On this page we study bounded sequences, that is, sequences whose terms cannot increase or decrease without bound, but instead remain within suitable numerical bounds. We shall examine the distinction between sequences that are bounded above, bounded below, and bounded, clarifying the mathematical meaning of each definition.
The notion of boundedness is fundamental in the study of numerical sequences, since it allows us to describe the overall behaviour of the terms of a sequence, regardless of whether the sequence converges or diverges.
Throughout the article we consider real sequences, that is, sequences of the form
\[ a:\mathbb{N}\to\mathbb{R}, \]
and we denote their terms by \(a_n\), as \(n\) ranges over \(\mathbb{N}\).
Throughout the text we assume that \(\mathbb{N}=\{0,1,2,\dots\}\).
Contents
- Sequences bounded above
- Sequences bounded below
- Bounded sequences
- Graphical interpretation of boundedness
- Examples of bounded and unbounded sequences
- First properties of bounded sequences
- Convergent sequences and bounded sequences
Sequences bounded above
A real sequence \((a_n)\) is said to be bounded above if there exists a real number \(M\) such that every term of the sequence is less than or equal to \(M\).
In symbols:
\[ \exists M\in\mathbb{R}:\forall n\in\mathbb{N},\quad a_n\le M. \]
The number \(M\) is called an upper bound of the sequence. To say that \((a_n)\) is bounded above is therefore to say that the set of its terms
\[ \{a_n:n\in\mathbb{N}\} \]
is a subset of \(\mathbb{R}\) that is bounded above.
It is important to note that an upper bound need not itself be a term of the sequence. It need only be a real number that lies above every term of the sequence.
Example. Consider the sequence
\[ a_n=\frac{1}{n+1}, \qquad n\in\mathbb{N}. \]
Since \(n\in\mathbb{N}\), we have \(n+1\ge 1\). Consequently:
\[ 0<\frac{1}{n+1}\le 1. \]
Hence, for every \(n\in\mathbb{N}\), we obtain
\[ a_n\le 1. \]
The sequence is therefore bounded above. For instance, \(M=1\) is an upper bound.
However, \(1\) is not the only upper bound. The numbers \(2\) and \(10\) are also upper bounds; more generally, every real number \(M\ge 1\) is an upper bound of the sequence.
Warning. To prove that a sequence is bounded above, there is no need to find the least upper bound. It suffices to find at least one real number \(M\) such that
\[ a_n\le M \]
for every \(n\in\mathbb{N}\).
Conversely, to prove that a sequence is not bounded above, one must show that no real number can be an upper bound. In symbols:
\[ \forall M\in\mathbb{R},\exists n\in\mathbb{N}: a_n>M. \]
This condition means that, whatever real number \(M\) is chosen, there is at least one term of the sequence that exceeds it.
An example of a sequence that is not bounded above
Consider the sequence
\[ a_n=n. \]
It is not bounded above. Indeed, given any \(M\in\mathbb{R}\), we can choose an index \(n\in\mathbb{N}\) such that
\[ n>M. \]
For such an index we have
\[ a_n=n>M. \]
Hence no real number \(M\) can be an upper bound of the sequence \((n)\).
Sequences bounded below
A real sequence \((a_n)\) is said to be bounded below if there exists a real number \(m\) such that every term of the sequence is greater than or equal to \(m\).
In symbols:
\[ \exists m\in\mathbb{R}:\forall n\in\mathbb{N},\quad a_n\ge m. \]
The number \(m\) is called a lower bound of the sequence. To say that \((a_n)\) is bounded below is therefore to say that the set of its terms
\[ \{a_n:n\in\mathbb{N}\} \]
is a subset of \(\mathbb{R}\) that is bounded below.
Here too, a lower bound need not be a term of the sequence. It need only be a real number that lies below every term of the sequence.
Example. Consider the sequence
\[ a_n=n^2+1, \qquad n\in\mathbb{N}. \]
Since \(n^2\ge 0\) for every \(n\in\mathbb{N}\), we have
\[ n^2+1\ge 1. \]
Hence, for every \(n\in\mathbb{N}\), we obtain
\[ a_n\ge 1. \]
The sequence is therefore bounded below. For instance, \(m=1\) is a lower bound.
Naturally, every real number \(m\le 1\) is likewise a lower bound of the sequence. Indeed, if all the terms are greater than or equal to \(1\), then they are certainly greater than or equal to any number smaller than \(1\).
