The limit theorem for monotone sequences is one of the fundamental results in the theory of limits of sequences. It states that a monotone sequence always has a limit, finite or infinite.
More precisely, an increasing sequence tends to its supremum, possibly equal to \(+\infty\), while a decreasing sequence tends to its infimum, possibly equal to \(-\infty\). In particular, a bounded monotone sequence is always convergent.
This result is of considerable importance, since it allows one to establish that a sequence has a limit without actually computing it. It is enough to check monotonicity and, whenever a finite limit is sought, boundedness as well.
Contents
- A review of monotone sequences
- The limit theorem for monotone sequences
- The case of an increasing sequence
- The case of a decreasing sequence
- Bounded monotone sequences
- Examples
A review of monotone sequences
Throughout, we assume that sequences are indexed by the positive natural numbers, that is, \(n\in\mathbb{N}\) with \(\mathbb{N}=\{1,2,3,\dots\}\).
A real sequence \((a_n)\) is said to be increasing if
\[ a_n\leq a_{n+1} \]
for every \(n\in\mathbb{N}\). In other words, each term is less than or equal to the one that follows it.
By contrast, a real sequence \((a_n)\) is said to be decreasing if
\[ a_n\geq a_{n+1} \]
for every \(n\in\mathbb{N}\). In this case each term is greater than or equal to the one that follows it.
In both cases we are using monotonicity in the weak sense: an increasing sequence may have equal consecutive terms, and the same is true of a decreasing sequence.
With this terminology, an increasing sequence is also called non-decreasing, while a decreasing sequence is also called non-increasing. When the inequalities are strict, one speaks instead of strictly increasing or strictly decreasing sequences.
A sequence is called monotone if it is either increasing or decreasing.
Monotonicity thus captures the orderly behavior of the terms of the sequence. On its own, however, it does not tell us whether the limit is finite or infinite. For instance, an increasing sequence may converge to a real number or diverge to \(+\infty\).
The limit theorem for monotone sequences
Let \((a_n)\) be a monotone real sequence.
If \((a_n)\) is increasing, then
\[ \lim_{n\to+\infty}a_n=\sup\{a_n:n\in\mathbb{N}\}, \]
where the supremum is understood in the extended sense and may be either a real number or \(+\infty\).
If \((a_n)\) is decreasing, then
\[ \lim_{n\to+\infty}a_n=\inf\{a_n:n\in\mathbb{N}\}, \]
where the infimum is understood in the extended sense and may be either a real number or \(-\infty\).
In compact form:
\[ \lim_{n\to+\infty}a_n = \begin{cases} \sup\{a_n:n\in\mathbb{N}\}, & \text{if } (a_n) \text{ is increasing},\\[4pt] \inf\{a_n:n\in\mathbb{N}\}, & \text{if } (a_n) \text{ is decreasing}. \end{cases} \]
This means that a monotone sequence can never fail to have a limit: its limit always exists, possibly as an infinite limit.
Since the limit of a sequence, when it exists, is unique, this value completely determines the limiting behavior of a monotone sequence.
For example, the sequence \(a_n=(-1)^n\) has no limit, but it is not monotone. Monotonicity is therefore a strong condition: it rules out persistent oscillation between distinct values.
The case of an increasing sequence
Suppose that \((a_n)\) is an increasing sequence. We distinguish two cases: the sequence may be bounded above or not bounded above.
Increasing and bounded above
Suppose that \((a_n)\) is increasing and bounded above. Then the set of its values
\[ \{a_n:n\in\mathbb{N}\} \]
is non-empty and bounded above. By the completeness property of the real numbers, its supremum exists. Set
\[ S=\sup\{a_n:n\in\mathbb{N}\}. \]
We want to show that
\[ \lim_{n\to+\infty}a_n=S. \]
By the definition of supremum, \(S\) is an upper bound of the set \(\{a_n:n\in\mathbb{N}\}\). Hence
\[ a_n\leq S \]
for every \(n\in\mathbb{N}\).
