2.3: Monotone Sequences

Definition \(\PageIndex{1}\)

A sequence \(\left\{a_{n}\right\}\) is called increasing if

\[a_{n} \leq a_{n+1} \text { for all } n \in \mathbb{N}.\]

It is called decreasing if

\[a_{n} \geq a_{n+1} \text { for all } n \in \mathbb{N}.\]

If \(\left\{a_{n}\right\}\) is increasing or decreasing, then it is called a monotone sequence.

The sequence is called strictly increasing (resp. strictly decreasing) if \(a_{n}<a_{n+1} \text { for all } n \in \mathbb{N}\) (resp. \(a_{n}>a_{n+1} \text { for all } n \in \mathbb{N}\).

It is easy to show by induction that if \(\left\{a_{n}\right\}\) is an increasing sequence, then \(a_{n} \leq a_{m}\) whenever \(n \leq m\).

Theorem \(\PageIndex{1}\) - Monotone Convergence Theorem.

Let \(\left\{a_{n}\right\}\) be a sequence of real numbers. The following hold:

1. If \(\left\{a_{n}\right\}\) is increasing and bounded above, then it is convergent.
2. If \(\left\{a_{n}\right\}\) is decreasing and bounded below, then it is convergent.

Proof

(a) Let \(\left\{a_{n}\right\}\) be an increasing sequence that is bounded above. Define

\(A=\left\{a_{n}: n \in \mathbb{N}\right\}\).

Then \(A\) is a subset of \(\mathbb{R}\) that is nonempty and bounded above and, hence, \(\sup A\) exists. Let \(\ell=\sup A\) and let \(\varepsilon>0\). By Proposition 1.5.1, there exists \(N \in \mathbb{N}\) such that

\(\ell-\varepsilon<a_{N} \leq \ell\).

Since \(\left\{a_{n}\right\}\) is increasing,

\(\ell-\varepsilon<a_{N} \leq a_{n} \text { for all } n \geq N\).

On the other hand, since \(\ell\) is an upper bound for \(A\), we have \(a_{n} \leq \ell\) for all \(n\). Thus,

\(\ell-\varepsilon<a_{n}<\ell+\varepsilon \text { for all } n \geq N\).

Therefore, \(\lim _{n \rightarrow \infty} a_{n}=\ell\).

(b) Let \(\left\{a_{n}\right\}\) be a decreasing sequence that is bounded below. Define

\(b_{n}=-a_{n}\).

Then \(\left\{b_{n}\right\}\) is increasing and bounded above (if \(M\) is a lower bound for \(\left\{a_{n}\right\}\), then \(-M\) is an upper bound for \(\left\{b_{n}\right\}\)). Let

\(\ell=\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty}\left(-a_{n}\right)\).

Then \(\left\{a_{n}\right\}\) converges to \(-\ell\) by Theorem 2.2.1. \(\square\)

Remark \(\PageIndex{2}\)

It follows from the proof of Theorem 2.3.1 that if \(\left\{a_{n}\right\}\) is increasing and bounded above, then

\[\lim _{n \rightarrow \infty} a_{n}=\sup \left\{a_{n}: n \in \mathbb{N}\right\}.\]

Similarly, if \(\left\{a_{n}\right\}\) is decreasing and bounded below, then

\[\lim _{n \rightarrow \infty} a_{n}=\inf \left\{a_{n}: n \in \mathbb{N}\right\}.\]

Example \(\PageIndex{1}\)

Given \(r \in \mathbb{R}\) with \(|r|<1\), define \(a_{n}=r^{n}\) for \(n \in \mathbb{N}\). Then

\[\lim _{n \rightarrow \infty} a_{n}=0.\]

Solution

This is clear if \(r=0\). Let us first consider the case where \(0<r<1\). Then \(0 \leq a_{n+1}=r a_{n} \leq a_{n}\) for all \(n\). Therefore, \(\left\{a_{n}\right\}\) is decreasing and bounded below. By Theorem 2.3.1, the sequence converges. Let

\[\ell=\lim _{n \rightarrow \infty} a_{n}.\]

Since \(a_{n+1}=r a_{n}\) for all \(n\), taking limits on both sides gives \(\ell = r \ell\). Thus, \((1-r) \ell=0\) and, hence, \(\ell = 0\). In the general case, we only need to consider the sequence defined by \(b_{n}=\left|a_{n}\right|\) for \(n \in \mathbb{N}\); see Exercise 2.1.3.

