Monotonicity of Real Sequences

Theorem
Let $\langle{a_n}\rangle: \mathbb D \to \R$ 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 $f: \Bbb X \to \R, x \mapsto f \left({x}\right)$ be a differentiable real function.

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


 * $f \left({n}\right) = a_n$

Then:


 * If $\forall x \in \Bbb X: D_x f \left({x}\right) \ge 0$, $\langle{a_n}\rangle$ is increasing


 * If $\forall x \in \Bbb X: D_x f \left({x}\right) > 0$, $\langle{a_n}\rangle$ is strictly increasing


 * If $\forall x \in \Bbb X: D_x f \left({x}\right) \le 0$, $\langle{a_n}\rangle$ is decreasing


 * If $\forall x \in \Bbb X: D_x f \left({x}\right) < 0$, $\langle{a_n}\rangle$ is strictly decreasing

where $D_x$ denotes differentiation w.r.t $x$.

Proof
Consider the case where $D_x f \left({x}\right) \ge 0$

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

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 Function is Riemann Integrable, $D_x f$ is integrable.

From Relative Sizes of Definite Integrals:


 * $\displaystyle \int_n^{n+1} D_x f\left({x}\right) \ \mathrm dx \ge 0$

From the Fundamental Theorem of Calculus:
 * $\displaystyle f \left({n+1}\right) - f \left({n}\right) = \int_n^{n+1} D_x f \left({x}\right) \ \mathrm dx \ge 0$

which implies:


 * $\displaystyle f \left({n+1}\right) \ge f \left({n}\right)$

By hypothesis:


 * $\forall n \in \mathbb D: f\left({n}\right) = a_n$

which implies, as $n \in \mathbb D$ was arbitrary:


 * $\forall n \in \mathbb D: a_{n+1} \ge a_n$

Hence the result, by the definition of monotone.

The proofs of the other cases are similar.

Also see

 * Restriction of Monotone Function is Monotone