# Monotonicity of Real Sequences

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## 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 $f$.

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

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.

$\blacksquare$