Example \(\PageIndex{2}\)

Consider the sequence \(\left\{a_{n}\right\}\) defined as follows:

\[a_{1}=2\]

\[a_{n+1}=\frac{a_{n}+5}{3} \text { for } n \geq 1\]

Solution

First we will show that the sequence is increasing. We proove by induction that for all \(n \in \mathbb{N}\), \(a_{n}<a_{n+1}\). Since \(a_{2}=\frac{a_{1}+5}{3}=\frac{7}{3}>2=a_{1}\), the statement is true for \(n=1\). Next, suppose \(a_{k}<a_{k+1}\) for some \(k \in \mathbb{N}\). Then \(a_{k}+5<a_{k+1}+5\) and \(\left(a_{k}+5\right) / 3<\left(a_{k+1}+5\right) / 3\). Therefore,

\[a_{k+1}=\frac{a_{k}+5}{3}<\frac{a_{k+1}+5}{3}=a_{k+2}.\]

It follows by induction that the sequence is increasing.

Next we prove that the sequence is bounded by \(3\). Again, we proceed by induction. The statement is clearly true for \(n=1\). Suppose that \(a_{k} \leq 3\) for some \(k \in \mathbb{N}\). Then

\[a_{k+1}=\frac{a_{k}+5}{3} \leq \frac{3+5}{3}=\frac{8}{3} \leq 3.\]

It follows that \(a_{n} \leq 3\) for all \(n \in \mathbb{N}\).

From the Monotone Convergence Theorem, we deduce that there is \(\ell \in \mathbb{R}\) such that \(\lim _{n \rightarrow \infty} a_{n}=\ell\). Since the subsequence \(\left\{a_{k+1}\right\}_{k=1}^{\infty}\) also converges to \(\ell\), taking limits on both sides of the equationin (2.7), we obtain

\[\ell=\frac{\ell+5}{3}.\]

Therefore, \(3 \ell=\ell+5\) and, hence, \(\ell=5 / 2\).

Example \(\PageIndex{3}\) - The number e.

Consider the sequence \(\left\{a_{n}\right\}\) given by

\[a_{n}=\left(1+\frac{1}{n}\right)^{n}, n \in \mathbb{N}.\]

By the binomial theorem,

\[\begin{aligned}
a_{n} &=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(\frac{1}{n}\right)^{k} \\
&=1+1+\frac{n(n-1)}{2 !} \frac{1}{n^{2}}+\frac{n(n-1)(n-2)}{3 !} \frac{1}{n^{3}}+\cdots+\frac{n(n-1) \cdots(n-(n-1))}{n !} \frac{1}{n^{n}} \\
&=1+1+\frac{1}{2 !}\left(1-\frac{1}{n}\right)+\frac{1}{3 !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots+\frac{1}{n !}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-1}{n}\right)
\end{aligned}.\]

The corresponding expression for \(a_{n+1}\) has one more term and each factor \(\left(1-\frac{k}{n}\right)\) is replaced by the larger factor \(\left(1-\frac{k}{n}\right)\). It is then clear that \(a_{n}<a_{n+1}\) for all \(n \in \mathbb{N}\). Thus, the sequence is increasing. Moreover,

\[\begin{aligned}
a_{n} & \leq 1+1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\
&<2+\frac{1}{1.2}+\frac{1}{2.3}+\cdots+\frac{1}{(n-1) \cdot n} \\
&=2+\sum_{k=1}^{n-1}\left(\frac{1}{k}-\frac{1}{k+1}\right)=3-\frac{1}{n}<3
\end{aligned}.\]

Hence the sequence is bounded above.

