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.