Monotonicity of Real Sequences

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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.

$\blacksquare$


Also see