5.2: CHAPTER 2

Exercise \(2.1.12\)

Answer

1. Suppose that \(\lim _{n \rightarrow \infty} a_{n}=\ell\). Then by Theorem 2.1.9, \[\lim _{n \rightarrow \infty} a_{2 n}=\ell \text { and } \lim _{n \rightarrow \infty} a_{2 n+1}=\ell .\] Now suppose that (5.2) is satisfied. Fix any \(\varepsilon > 0\). Choose \(N_{1} \in \mathbb{N}\) such that \[\left|a_{2 n}-\ell\right|<\varepsilon \text { whenever } n \geq N_{1} ,\] and choose \(N_{2} \in \mathbb{N}\) such that \[\left|a_{2 n+1}-\ell\right|<\varepsilon \text { whenever } n \geq N_{2} .\] Let \(N=\max \left\{2 N_{1}, 2 N_{2}+1\right\}\). Then \[\left|a_{n}-\ell\right|<\varepsilon \text { whenever } n \geq N .\] Therefore, \(\lim _{n \rightarrow \infty} a_{n}=\ell\).

This problem is sometimes very helpful to show that a limit exists. For example, consider the sequence defined by

\[x_{1}=1 / 2 ,\] \[x_{n+1}=\frac{1}{2+x_{n}} \text { for } n \in \mathbb{N} . \nonumber\]

We will see later that \(\left\{x_{2 n+1}\right\}\) and \(\left\{x_{2 n}\right\}\) both converge to \(\{x_n\}\), so we can conclude that \(\left\{x_{n}\right\}\) converges to \(\left\{x_{n}\right\}\).

2. Use a similar method to the solution of part (a).

Exercise \(2.1.8\)

Answer

Consider the case where \(\ell > 0\). By the definition of limit, we can find \(n_{1} \in \mathbb{N}\) such that \[\left|a_{n}\right|>\ell / 2 \text { for all } n \geq n_{1} .\] Given any \(\varepsilon > 0\), we can find \(n_{2} \in \mathbb{N}\) such that \[\left|a_{n}-\ell\right|<\frac{\ell \varepsilon}{4} \text { for all } n \geq n_{2} .\] Choose \(n_{0}=\max \left\{n_{1}, n_{2}\right\}\). For any \(n \geq n_{0}\), one has \[\left|\frac{a_{n+1}}{a_{n}}-1\right|=\frac{\left|a_{n}-a_{n+1}\right|}{\left|a_{n}\right|} \leq \frac{\left|a_{n}-\ell\right|+\left|a_{n+1}-\ell\right|}{\left|a_{n}\right|}<\frac{\frac{\ell \varepsilon}{4}+\frac{\ell \varepsilon}{4}}{\frac{\ell}{2}}=\varepsilon .\] Therefore, \(\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=1\). If \(\ell < 0\), consider the sequence \(\left\{-a_{n}\right\}\).

The conclusion is no longer true if \(\ell = 0\). A counterexample is \(a_{n}=\lambda^{n}\) where \(\lambda \in (0, 1)\).

Exercise \(2.2.3\)

Answer

1. The limit is calculated as follows: \[\begin{aligned}
   \lim _{n \rightarrow \infty}\left(\sqrt{n^{2}+n}-n\right) &=\lim _{n \rightarrow \infty} \frac{\left(\sqrt{n^{2}+n}-n\right)\left(\sqrt{n^{2}+n}+n\right)}{\sqrt{n^{2}+n}+n} \\
   &=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}+n} \\
   &=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}(1+1 / n)}+n} \\
   &=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+1 / n}+1}=1 / 2 \text {.}
   \end{aligned}\]
2. The limit is calculated as follows: \[\begin{aligned}
   \lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-n\right) &=\lim _{n \rightarrow \infty} \frac{\left(\sqrt[3]{n^{3}+3 n^{2}}-n\right)\left(\sqrt[3]{\left(n^{3}+3 n^{2}\right)^{2}}+n \sqrt[3]{n^{3}+3 n^{2}}+n^{2}\right)}{\left.\sqrt[3]{\left(n^{3}+3 n^{2}\right)^{2}}+n \sqrt[3]{n^{3}+3 n^{2}}+n^{2}\right)} \\
   &=\lim _{n \rightarrow \infty} \frac{3 n^{2}}{\sqrt[3]{\left(n^{3}+3 n^{2}\right)^{2}}+n \sqrt[3]{n^{3}+3 n^{2}}+n^{2}} \\
   &=\lim _{n \rightarrow \infty} \frac{3 n^{2}}{\sqrt[3]{n^{6}(1+3 / n)^{2}}+n \sqrt[3]{n^{3}(1+3 / n)}+n^{2}} \\
   &=\lim _{n \rightarrow \infty} \frac{3 n^{2}}{n^{2}\left(\sqrt[3]{(1+3 / n)^{2}}+\sqrt[3]{(1+3 / n)}+1\right)} \\
   &=\lim _{n \rightarrow \infty} \frac{3}{\left(\sqrt[3]{(1+3 / n)^{2}}+\sqrt[3]{(1+3 / n)}+1\right)}=1 \text {.}
   \end{aligned}\]
3. We use the result in part (a) and part (b) to obtain \[\begin{aligned}
   \lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-\sqrt{n^{2}+1}\right) &=\lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-n+n-\sqrt{n^{2}+1}\right) \\
   &=\lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-n\right)+\lim _{n \rightarrow \infty}\left(n-\sqrt{n^{2}+1}\right)=1-1 / 2=1 / 2 \text {.}
   \end{aligned}\] Using a similar technique, we can find the following limit: \[\lim _{n \rightarrow \infty}\left(\sqrt[3]{a n^{3}+b n^{2}+c n+d}-\sqrt{\alpha n^{2}+\beta n+\gamma}\right) ,\] where \(a > 0\) and \(\alpha > 0\).

