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**Example 23**.: Consider the sequence \(\left\{\frac{\sin n}{n}\right\}\). Since \(|\sin n|\leq 1\), we know
\[-\frac{1}{n}\leq\frac{\sin n}{n}\leq\frac{1}{n}.\]
Both the right- and left-hand sequences converge to zero as \(n\to\infty\); the sequence in the middle is “squeezed” to the same limit.
**Remark 10**.: The squeeze principle can be used to conclude more convergence properties for sequences. For example, if \(x_{n}\to L\), then \(|x_{n}|\to|L|\) too. Notice that \(x_{n}\to L\) implies \(x_{n}-L\to 0\). The _reverse triangle inequality_ implies
\[0\leq||x_{n}|-|L||\leq|x_{n}-L|.\]
The right-hand sequence tends to zero so that by the squeeze principle \(|x_{n}|\to|L|\). Similarly if \(|x_{n}|\to 0\) and \(\{y_{n}\}\) is another _bounded_ sequence, then \(x_{n}y_{n}\to 0\) as well. This follows from the inequality
\[0\leq|x_{n}y_{n}|\leq M|x_{n}|,\]
where the boundedness of \(\{y_{n}\}\) implies we can choose \(M\) so that \(|y_{n}|\leq M\) for all \(n\), and \(|x_{n}y_{n}|\to 0\) if and only if \(x_{n}y_{n}\to 0\).
We showed in Theorem 10, part b) that convergent sequences are bounded. The converse of this statement is not true. However, if a bounded sequence is monotone, then in fact it converges. Recall that a sequence is called _monotonic_ if it is either nondecreasing or nonincreasing; that is,
\[x_{1}\leq x_{2}\leq x_{3}\leq\ldots\qquad\text{or}\qquad x_{1}\geq x_{2}\geq x_ {3}\geq\ldots.\]
The following Monotone Convergence Theorem is very useful because it simply asserts the convergence of a sequence without explicitly finding its actual limit.
**Theorem 13** (Monotone Convergence Theorem).: _Every monotonic and bounded sequence of real numbers is convergent._
Proof
: Let \(\{x_{n}\}\) be monotone and bounded. Suppose \(\{x_{n}\}\) is nondecreasing (the nonincreasing case is handled similarly). Then by hypothesis its range \(\{x_{n}:\quad n\in\mathbb{N}\}\) is bounded, so we can set \(\lambda=\sup\{x_{n}:\ n\in\mathbb{N}\}\). We need a candidate for the limit, and it is reasonable to claim \(\lim\limits_{n\to\infty}x_{n}=\lambda\). To prove this, let \(\epsilon>0\). Since \(\lambda\) is an upper bound for \(\{x_{n}:\ n\in\mathbb{N}\}\), we have \(x_{n}\leq\lambda\) for all \(n\in\mathbb{N}\), but \(\lambda-\epsilon\) is not an upper bound for each \(\epsilon>0\), i.e., there is a point in the sequence \(x_{N}\) such that \(x_{N}>\lambda-\epsilon\) (Figure 17).
Since the sequence \(\{x_{n}\}\) is nondecreasing, it follows that \(x_{n}\geq\lambda-\epsilon\) for all \(n\geq N\). Combining this with the inequality \(x_{n}\leq\lambda\),
\[\lambda-\epsilon implies \(|x_{n}-\lambda|<\epsilon\) for all \(n\in\mathbb{N}\), which gives \(x_{n}\to\lambda\) as desired. Later we will see the Monotone Convergence Theorem will be a great help for the study of infinite series. In the coming sections we will consider certain properties of the real line that may be defined either in terms of subsets of \(\mathbb{R}\) or in terms of sequences. The Nested Interval Theorem is useful in this regard. In our discussion of sequences so far we considered sequences of real numbers. However, one can have sequences in \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\) or in general in the \(n\)-dimensional vector space \(\mathbb{R}^{n}\) as well. In the following, we consider sequences in \(\mathbb{R}^{n}\) and show the connection of sequences in \(\mathbb{R}\) and \(\mathbb{R}^{n}\). **Definition 12**.: **Euclidean \(n\)-space**, denoted \(\mathbb{R}^{n}\), consists of all ordered \(n\)-tuples of real numbers. \[\mathbb{R}^{n}=\{(x_{1},\ldots,x_{n}):\quad x_{1},\ldots,x_{n}\in\mathbb{R}\}.