# Monotonicity of Real Sequences

## Theorem

Let $\sequence {a_n}: \mathbb D \to \R$ be a real sequence, where $\mathbb D$ is a subset of $\N$.

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

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

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

$\map f n = a_n$

Then:

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

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

Then:

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

Hence the result, by the definition of monotone.

The proofs of the other cases are similar.

$\blacksquare$