Exercise \(2.3.1\)

Answer

1. Clearly, \(a_{1} < 2\). Suppose that \(a_{k}<2\) for \(k \in \mathbb{N}\). Then \[a_{k+1}=\sqrt{2+a_{k}}<\sqrt{2+2}=2 .\] By induction, \(a_{n}<2\) for all \(n \in \mathbb{N}\).
2. Clearly, \(a_{1}=\sqrt{2}<\sqrt{2+\sqrt{2}}=a_{2}\). Suppose that \(a_{k}<a_{k+1}\) for \(k \in \mathbb{N}\). Then \[a_{k}+2<a_{k+1}+2 ,\] which implies \[\sqrt{a_{k}+2}<\sqrt{a_{k+1}+2} .\] Thus, \(a_{k+1}<a_{k+2}\). By induction, \(a_{n}<a_{n+1}\) for all \(n \in \mathbb{N}\). Therefore, \(\left\{a_{n}\right\}\) is an increasing sequence.
3. By the monotone convergence theorem, \(\lim _{n \rightarrow \infty} a_{n}\) exists. Let \(\ell=\lim _{n \rightarrow \infty} a_{n}\). Since \(a_{n+1}= \sqrt{2+a_{n}}\) and \(\lim _{n \rightarrow \infty} a_{n+1}=\ell\), we have \[\ell=\sqrt{2+\ell} \text { or } \ell^{2}=2+\ell .\] Solving this quadratic equation yields \(\ell = -1\) or \(\ell = 2\). Therefore, \(\lim _{n \rightarrow \infty} a_{n}=2\). Define a more general sequence as follows: \[\begin{aligned}
   a_{1} &=c>0 \text {,}\\
   a_{n+1} &=\sqrt{c+a_{n}} \quad \text { for } n \in \mathbb{N} \text {.}
   \end{aligned}\] We can prove that \(\left\{a_{n}\right\}\) is monotone increasing and bounded above by \(\frac{1+\sqrt{1+4 c}}{2}\). In fact, \(\left\{a_{n}\right\}\) converges to this limit. The number \(\frac{1+\sqrt{1+4 c}}{2}\) is obtained by solving the equation \(\ell=\sqrt{c+\ell}\), where \(\ell > 0\).

Exercise \(2.3.2\)

Answer

1. The limit is \(3\).
2. The limit is \(3\).
3. We use the well-known inequality \[\frac{a+b+c}{3} \geq \sqrt[3]{a b c} \text { for } a, b, c \geq 0 .\] By induction, we see that \(a_{n}>0\) for all \(n \in \mathbb{N}\). Moreover, \[a_{n+1}=\frac{1}{3}\left(2 a_{n}+\frac{1}{a_{n}^{2}}\right)=\frac{1}{3}\left(a_{n}+a_{n}+\frac{1}{a_{n}^{2}}\right) \geq \frac{1}{3} \sqrt[3]{a_{n} \cdot a_{n} \cdot \frac{1}{a_{n}^{2}}}=1 .\] We also have, for \(n \geq 2\), \[a_{n+1}-a_{n}=\frac{1}{3}\left(2 a_{n}+\frac{1}{a_{n}^{2}}\right)-a_{n}=\frac{-a_{n}^{3}+1}{3 a_{n}^{2}}=\frac{-\left(a_{n}-1\right)\left(a_{n}^{2}+a_{n}+1\right)}{3 a_{n}^{2}}<0 .\] Thus, \(\left\{a_{n}\right\}\) is monotone deceasing (for \(n \geq 2\)) and bounded below. We can show that \(\lim _{n \rightarrow \infty} a_{n}=1\).
4. Use the inequality \(\frac{a+b}{2} \geq \sqrt{a b}\) for \(a, b \geq 0\) to show that \(a_{n+1} \geq \sqrt{b}\) for all \(n \in \mathbb{N}\). Then follow part 3 to show that \(\left\{a_{n}\right\}\) is monotone decreasing. The limit is \(\sqrt{b}\).