Solution

By the monotone Convergence Theorem, \(\lim _{n \rightarrow \infty} a_{n}\) exists and is denoted by \(e\). In fact, \(e\) is an irrational number and \(e \approx 2.71828\).

The following fundamental result is an application of the Monotone Convergence Theorem.

Theorem \(\PageIndex{3}\) - Nested Intervals Theorem.

Let \(\left\{I_{n}\right\}_{n=1}^{\infty}\) be a sequence of nonempty closed bounded intervals satisfying \(I_{n+1} \subset I_{n}\) for all \(n \in \mathbb{N}\). Then the following hold:

1. \(\bigcap_{n=1}^{\infty} I_{n} \neq \emptyset\).
2. If, in addition, the lengths of the intervals \(I_{n}\) converge to zero, then \(\bigcap_{n=1}^{\infty} I_{n}\) consists of a single point.

Proof

Let \(\left\{I_{n}\right\}\) be as in the statement with \(I_{n}=\left[a_{n}, b_{n}\right]\). In particular, \(a_{n} \leq b_{n}\) for all \(n \in \mathbb{N}\). Given that \(I_{n+1} \subset I_{n}\), we have \(a_{n} \leq a_{n+1}\) and \(b_{n+1} \leq b_{n}\) for all \(n \in \mathbb{N}\). This shows that \(\left\{a_{n}\right\}\) is an increasing sequence bounded above by \(b_{1}\) and \(\left\{b_{n}\right\}\) is a decreasing sequence bounded below by \(a_{1}\). By the Monotone Convergence Theorem (Theorem 2.3.1), there exist \(a, b \in \mathbb{R}\) such that \(\lim _{n \rightarrow \infty} a_{n}=a\) and \(\lim _{n \rightarrow \infty} b_{n}=b\). Since \(a_{n} \leq b_{n}\) for all \(n\), by Theorem 2.1.5, we get \(a \leq b\). Now, we also have \(a_{n} \leq a\) and \(b \leq b_{n}\) for all \(n \in \mathbb{N}\) (since \(\left\{a_{n}\right\}\) is increasing and \(\left\{b_{n}\right\}\) is decreasing). This shows that if \(a \leq x \leq b\), then \(x \in I_{n}\) for all \(n \in \mathbb{N}\). Thus, \([a, b] \subset \bigcap_{n=1}^{\infty} I_{n}\). It follows that \(\bigcap_{n=1}^{\infty} I_{n} \neq \emptyset\). This proves part (a).

Now note also that \(\bigcap_{n=1}^{\infty} I_{n} \subset[a, b]\). Indeed, if \(x \in \bigcap_{n=1}^{\infty} I_{n}\), then \(x \in I_{n}\) for all \(n\). Therefore, \(a_{n} \leq x \leq b_{n}\) for all \(n\). Using Theorem 2.1.5, we conclude \(a \leq x \leq b\). Thus, \(x \in[a, b]\). This proves the desired inclusion and, hence, \(\bigcap_{n=1}^{\infty} I_{n}=[a, b]\).

We now prove part (b). Suppose the lengths of the intervals \(I_{n}\) converge to zero. This means \(b_{n}-a_{n} \rightarrow 0\) as \(n \rightarrow \infty\). Then \(b=\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty}\left[\left(b_{n}-a_{n}\right)+a_{n}\right]=a\). It follows that \(\bigcap_{n=1}^{\infty} I_{n}=\{a\}\) as desired. \(\square\)

When a monotone sequence is not bounded, it does not converge. However, the behavior follows a clear pattern. To make this precise we provide the following definition.

