# Monotonicity of Real Sequences

## Theorem

Let $\mathbb D$ be a subset of $\N$.

Let $\sequence {a_n}: \mathbb D \to \R$ be a real sequence.

Let $\mathbb X$ be a real interval such that $\mathbb D \subseteq \mathbb X$.

Let $f: \mathbb X \to \R, x \mapsto \map f x$ be a differentiable real function.

Suppose that for every $n \in \mathbb D$:

$\map f n = a_n$

Then:

If $\forall x \in \mathbb X: D_x \map f x \ge 0$, $\sequence {a_n}$ is increasing
If $\forall x \in \mathbb X: D_x \map f x > 0$, $\sequence {a_n}$ is strictly increasing
If $\forall x \in \mathbb X: D_x \map f x \le 0$, $\sequence {a_n}$ is decreasing
If $\forall x \in \mathbb X: D_x \map f x < 0$, $\sequence {a_n}$ is strictly decreasing

where $D_x$ denotes differentiation with respect to $x$.

## Proof

Consider the case where $D_x \map f x \ge 0$

Let $n \in \N$ be in the domain of $\sequence {a_n}$.

From Derivative of Monotone Function, the sign of $D_x f$ is indicative of the monotonicity of $f$.

Hence:

 $\ds \map f {n + 1} - \map f n$ $=$ $\ds \int_n^{n + 1} D_x \map f x \rd x \ge 0$ Fundamental Theorem of Calculus $\ds$ $\ge$ $\ds 0$ Relative Sizes of Definite Integrals $\ds \leadsto \ \$ $\ds \map f {n + 1}$ $\ge$ $\ds \map f n$

Then:

 $\ds \forall n \in \mathbb D: \,$ $\ds \map f n$ $=$ $\ds a_n$ by hypothesis $\ds \leadsto \ \$ $\ds \forall n \in \mathbb D: \,$ $\ds a_{n + 1}$ $\ge$ $\ds a_n$ as $n \in \mathbb D$ was arbitrary

Hence the result, by the definition of monotone.

The proofs of the other cases are similar.

$\blacksquare$