Exercise \(2.3.3\)

Answer

1. Let \(\left\{a_{n}\right\}\) be the given sequence. Observe that \(a_{n+1}=\sqrt{2 a_{n}}\). Then show that \(\left\{a_{n}\right\}\) is monotone increasing and bounded above. The limit is \(2\).
2. Let \(\left\{a_{n}\right\}\) be the given sequence. Then \[a_{n+1}=\frac{1}{2+a_{n}} .\] Show that \(\left\{a_{2 n+1}\right\}\) is monotone decreasing and bounded below; \(\left\{a_{2 n}\right\}\) is monotone increasing and bounded above. Thus \(\left\{a_{n}\right\}\) converges by Exercise 2.1.12. The limit is \(\sqrt{2} - 1\).

Exercise \(2.3.5\)

Answer

Observe that \[b_{n+1}=\frac{a_{n}+b_{n}}{2} \geq \sqrt{a_{n} b_{n}}=a_{n+1} \text { for all } n \in \mathbb{N} .\] Thus, \[a_{n+1}=\sqrt{a_{n} b_{n}} \geq \sqrt{a_{n} a_{n}}=a_{n} \text { for all } n \in \mathbb{N} ,\] \[b_{n+1}=\frac{a_{n}+b_{n}}{2} \leq \frac{b_{n}+b_{n}}{2}=b_{n} \text { for all } n \in \mathbb{N} .\] It follows that \(\left\{a_{n}\right\}\) is monotone increasing and bounded above by \(b_{1}\), and \(\left\{b_{n}\right\}\) is decreasing and bounded below by \(a_{1}\). Let \(x=\lim _{n \rightarrow \infty} a_{n}\) and \(y=\lim _{n \rightarrow \infty} b_{n}\). Then \[x=\sqrt{x y} \text { and } y=\frac{x+y}{2} .\] Therefore, \(x = y\).

Exercise \(2.4.1\).

Answer

Here we use the fact that in \(\mathbb{R}\) a sequence is a Cauchy sequence if and only if it is convergent.

1. Not a Cauchy sequence. See Example 2.1.7.
2. A Cauchy sequence. This sequence converges to \(0\).
3. A Cauchy sequence. This sequence converges to \(1\).
4. A Cauchy sequence. This sequence converges to \(0\) (see Exercise 2.1.5).

Exercise \(2.5.4\)

Answer

1. Define \[\alpha_{n}=\sup _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\sup _{k \geq n} a_{k}, \gamma_{n}=\sup _{k \geq n} b_{k} .\] By the definition, \[\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=\lim _{n \rightarrow \infty} \alpha_{n}, \limsup _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \beta_{n}, \limsup _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty} \gamma_{n} .\] By Exercise 2.5.3, \[\alpha_{n} \leq \beta_{n}+\gamma_{n} \text { for all } n \in \mathbb{N} .\] This implies \[\lim _{n \rightarrow \infty} \alpha_{n} \leq \lim _{n \rightarrow \infty} \beta_{n}+\lim _{n \rightarrow \infty} \gamma_{n} \text { for all } n \in \mathbb{N} .\] Therefore, \[\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n} .\] This conclusion remains valid for unbounded sequences provided that the right-hand side is well-defined. Note that the right-hand side is not well-defined, for example, when \(\limsup _{n \rightarrow \infty} a_{n}=\infty\) and \(\limsup _{n \rightarrow \infty} b_{n}=-\infty\).
2. Define \[\alpha_{n}=\inf _{k \geq n}\left(a_{n}+b_{n}\right), \beta_{n}=\inf _{k \geq n} a_{k}, \gamma_{n}=\inf _{k \geq n} b_{k} .\] Proceed as in part (a), but use part (b) of Exercise 2.5.3.
3. Consider \(a_{n}=(-1)^{n}\) and \(b_{n}=(-1)^{n+1}\).

Exercise \(2.6.3\)

Answer

Suppose \(A\) and \(B\) are compact subsets of \(\mathbb{R}\). Then, by Theorem 2.6.5, \(A\) and \(B\) are closed and bounded. From Theorem 2.6.2(c) we get that \(A \cup B\) is closed. Moreover, let \(M_{A}, m_{A}\), \(M_{B}, M_{B}\) be upper and lower bounds for \(A\) and \(B\), respectively. Then \(M=\max \left\{M_{A}, M_{B}\right\}\) and \(m=\min \left\{m_{A}, m_{B}\right\}\) are upper and lower bounds for \(A \cup B\). In particular, \(A \cup B\) is bounded. We have shown that \(A \cup B\) is both closed and bounded. It now follows from Theorem 2.6.5 that \(A \cup B\) is compact.