Moreover, again by the definition of supremum, for every \(\varepsilon>0\) the number \(S-\varepsilon\) is not an upper bound of the set. Consequently there exists an index \(k\in\mathbb{N}\) such that
\[ S-\varepsilon<a_k. \]
Since the sequence is increasing, for every \(n\geq k\) we have
\[ a_k\leq a_n. \]
Therefore, for every \(n\geq k\),
\[ S-\varepsilon<a_k\leq a_n\leq S. \]
From this chain of inequalities it follows that
\[ 0\leq S-a_n<\varepsilon. \]
Hence
\[ |a_n-S|<\varepsilon \]
for every \(n\geq k\). By the definition of limit,
\[ \lim_{n\to+\infty}a_n=S. \]
Thus an increasing sequence that is bounded above converges to its supremum.
Increasing and unbounded above
Now suppose that \((a_n)\) is increasing and unbounded above. We want to show that
\[ \lim_{n\to+\infty}a_n=+\infty. \]
Since the sequence is not bounded above, for every \(M>0\) there exists an index \(\nu\in\mathbb{N}\) such that
\[ a_\nu>M. \]
Since \((a_n)\) is increasing, for every \(n\geq \nu\) we have
\[ a_n\geq a_\nu>M. \]
Hence, for every \(M>0\), there exists \(\nu\in\mathbb{N}\) such that, for every \(n\geq\nu\),
\[ a_n>M. \]
By the definition of divergence to \(+\infty\),
\[ \lim_{n\to+\infty}a_n=+\infty. \]
Thus an increasing sequence that is unbounded above diverges to \(+\infty\).
The case of a decreasing sequence
Suppose that \((a_n)\) is a decreasing sequence. Here too we distinguish two possibilities: the sequence may be bounded below or not bounded below.
Decreasing and bounded below
Suppose that \((a_n)\) is decreasing and bounded below. Then the set of its values
\[ \{a_n:n\in\mathbb{N}\} \]
is non-empty and bounded below. By the completeness property of the real numbers, its infimum exists. Set
\[ L=\inf\{a_n:n\in\mathbb{N}\}. \]
We want to show that
\[ \lim_{n\to+\infty}a_n=L. \]
By the definition of infimum, \(L\) is a lower bound of the set \(\{a_n:n\in\mathbb{N}\}\). Hence
\[ L\leq a_n \]
for every \(n\in\mathbb{N}\).
Moreover, by the definition of infimum, for every \(\varepsilon>0\) the number \(L+\varepsilon\) is not a lower bound of the set. Consequently there exists an index \(k\in\mathbb{N}\) such that
\[ a_k<L+\varepsilon. \]
Since the sequence is decreasing, for every \(n\geq k\) we have
\[ a_n\leq a_k. \]
Therefore, for every \(n\geq k\),
\[ L\leq a_n\leq a_k<L+\varepsilon. \]
From this chain of inequalities it follows that
\[ 0\leq a_n-L<\varepsilon. \]
Hence
\[ |a_n-L|<\varepsilon \]
for every \(n\geq k\). By the definition of limit,
\[ \lim_{n\to+\infty}a_n=L. \]
Thus a decreasing sequence that is bounded below converges to its infimum.
Decreasing and unbounded below
Now suppose that \((a_n)\) is decreasing and unbounded below. We want to show that
\[ \lim_{n\to+\infty}a_n=-\infty. \]
Since the sequence is not bounded below, for every \(M>0\) there exists an index \(\nu\in\mathbb{N}\) such that
\[ a_\nu<-M. \]
Since \((a_n)\) is decreasing, for every \(n\geq \nu\) we have
\[ a_n\leq a_\nu<-M. \]
Hence, for every \(M>0\), there exists \(\nu\in\mathbb{N}\) such that, for every \(n\geq\nu\),
\[ a_n<-M. \]
By the definition of divergence to \(-\infty\),
\[ \lim_{n\to+\infty}a_n=-\infty. \]
Thus a decreasing sequence that is unbounded below diverges to \(-\infty\).
Bounded monotone sequences
The previous theorem yields a frequently used criterion.