\] Figure 17: Monotone and bounded sequence Clearly \(\mathbb{R}^{n}=\mathbb{R}\times\cdots\times\mathbb{R}\) (\(n\)-times) is the Cartesian product of \(\mathbb{R}\) with itself \(n\)-times. Elements of \(\mathbb{R}^{n}\) are usually denoted by single letters that stand for \(n\)-tuples such as \(x=(x_{1},\ldots,x_{n})\), and we speak of \(x\) as a **point** in \(\mathbb{R}^{n}\). Addition and scalar multiplication of \(n\)-tuples are defined as \[(x_{1},\ldots,x_{n})+(y_{1},\ldots,y_{n})=(x_{1}+y_{1},\ldots,x_{n}+y_{n})\] and \[a(x_{1},\ldots,x_{n})=(ax_{1},\ldots,ax_{n})\quad\text{for}\quad a\in\mathbb{R}.\] The **distance** or the **norm** between two elements of \(x,y\in\mathbb{R}^{n}\) is a real number \[||x-y||=\left(\sum_{i=1}^{n}(x_{i}-y_{i})^{2}\right)^{\frac{1}{2}}.\] **Proposition 1**.: _If \(x=(x_{1},\ldots,x_{n})\) and \(y=(y_{1},\ldots,y_{n})\) are vectors in \(\mathbb{R}^{n}\) and we let \(\rho(x,y)=\max\{|x_{1}-y_{1}|,|x_{2}-y_{2}|,\ldots,|x_{n}-y_{n}|\}\), then_ \[\rho(x,y)\leq||x-y||\leq\sqrt{n}\rho(x,y).\] : Since \[|x_{i}-y_{i}|=\sqrt{|x_{i}-y_{i}|^{2}}\leq\sqrt{\sum_{i=1}^{n}|x_{i}-y_{i}|^{2 }}=||x-y||,\] we have \(\rho(x,y)\leq||x-y||\), and \[||x-y||=\sqrt{\sum_{i=1}^{n}|x_{i}-y_{i}|^{2}}\leq\sqrt{\sum_{i=1}^{n}\max_{i }|x_{i}-y_{i}|^{2}}\leq\sqrt{n\rho(x,y)^{2}}=\sqrt{n}\rho(x,y)\] gives the right-hand inequality. We will use the above proposition to prove the following theorem. **Theorem 14** (Convergence in \(\mathbb{R}^{n}\)).: _Suppose \(\{x_{k}\}\) is a sequence in \(\mathbb{R}^{n}\) and \(x_{k}=(x_{k}^{1},x_{k}^{2},\ldots,x_{k}^{n})\) for \(k\geq 1\). Then \(\{x_{k}\}\) converges to \(x\) in \(\mathbb{R}^{n}\) if and only if each sequence of coordinates converges to the corresponding coordinate of \(x\) as a sequence in \(\mathbb{R}\). That is,_ \[x_{k}\to x\quad\text{in}\ \,\mathbb{R}^{n}\quad\text{if and only if}\quad\lim_{k\to\infty}x_{k}^{i}\to x^{i}\quad\text{in}\ \,\mathbb{R}\ \text{ for each}\ i=1,2,\ldots,n.\] Note that this theorem can be written compactly as : Let \(\rho(x,y)=\max\{|x_{1}-y_{1}|,|x_{2}-y_{2}|,\ldots,|x_{n}-y_{n}|\}\), then \[\rho(x,y)\leq||x-y||\leq\sqrt{n}\rho(x,y)\] by the above proposition. Suppose \(\{x_{k}\}\) converges to \(x\) in \(\mathbb{R}^{n}\) and \(1\leq j\leq n\). Let \(\epsilon>0\) be given; then there is an \(N\) such that \(||x_{k}-x||<\epsilon\) whenever \(k\geq N\). Therefore, for such \(k\) we have \[|x^{j}-x_{k}^{j}|\leq\rho(x,x_{k})\leq||x-x_{k}||<\epsilon,\] and thus \(x_{k}^{j}\to x^{j}\) in \(\mathbb{R}\). To prove the converse suppose that for each \(j=1,2,\ldots,n\), we have \(\lim\limits_{k\to\infty}x_{k}^{j}\to x^{j}\) in \(\mathbb{R}\). Let \(\epsilon>0\). For any \(\epsilon_{*}>0\), there exist integers \(N_{1},N_{2},\ldots,N_{n}\) such that \[|x^{1}-x_{k}^{1}|<\epsilon_{*}\quad\text{whenever}\quad k\geq N_{1}\] \[|x^{2}-x_{k}^{2}|<\epsilon_{*}\quad\text{whenever}\quad k\geq N_{2}\] \[\vdots\qquad\qquad\qquad\vdots\] \[|x^{n}-x_{k}^{n}|<\epsilon_{*}\quad\text{whenever}\quad k\geq N_{n}.\] Now set \(N=\max\{N_{1},N_{2},\ldots,N_{n}\}\). Then for \(k\geq N\), \[||x-x_{k}||\leq\sqrt{n}\rho(x,x_{k})\leq\sqrt{n}\epsilon_{*}.\] Thus by taking \(\epsilon_{*}=\dfrac{\epsilon}{\sqrt{n}}\), we obtain
Proof
Proof
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