Remark. To prove that a sequence is bounded below, there is no need to find the greatest lower bound. It suffices to find at least one real number \(m\) such that
\[ a_n\ge m \]
for every \(n\in\mathbb{N}\).
On the other hand, to prove that a sequence is not bounded below, one must show that no real number can be a lower bound. In symbols:
\[ \forall m\in\mathbb{R},\exists n\in\mathbb{N}: a_n<m. \]
This condition means that, whatever real number \(m\) is chosen, there is at least one term of the sequence that falls below \(m\).
An example of a sequence that is not bounded below
Consider the sequence
\[ a_n=-n. \]
It is not bounded below. Indeed, given any \(m\in\mathbb{R}\), we can choose an index \(n\in\mathbb{N}\) such that
\[ n>|m|+1. \]
Then \(n>-m\), and hence, multiplying by \(-1\), we obtain
\[ -n<m. \]
Since \(a_n=-n\), it follows that
\[ a_n<m. \]
Hence no real number \(m\) can be a lower bound of the sequence \((-n)\).
Bounded sequences
A real sequence \((a_n)\) is said to be bounded if it is both bounded above and bounded below.
In other words, \((a_n)\) is bounded if there exist two real numbers \(m\) and \(M\) such that
\[ m\le a_n\le M \]
for every \(n\in\mathbb{N}\).
In symbols:
\[ \exists m,M\in\mathbb{R}:\forall n\in\mathbb{N},\quad m\le a_n\le M. \]
The number \(m\) is a lower bound, while the number \(M\) is an upper bound. The sequence is thus contained, term by term, within the closed interval \([m,M]\).
Example. Consider the sequence
\[ a_n=\frac{(-1)^n}{n+1}, \qquad n\in\mathbb{N}. \]
For every \(n\in\mathbb{N}\), we have
\[ |(-1)^n|=1 \]
and hence
\[ |a_n|=\left|\frac{(-1)^n}{n+1}\right|=\frac{1}{n+1}. \]
Since \(n+1\ge 1\), it follows that
\[ \frac{1}{n+1}\le 1. \]
Therefore
\[ |a_n|\le 1. \]
From the inequality \(|a_n|\le 1\) it follows that
\[ -1\le a_n\le 1. \]
The sequence is therefore bounded.
Characterisation via the absolute value
A real sequence \((a_n)\) is bounded if and only if there exists a real number \(K>0\) such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
Proof. Suppose first that the sequence \((a_n)\) is bounded. Then there exist \(m,M\in\mathbb{R}\) such that
\[ m\le a_n\le M \]
for every \(n\in\mathbb{N}\).
Consider
\[ K=\max\{1,|m|,|M|\}. \]
Since \(m\le a_n\le M\), every term \(a_n\) belongs to the interval \([m,M]\). Consequently its absolute value is less than or equal to the larger of \(|m|\) and \(|M|\), and hence also less than or equal to \(K\). Therefore
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
Conversely, suppose that there exists \(K>0\) such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
By the definition of absolute value, this inequality is equivalent to
\[ -K\le a_n\le K. \]
Hence \(-K\) is a lower bound and \(K\) is an upper bound of the sequence. The sequence is therefore bounded.
Warning. A sequence may be bounded above without being bounded below, or bounded below without being bounded above.
For example, the sequence
\[ a_n=-n \]
is bounded above, since
\[ -n\le 0 \]
for every \(n\in\mathbb{N}\), but it is not bounded below.
Similarly, the sequence
\[ b_n=n \]
is bounded below, since
\[ n\ge 0 \]
for every \(n\in\mathbb{N}\), but it is not bounded above.
Hence, in order to be bounded, a sequence must be controlled both from above and from below.
Graphical interpretation of boundedness
Graphically, a real sequence \((a_n)\) can be represented by means of the points
\[ (n,a_n), \qquad n\in\mathbb{N}. \]
In this representation, the index \(n\) is placed on the horizontal axis, while the term \(a_n\) is placed on the vertical axis.
To say that a sequence is bounded above means that all of its points lie on or below a certain horizontal line.
Indeed, if there exists \(M\in\mathbb{R}\) such that
\[ a_n\le M \]
for every \(n\in\mathbb{N}\), then all the points \((n,a_n)\) lie below the horizontal line of equation
\[ y=M. \]
Similarly, to say that a sequence is bounded below means that all of its points lie on or above a certain horizontal line.
Indeed, if there exists \(m\in\mathbb{R}\) such that
\[ a_n\ge m \]
for every \(n\in\mathbb{N}\), then all the points \((n,a_n)\) lie above the horizontal line of equation
\[ y=m. \]
Finally, to say that a sequence is bounded means that all of its points lie between two horizontal lines.
