# 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 $f$.

Because Differentiable Function is Continuous and Continuous Real Function is Darboux Integrable, $D_x f$ is integrable.

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.

$\blacksquare$