Definition \(\PageIndex{2}\)

A sequence \(\left\{a_{n}\right\}\) is said to diverge to \(\infty\) if for every \(M \in \mathbb{R}\), there exists \(N \in \mathbb{N}\) such that

\[a_{n}>M \text { for all } n \geq N.\]

In this case, we write \(\lim _{n \rightarrow \infty} a_{n}=\infty\). Similarly, we say that \(\left\{a_{n}\right\}\) diverges to \(-\infty\) and write \(\lim _{n \rightarrow \infty} a_{n}=-\infty\) if for every \(M \in \mathbb{R}\), there exists \(N \in \mathbb{N}\) such that

\[a_{n}<M \text { for all } n \geq N.\]

Remark \(\PageIndex{4}\)

We should not confuse a sequence that diverges to \(\infty\) (that is, one that satisfies the previous definition), with a divergent sequence (that is, one that does not converge).

Example \(\PageIndex{4}\)

Consider the sequence \(a_{n}=\frac{n^{2}+1}{5 n}\). We will show, using Definition 2.3.2, that \(\lim _{n \rightarrow \infty} a_{n}=\infty\).

Solution

Let \(M \in \mathbb{R}\). Note that

\[\frac{n^{2}+1}{5 n}=\frac{n}{5}+\frac{1}{5 n} \geq \frac{n}{5}.\]

Choose \(N>5M\). Then, if \(n \geq N\), we have

\[a_{n} \geq \frac{n}{5} \geq \frac{N}{5}>M.\]

The following result completes the description of the behavior of monotone sequences.

Theorem \(\PageIndex{5}\)

If a sequence \(\left\{a_{n}\right\}\) is increasing and not bounded above, then

\[\lim _{n \rightarrow \infty} a_{n}=\infty.\]

Similarly, if \(\left\{a_{n}\right\}\) is decreasing and not bounded below, then

\[\lim _{n \rightarrow \infty} a_{n}=-\infty\]

Proof

Fix any real number \(M\). Since \(\left\{a_{n}\right\}\) is not bounded above, there exists \(N \in \mathbb{N}\) such that \(a_{N} \geq M\). Then

\[a_{n} \geq a_{N} \geq M \text { for all } n \geq N\]

because \(\left\{a_{n}\right\}\) is increasing. Therefore, \(\lim _{n \rightarrow \infty} a_{n}=\infty\). The proof for the second case is similar. \(\square\)

Theorem \(\PageIndex{6}\)

Let \(\left\{a_{n}\right\}\), \(\left\{b_{n}\right\}\), and \(\left\{c_{n}\right\}\) be sequences of real numbers and let \(k\) be a constant. Suppose

\[\lim _{n \rightarrow \infty} a_{n}=\infty, \lim _{n \rightarrow \infty} b_{n}=\infty, \text { and } \lim _{n \rightarrow \infty} c_{n}=-\infty\]

Then

1. \(\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\infty\);
2. \(\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=\infty\);
3. \(\lim _{n \rightarrow \infty}\left(a_{n} c_{n}\right)=-\infty\);
4. \(\lim _{n \rightarrow \infty} k a_{n}=\infty\) if \(k>0\), and \(\lim _{n \rightarrow \infty} k a_{n}=-\infty\) if \(k<0\);
5. \(\lim _{n \rightarrow \infty} \frac{1}{a_{n}}=0\). (Here we assume \(a_{n} \neq 0\) for all \(n\).)

Proof

We provide proofs for (a) and (e) and leave the others as exercises.

(a) Fix any \(M \in \mathbb{R}\). Since \(\lim _{n \rightarrow \infty} a_{n}=\infty\), there exists \(N_{1} \in \mathbb{N}\) such that

\[a_{n} \geq \frac{M}{2} \text { for all } n \geq N_{1}.\]

Similarly, there exists \(N_{2} \in \mathbb{N}\) such that

\[b_{n} \geq \frac{M}{2} \text { for all } n \geq N_{1}\]

Let \(N=\max \left\{N_{1}, N_{2}\right\}\). Then it is clear that

\[a_{n}+b_{n} \geq M \text { for all } n \geq N.\]

This implies (a).

