. Find limit candidates by solving $a=f(a)$
2. Prove $(a_n)$ is bounded (often: show it stays above/below the candidate)
3. Prove monotonicity (often follows from the bound)
4. Monotone + bounded $⟹$ convergent, and step 1 identifies the limit

Setting $\lim a_n=\lim a_{n+1}=a$ is only valid after convergence is established.