More precisely, if there exist \(m,M\in\mathbb{R}\) such that
\[ m\le a_n\le M \]
for every \(n\in\mathbb{N}\), then all the points \((n,a_n)\) are contained in the horizontal strip lying between the lines
\[ y=m \qquad \text{and} \qquad y=M. \]
This interpretation is useful because it makes the meaning of boundedness immediately apparent: a bounded sequence can neither rise indefinitely towards \(+\infty\) nor fall indefinitely towards \(-\infty\).
Examples of bounded and unbounded sequences
We now consider some basic examples, useful for correctly distinguishing the various forms of boundedness.
A bounded sequence
Consider the sequence
\[ a_n=(-1)^n. \]
Its terms take only the values \(1\) and \(-1\). Indeed, for every \(n\in\mathbb{N}\), we have
\[ (-1)^n\in\{-1,1\}. \]
Consequently
\[ -1\le (-1)^n\le 1 \]
for every \(n\in\mathbb{N}\).
Hence the sequence \(((-1)^n)\) is bounded.
A sequence bounded above but not below
Consider the sequence
\[ a_n=-n^2. \]
Since \(n^2\ge 0\) for every \(n\in\mathbb{N}\), we have
\[ -n^2\le 0. \]
Hence \(0\) is an upper bound of the sequence, and therefore \((a_n)\) is bounded above.
However, the sequence is not bounded below. Indeed, given any \(m\in\mathbb{R}\), we can choose \(n\in\mathbb{N}\) large enough that
\[ n^2>|m|+1. \]
Then \(n^2>-m\), and hence, multiplying by \(-1\), we obtain
\[ -n^2<m. \]
Thus there is a term of the sequence smaller than \(m\). Since this holds for every \(m\in\mathbb{R}\), the sequence is not bounded below.
A sequence bounded below but not above
Consider the sequence
\[ a_n=n^2. \]
Since
\[ n^2\ge 0 \]
for every \(n\in\mathbb{N}\), the sequence is bounded below. For instance, \(0\) is a lower bound.
The sequence, however, is not bounded above. Indeed, given any \(M\in\mathbb{R}\), we can choose \(n\in\mathbb{N}\) large enough that
\[ n^2>M. \]
Hence no real number \(M\) can be an upper bound of the sequence.
A sequence bounded neither above nor below
Consider the sequence
\[ a_n=(-1)^n n. \]
The terms of the sequence change sign according to the parity of \(n\). For even indices one obtains ever larger positive values, while for odd indices one obtains ever smaller negative values.
The sequence is not bounded above. Indeed, given any \(M\in\mathbb{R}\), we may take an even index \(n\) large enough that
\[ n>M. \]
For such an index, since \(n\) is even, we have \((-1)^n=1\), so that
\[ a_n=(-1)^n n=n>M. \]
Hence no real number \(M\) can be an upper bound.
Moreover, the sequence is not bounded below. Indeed, given any \(m\in\mathbb{R}\), we may take an odd index \(n\) large enough that
\[ -n<m. \]
For such an index, since \(n\) is odd, we have \((-1)^n=-1\), so that
\[ a_n=(-1)^n n=-n<m. \]
Hence no real number \(m\) can be a lower bound.
The sequence \(((-1)^n n)\) is therefore bounded neither above nor below.
First properties of bounded sequences
Bounded sequences enjoy a number of fundamental properties, useful in the study of operations on sequences and in the computation of limits.
In this section we consider real sequences defined on \(\mathbb{N}\).
Sum of bounded sequences
If \((a_n)\) and \((b_n)\) are two bounded sequences, then the sum sequence
\[ (a_n+b_n) \]
is also bounded.
Indeed, since \((a_n)\) is bounded, there exists \(A>0\) such that
\[ |a_n|\le A \]
for every \(n\in\mathbb{N}\). Similarly, since \((b_n)\) is bounded, there exists \(B>0\) such that
\[ |b_n|\le B \]
for every \(n\in\mathbb{N}\).
Using the triangle inequality, we obtain
\[ |a_n+b_n|\le |a_n|+|b_n|. \]
Since \(|a_n|\le A\) and \(|b_n|\le B\), it follows that
\[ |a_n+b_n|\le A+B. \]
Hence the sequence \((a_n+b_n)\) is bounded.
Product of bounded sequences
If \((a_n)\) and \((b_n)\) are two bounded sequences, then the product sequence
\[ (a_n b_n) \]
is also bounded.
