Bounded Sequences
A sequence is called bounded if there exists a real number \( M > 0 \) such that, for every \( n \in \mathbb{N} \), one has \[ |a_n| \le M \]
Boundedness can be described in two complementary ways.
- A sequence is bounded above if there exists a real number \( M \) such that \[ a_n \le M \quad \text{for all } n \] In this case, the sequence admits an upper bound, meaning none of its terms exceed a fixed value.
- A sequence is bounded below if there exists a real number \( m \) such that \[ a_n \ge m \quad \text{for all } n \] In this case, the sequence admits a lower bound, meaning all its terms remain greater than or equal to a fixed value.
A sequence is bounded if and only if it is both bounded above and bounded below.
A sequence that fails to be bounded is called an unbounded sequence.
Examples
Example 1
This sequence is bounded above.
\[ a_n = \frac{1}{n} \]
Indeed, every term satisfies $ a_n \le 1 $, so $ M = 1 $ is an upper bound.
\[ 1,\ \frac{1}{2},\ \frac{1}{3},\ \frac{1}{4},\ \dots \]
Example 2
This sequence is bounded below.
\[ a_n = n \]
Every term satisfies $ a_n \ge 1 $, so $ m = 1 $ is a lower bound.
\[ 1,\ 2,\ 3,\ 4,\ \dots \]
Example 3
This sequence is bounded.
\[ a_n = (-1)^n \]
Its values lie in the interval \( [-1, 1] \), hence it is bounded both above and below.
\[ -1,\ 1,\ -1,\ 1,\ \dots \]
Example 4
This sequence is unbounded.
\[ a_n = n^2 \]
In this case, the terms grow without bound as \( n \to \infty \).
\[ 1,\ 4,\ 9,\ 16,\ \dots \]
Every convergent sequence is bounded
Any sequence that has a finite limit is necessarily bounded.
Thus, all convergent sequences are bounded.
Note. There are also examples of bounded sequences that are not regular. For instance, the following sequence oscillates between -1 and +1. It neither converges nor diverges, yet it remains bounded. $$ a_n = ( -1 )^n $$
Proof
Suppose a sequence an converges to l:
$$ \lim_{ n \rightarrow \infty } a_n = l $$
Let’s choose ε equal to 1:
$$ \epsilon = 1 $$
By the definition of the limit, there exists a value v such that:
$$ |a_n - l| < 1 \quad \forall n > v $$
Adding |l| to both sides yields:
$$ |(a_n - l) + l| < 1 + |l| $$
Recognizing that an = |an + l - l|, we can write:
$$ |a_n| < 1 + |l| $$
Therefore, we can conclude that:
$$ |a_n| < M $$
where M is defined as the largest value among the absolute values of the terms of the sequence and 1 + |l|:
$$ M = \max \{ |a_1|, |a_2|, \dots , |a_n|, 1 + |l| \} $$
And so on.