If a sequence is increasing and bounded above, then it converges, and its limit is its supremum:
\[ \lim_{n\to+\infty}a_n=\sup\{a_n:n\in\mathbb{N}\}. \]
If a sequence is decreasing and bounded below, then it converges, and its limit is its infimum:
\[ \lim_{n\to+\infty}a_n=\inf\{a_n:n\in\mathbb{N}\}. \]
In particular, every bounded monotone real sequence is convergent.
This criterion is often called the monotone convergence theorem for sequences. It is useful because it establishes the existence of a limit without requiring its explicit value to be known in advance.
Examples
Example 1. Consider the sequence
\[ a_n=1-\frac{1}{n}. \]
This sequence is increasing, since
\[ a_{n+1}=1-\frac{1}{n+1} \]
and, because
\[ \frac{1}{n+1}<\frac{1}{n}, \]
we have
\[ 1-\frac{1}{n+1}>1-\frac{1}{n}. \]
Hence
\[ a_{n+1}>a_n. \]
Moreover, for every \(n\in\mathbb{N}\),
\[ a_n<1. \]
The sequence is therefore increasing and bounded above. By the limit theorem for monotone sequences, it converges to its supremum.
In this case
\[ \sup\{a_n:n\in\mathbb{N}\}=1. \]
Indeed,
\[ 1-\frac{1}{n}<1 \]
for every \(n\in\mathbb{N}\), so that \(1\) is an upper bound. Moreover, for every \(\varepsilon>0\) there exists \(n\in\mathbb{N}\) such that
\[ 1-\varepsilon<1-\frac{1}{n}. \]
This inequality is equivalent to
\[ \frac{1}{n}<\varepsilon, \]
which holds for \(n\) sufficiently large. Hence \(1\) is the least upper bound.
Therefore
\[ \lim_{n\to+\infty}\left(1-\frac{1}{n}\right)=1. \]
Example 2. Consider the sequence
\[ b_n=\frac{1}{n}. \]
This sequence is decreasing, since
\[ \frac{1}{n+1}<\frac{1}{n} \]
for every \(n\in\mathbb{N}\). It is also bounded below, since
\[ b_n>0 \]
for every \(n\in\mathbb{N}\).
Thus \((b_n)\) is decreasing and bounded below. By the limit theorem for monotone sequences, it converges to its infimum.
In this case
\[ \inf\{b_n:n\in\mathbb{N}\}=0. \]
Indeed, \(0\) is a lower bound of the sequence, since
\[ \frac{1}{n}>0 \]
for every \(n\in\mathbb{N}\). Moreover, for every \(\varepsilon>0\) there exists \(n\in\mathbb{N}\) such that
\[ \frac{1}{n}<\varepsilon. \]
Hence the terms of the sequence become arbitrarily close to \(0\) from above. Consequently \(0\) is the greatest lower bound.
Therefore
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Example 3. Consider the sequence
\[ c_n=n. \]
This sequence is increasing but unbounded above. Indeed, for every \(M>0\) there exists \(n\in\mathbb{N}\) such that
\[ n>M. \]
Hence, by the limit theorem for monotone sequences,
\[ \lim_{n\to+\infty}n=+\infty. \]
Example 4. Consider the sequence
\[ d_n=-n. \]
This sequence is decreasing and unbounded below. Indeed, for every \(M>0\) there exists \(n\in\mathbb{N}\) such that
\[ -n<-M. \]
Hence, by the limit theorem for monotone sequences,
\[ \lim_{n\to+\infty}(-n)=-\infty. \]
These examples show that monotonicity always guarantees the existence of a limit, but not that the limit is finite. To obtain convergence to a real number, one also needs the appropriate boundedness: above for increasing sequences, below for decreasing ones.
The theorem depends essentially on the completeness of the real numbers. Indeed, in \(\mathbb{Q}\) a bounded monotone sequence may fail to converge to a rational number. For example, the sequence of finite decimal approximations of \(\sqrt{2}\),
\[ 1,\ 1.4,\ 1.41,\ 1.414,\ 1.4142,\ \ldots \]
is increasing and bounded in \(\mathbb{Q}\), yet does not converge in \(\mathbb{Q}\), because its limit in \(\mathbb{R}\) is \(\sqrt{2}\), which is irrational.