Indeed, since \((a_n)\) is bounded, there exists \(A>0\) such that
\[ |a_n|\le A \]
for every \(n\in\mathbb{N}\). Since \((b_n)\) is bounded, there exists \(B>0\) such that
\[ |b_n|\le B \]
for every \(n\in\mathbb{N}\).
Then, for every \(n\in\mathbb{N}\), we have
\[ |a_n b_n|=|a_n||b_n|\le AB. \]
Hence the sequence \((a_n b_n)\) is bounded.
Product of a bounded sequence by a constant
If \((a_n)\) is a bounded sequence and \(c\in\mathbb{R}\), then the sequence
\[ (c a_n) \]
is also bounded.
Indeed, since \((a_n)\) is bounded, there exists \(A>0\) such that
\[ |a_n|\le A \]
for every \(n\in\mathbb{N}\).
If \(c=0\), then \(c a_n=0\) for every \(n\in\mathbb{N}\), and hence the sequence \((c a_n)\) is bounded.
If, on the other hand, \(c\ne 0\), we have
\[ |c a_n|=|c||a_n|\le |c|A. \]
In this case too the sequence \((c a_n)\) is bounded.
A bounded sequence need not be convergent
Boundedness alone does not imply convergence.
For example, the sequence
\[ a_n=(-1)^n \]
is bounded, since
\[ -1\le (-1)^n\le 1 \]
for every \(n\in\mathbb{N}\).
Nevertheless, it is not convergent. Indeed, its terms oscillate endlessly between \(1\) and \(-1\), without ever settling down towards a single real number.
Hence a sequence may be bounded without possessing a finite limit.
An unbounded sequence cannot be convergent
If a real sequence is not bounded, then it cannot converge to a real number.
This statement is the contrapositive of the theorem that every convergent sequence is bounded, which we shall prove in the next section.
In other words, boundedness is a necessary condition for convergence, but it is not a sufficient condition.
Convergent sequences and bounded sequences
The most important link between boundedness and convergence is the following: every convergent real sequence is bounded.
Theorem. If a real sequence \((a_n)\) converges to a real number \(\ell\), then \((a_n)\) is bounded.
Proof. Suppose that
\[ a_n\to \ell. \]
By the definition of a finite limit, for every \(\varepsilon>0\) there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |a_n-\ell|<\varepsilon. \]
We choose, in particular,
\[ \varepsilon=1. \]
Then there exists \(N\in\mathbb{N}\) such that, for every \(n\ge N\),
\[ |a_n-\ell|<1. \]
From this inequality it follows that, for every \(n\ge N\),
\[ \ell-1<a_n<\ell+1. \]
Hence all the terms of the sequence from the index \(N\) onwards lie between \(\ell-1\) and \(\ell+1\).
We may assume, increasing \(N\) if necessary, that \(N\ge 1\). It remains only to consider the preceding terms:
\[ a_0,a_1,\dots,a_{N-1}. \]
These are finite in number. Since a finite set of real numbers is always bounded, we may define
\[ K=\max\{1,|a_0|,|a_1|,\dots,|a_{N-1}|,|\ell|+1\}. \]
Then \(K>0\) and, by construction, we have
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
Hence the sequence \((a_n)\) is bounded.
Remark. The theorem just proved asserts that convergence implies boundedness:
\[ a_n\to \ell \quad \Longrightarrow \quad (a_n)\ \text{is bounded}. \]
The converse, however, is not true in general.
Indeed, as already observed, the sequence
\[ a_n=(-1)^n \]
is bounded, but not convergent.
We may therefore say that boundedness is a necessary condition for convergence, but not a sufficient condition.
A special case: monotone sequences
For monotone sequences, on the other hand, boundedness plays an even stronger role.
If a sequence is increasing and bounded above, then it converges. Similarly, if a sequence is decreasing and bounded below, then it converges.
This result is known as the monotone convergence theorem.
Hence, in general, a bounded sequence need not converge; however, if monotonicity is added to boundedness, one obtains a very important criterion for convergence.
Bounded sequences enable us to describe the overall behaviour of the terms of a sequence. A sequence may be bounded above, bounded below, or bounded on both sides.
In particular, a real sequence \((a_n)\) is bounded if there exists \(K>0\) such that
\[ |a_n|\le K \]
for every \(n\in\mathbb{N}\).
This means that all the terms of the sequence remain contained within a finite interval. Boundedness is a fundamental property because every convergent sequence is bounded, even though a bounded sequence is not necessarily convergent.