The limit theorem for monotone sequences is one of the fundamental results about limits of sequences. It states that a monotone sequence always has a limit, either finite or infinite.
More precisely, an increasing sequence tends to its supremum, while a decreasing sequence tends to its infimum. In particular, a bounded monotone sequence is always convergent.
This result is especially useful because it lets us establish that the limit of a sequence exists without having to compute it explicitly. It is enough to verify monotonicity and, when a finite limit is sought, boundedness.
Contents
- A review of monotone sequences
- 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 natural numbers starting at \(1\), that is, \(n\geq1\).
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 next.
A real sequence \((a_n)\) is said instead 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 next.
In both definitions monotonicity is taken in the weak sense: an increasing sequence may have equal consecutive terms, and likewise a decreasing sequence may have equal consecutive terms.
With this terminology, an increasing sequence is also called nondecreasing, while a decreasing sequence is also called nonincreasing. When the inequalities are strict, one speaks instead of strictly increasing or strictly decreasing sequences.
A sequence is said to be monotone if it is either increasing or decreasing.
Monotonicity thus describes the orderly behavior of the terms of a sequence. By itself, however, it does not tell us whether the limit is finite or infinite. For instance, an increasing sequence may either converge to a real number or diverge to \(+\infty\).
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 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 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 cannot fail to have a limit in the extended sense: 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 the 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 unbounded 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 nonempty and bounded above. By the completeness property of the real numbers, its supremum exists. Set
\[ S=\sup\{a_n:n\in\mathbb{N}\}. \]
We claim 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 is 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 claim that
\[ \lim_{n\to+\infty}a_n=+\infty. \]
Since the sequence is not bounded above, for every \(M>0\) there is an index \(\nu\in\mathbb{N}\) such that
\[ a_\nu>M. \]
Because \((a_n)\) is increasing, for every \(n\geq \nu\) we have
\[ a_n\geq a_\nu>M. \]
Hence, for every \(M>0\), there is \(\nu\in\mathbb{N}\) 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 unbounded 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 nonempty and bounded below. By the completeness property of the real numbers, its infimum exists. Set
\[ L=\inf\{a_n:n\in\mathbb{N}\}. \]
We claim 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 is 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 claim that
\[ \lim_{n\to+\infty}a_n=-\infty. \]
Since the sequence is not bounded below, for every \(M>0\) there is an index \(\nu\in\mathbb{N}\) such that
\[ a_\nu<-M. \]
Because \((a_n)\) is decreasing, for every \(n\geq \nu\) we have
\[ a_n\leq a_\nu<-M. \]
Hence, for every \(M>0\), there is \(\nu\in\mathbb{N}\) 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
From the preceding theorem we obtain a criterion that is used very often.
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 commonly called the monotone convergence theorem for sequences. It is useful because it allows one to establish the existence of a limit without immediately knowing its explicit value.
Examples
Example 1. Consider the sequence
\[ a_n=1-\frac{1}{n}. \]
The 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 \(n\in\mathbb{N}\). Moreover, for every \(\varepsilon>0\) there is \(n\in\mathbb{N}\) such that
\[ 1-\varepsilon<1-\frac{1}{n}. \]
This inequality is equivalent to
\[ \frac{1}{n}<\varepsilon, \]
which holds for all sufficiently large \(n\). Hence \(1\) is the least upper bound.
Therefore
\[ \lim_{n\to+\infty}\left(1-\frac{1}{n}\right)=1. \]
Example 2. Consider the sequence
\[ b_n=\frac{1}{n}. \]
The 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 \(n\in\mathbb{N}\).
Hence \((b_n)\) 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 is \(n\in\mathbb{N}\) such that
\[ \frac{1}{n}<\varepsilon. \]
Thus 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
\[ c_n=n. \]
The sequence is increasing but not bounded above. Indeed, for every \(M>0\) there is \(n\in\mathbb{N}\) such that
\[ n>M. \]
Hence, by the limit theorem for monotone sequences,
\[ \lim_{n\to+\infty}n=+\infty. \]
Example 4. Consider the sequence
\[ d_n=-n. \]
The sequence is decreasing and not bounded below. Indeed, for every \(M>0\) there is \(n\in\mathbb{N}\) 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 the 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 sequences.
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 it does not converge in \(\mathbb{Q}\), because its limit in \(\mathbb{R}\) is \(\sqrt{2}\), which is irrational.