\( \definecolor{colordef}{RGB}{249,49,84} \definecolor{colorprop}{RGB}{18,102,241} \)

Monotonicity of Sequences

Definition

Definition Increasing and Decreasing Sequences
Let \((u_n)\) be a sequence.
  • The sequence \((u_n)\) is increasing from index \(n_0\) if for all \(n \ge n_0\), \(\boldsymbol{u_{n+1} \ge u_n}\).
  • The sequence \((u_n)\) is decreasing from index \(n_0\) if for all \(n \ge n_0\), \(\boldsymbol{u_{n+1} \le u_n}\).
  • A sequence that is either increasing or decreasing is said to be monotonic.
Note
  • Just like with functions, if we replace the non-strict inequalities with strict ones, we speak of a strictly increasing or strictly decreasing sequence.
  • Some sequences are not monotonic, such as the sequence \((u_n)\) defined by \(u_n = (-1)^n\), which alternates between \(1\) and \(-1\).

Studying the Variations of a Sequence

Method Studying the Sign of the Difference \(u_{n+1}-u_n\)
To study the monotonicity of a sequence \((u_n)\), we can determine the sign of the difference between two consecutive terms:
  • If for all \(n\geq n_0\), \(\textcolor{colorprop}{u_{n+1} - u_n \ge 0}\), the sequence is increasing from index \(n_0\).
  • If for all \(n\geq n_0\), \(\textcolor{colorprop}{u_{n+1} - u_n \le 0}\), the sequence is decreasing from index \(n_0\).
Example
Study the variations of the sequence \((u_n)\) defined by \(u_n = 2n - 3\).

Let \(n \in \mathbb{N}\).$$\begin{aligned}u_{n+1} - u_n &= [2(n+1) - 3] - [2n - 3] \\ &= 2n + 2 - 3 - 2n + 3 \\ &= 2\end{aligned}$$Since \(2 > 0\), the sequence \((u_n)\) is strictly increasing.

Method Comparing the Ratio \(\frac{u_{n+1}}{u_n}\) to 1
For a sequence with strictly positive terms (\(u_n > 0\)):
  • If for all \(n\geq n_0\), \(\textcolor{colorprop}{\frac{u_{n+1}}{u_n} \geq 1}\), the sequence is increasing from index \(n_0\)..
  • If for all \(n\geq n_0\), \(\textcolor{colorprop}{\frac{u_{n+1}}{u_n} \leq 1}\), the sequence is decreasing from index \(n_0\)..
Example
Study the variations of the sequence \((u_n)\) defined by \(u_n = 2^n\).

Let \(n \in \mathbb{N}\).
All terms are positive (\(2^n > 0\)).$$\begin{aligned} \frac{u_{n+1}}{u_n} &= \frac{2^{n+1}}{2^n}\\ &= 2^{n+1-n}\\ &= 2 \\ \end{aligned}$$Since \(2 > 1\), the sequence \((u_n)\) is strictly increasing.