(e) For any \(\varepsilon>0\), let \(M=\frac{1}{\varepsilon}\). Since \(\lim _{n \rightarrow \infty} a_{n}=\infty\), there exists \(N \in \mathbb{N}\) such that

\[a_{n}>\frac{1}{\varepsilon} \text { for all } n \geq N\]

This implies that for \(n \geq N\),

\[\left|\frac{1}{a_{n}}-0\right|=\frac{1}{a_{n}}<\varepsilon.\]

Thus, (e) holds. \(\square\)

The proof of the comparison theorem below follows directly from Definition 2.3.2 (see also Theorem 2.1.5).

Theorem \(\PageIndex{7}\)

Suppose \(a_{n} \leq b_{n}\) for all \(n \in \mathbb{N}\).

1. If \(\lim _{n \rightarrow \infty} a_{n}=\infty\), then \(\lim _{n \rightarrow \infty} b_{n}=\infty\).
2. If \(\lim _{n \rightarrow \infty} b_{n}=-\infty\), then \(\lim _{n \rightarrow \infty} a_{n}=-\infty\).

Proof

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Exercise \(\PageIndex{1}\)

Let \(a_{1}=\sqrt{2}\). Define

\[a_{n+1}=\sqrt{a_{n}+2} \text { for } n \geq 1\]

1. Prove that \(a_{n}<2\) for all \(n \in \mathbb{N}\).
2. Prove that \(\left\{a_{n}\right\}\) is an increasing sequence.
3. Prove that \(\lim _{n \rightarrow \infty} a_{n}=2\).

Answer

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Exercise \(\PageIndex{2}\)

Prove that each of the following sequences is convergent and find its limit.

1. \(a_{1}=1\) and \(a_{n+1}=\frac{a_{n}+3}{2}\) for \(n \geq 1\).
2. \(a_{1}=\sqrt{6}\) and \(a_{n+1}=\sqrt{a_{n}+6}\) for \(n \geq 1\).
3. \(a_{n+1}=\frac{1}{3}\left(2 a_{n}+\frac{1}{a_{n}^{2}}\right), n \geq 1, a_{1}>0\).
4. \(a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{b}{a_{n}}\right), b>0\).

Answer

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Exercise \(\PageIndex{3}\)

Prove that each of the following sequences is convergent and find its limit.

1. \(\sqrt{2} ; \sqrt{2 \sqrt{2}} ; \sqrt{2 \sqrt{2 \sqrt{2}}} ; \cdots\)
2. \(1 / 2 ; \frac{1}{2+1 / 2} ; \frac{1}{2+\frac{1}{2+1 / 2}} ; \cdots\)

Answer

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Exercise \(\PageIndex{4}\)

Prove that the following sequence is convergent:

\(a_{n}=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !}, n \in \mathbb{N}\).

Answer

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Exercise \(\PageIndex{5}\)

Let \(a\) and \(b\) be two positive real numbers with \(a<b\). Define \(a_{1}=a, b_{1}=b\), and

\[a_{n+1}=\sqrt{a_{n} b_{n}} \text { and } b_{n+1}=\frac{a_{n}+b_{n}}{2} \text { for } n \geq 1.\]

Show that \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are convergent to the same limit.

Answer

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Exercise \(\PageIndex{6}\)

Prove the following using Definition 2.3.2.

1. \(\lim _{n \rightarrow \infty} \frac{2 n^{2}+n+1}{n-2}=\infty\).
2. \(\lim _{n \rightarrow \infty} \frac{1-3 n^{2}}{n+2}=-\infty\).

Answer

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Exercise \(\PageIndex{7}\)

Prove parts (b), (c), and (d) of Theorem 2.3.6.

Answer

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Exercise \(\PageIndex{8}\)

Prove Theorem 2.3.7.

Answer

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