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## Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 3

#### Exercise 3.1

Solution: Let $\e=0.001$. Because $(a_n)$ converges to $0.001$, there exist $N$ such that for all $n> N$ we have$|a_n-0.001| < 0.001,$$\Longrightarrow -0.001 < a_n- 0.001 < 0.001,$$\Longrightarrow 0< a_n < 0.002.$Hence for all $n> N$, the values of $a_n$ are positive.

#### Exercise 3.2

Solution:

(a) For any $\e >0$. Let $N=\dfrac{1}{\e^2}$, then for any $n > N$, we have $\frac{1}{n}< \frac{1}{N}=\e^2\Longrightarrow \frac{1}{\sqrt{n}} < \e.$Hence $$|a_n-7|=\frac{1}{\sqrt{n}} < \e$$ for all $n >N$. By the definition of sequence convergence, the sequence $(a_n)$ converges to 7.

(b) For any $\e > 0$. Let $N=\dfrac{8}{9\e}$, then for any $n> N$, we have$\frac{8}{9n} < \frac{8}{9N}=\e.$Hence \begin{align*}\left|a_n-\frac{2}{3}\right|&=\left|\frac{2n-2}{3n+1}-\frac23\right|\\ &= \frac{|3(2n-2)-2(3n+1)|}{3(3n+1)}\\ &=\frac{8}{3(3n+1)} < \frac{8}{9n} <  \e\end{align*}for all $n > N$. Hence the sequence $(a_n)$ converges to $2/3$.

(c) For any $\e >0$. Let $N=\dfrac{1}{\e^2}$, then for any $n > N$, we have $\frac{1}{n}< \frac{1}{N}=\e^2\Longrightarrow \frac{1}{\sqrt{n}} < \e.$In particular, we also have$\frac{1}{\sqrt{n+\sqrt n+13}} < \frac{1}{\sqrt n} < \e.$Hence $$|a_n-7|=\frac{1}{\sqrt{n+\sqrt n+13}} < \e$$ for all $n >N$. By the definition of sequence convergence, the sequence $(a_n)$ converges to 7.

#### Exercise 3.3

Solution: An example is the sequence $(a_n)$ given by $a_n=-\dfrac{1}{n}$.

Clearly, all $a_n$ are negative. We show that $a_n\to 0$. For any $\e >0$, then for any $n>\dfrac{1}{\e}$, we have$|a_n|=\frac{1}{n} < \e.$Hence the sequence $(a_n)$ converges to zero.

#### Exercise 3.4

Solution:

(a) Let $a_n=n$ and $a=1$. Then for any $\e >0$, taking $n=1$, we have $|a_1-1|=0 < \e$.

(b) Let $a_1=1$, $a_2=2$, $a_n=3$ for all $n\geqslant 3$, and $a=2$. Then for any $\e >0$, taking $N=1$ and $n=2$, we have $|a_2-2|=0 < \e$.

(c) Let $a_{2n-1}=1$, $a_{2n}=2$, and $a=2$. For all $N\in\mathbb N$, taking $n=2N$, we have $n > N$ and $|a_{n}-a|=0 < \e$.

(d) Let $a_1=1$, $a_2=2$, $a_n=1$ for all $n\geqslant 3$, and $a=2$. Then for $\e =1/2$, taking $N=1$ and $n=2$, we have $|a_2-2|=0 < \e$. But it does not work for Nonverges-type-3, since for $N=2$, there is no $n > N$ such that $|a_n-a| < 1/2$ Since $|a_n-a|=1$.

#### Exercise 3.5

Solution: Because this set consists of all rational numbers in $(0,1)$. Therefore the only possibility for some L is inside of $[0,1]$.

For any $L\in [0,1]$, we can find a subsequence whose limit is equal to $L$. Let us prove it.

If $L=0$, the subsequence $\left(\dfrac{1}{n}\right)$ is convergent to zero.

If $L=1$, the subsequence $\left(\dfrac{n}{n+1}\right)$ is convergent to one.

If $L\in (0,1)$, then for any positive $n$ such that $n > \dfrac{1}{L}$ and $n > \dfrac{1}{1-L}$. We have$\frac{n-1}{n}> L > \frac{1}{n}.$Therefore we can find a positive integer $k_n\leqslant n-1$ such that $$\frac{k_n}{n}\leqslant L\leqslant \frac{k_n+1}{n}.$$Then the subsequence $\left(\dfrac{k_n}{n}\right)$ with $n$ starting from sufficiently large $n$ is convergent to $L$. This is because $$\left|\frac{k_n}{n}-x\right| \leqslant \frac{1}{n}.$$

Remark:  The statement is that the set of all rational numbers in $(0,1)$ is dense in $[0,1]$.

#### Exercise 3.6

Solution:

(a) Impossible. Suppose $(a_n)$ converges to $3.5$. By Definition 3.7, there exists $N$ such that $a_n$ is in the $0.5$-neighborhood of $3.5$ for all $n> N$. The $0.5$-neighborhood of $3.5$ is $(3,4)$ which contains no integers. But all $a_n$ are integers. Hence no $a_n$ is in the $0.5$-neighborhood of $3.5$. Thus we obtain a contradiction. Therefore $a_n$ does not converge to 3.5.

(b) Let $\e=1$. For any $n\ne m$, by assumption $|a_n-a_m|$ is a nonzero positive integer. Hence $|a_n-a_m|\geqslant 1=\e$ for all $n\ne m$. In particular, there does not exist $N$ such that $|a_n-a_m| < \e$ for all $n,m > N$. Hence $(a_n)$ is not Cauchy and hence does not converge by Theorem 3.42.

(c) Let $\e =1/2$, since $(a_n)$ converges, by Theorem 3.42 $(a_n)$ is Cauchy. Hence there exists $N$ such that $|a_n-a_m|<\e =1/2$ for all $n,m > N$. But $|a_n-a_m|$ is a nonnegative integer. Hence the only possibility is that $|a_n-a_m|=0$ for all $n,m > N$. This implies $a_n=a_m$ for all $n,m > N$. In other words, from some point on, all $a_n$ are the same.

#### Exercise 3.7

Solution:

(a) For any $\e>0$, since $(a_n)$ converges to $a$, there exists $N_1$ such that for all $n>N_1$, we have $$|a_n-a| < \e /2.$$Similarly, since $(b_n)$ converges to $b$, there exists $N_2$ such that for all $n>N_2$, we have $$|b_n-b| < \e /2.$$Let $N=\max\{N_1,N_2\}$. Then for all $n> N$, we have $|a_n-a| < \e /2,\quad |b_n-b| < \e /2.$By the triangle inequality, we have\begin{align*}|(a_n+b_n)-(a+b)|&=|(a_n-a)+(b_n-b)|\\ &\leqslant  |a_n-a|+|b_n-b|\\ & < \e /2 +\e/2 =\e. \end{align*}Hence $(a_n+b_n)$ converges to $a+b$.

(b) If $c=0$, it is clear that $ca_n=0$ for all $n$, hence $(c\cdot a_n)$ converges to $0=c\cdot a$.

If $c\ne 0$, since $(a_n)$ converges to $a$, there exists $N$ such that for all $n>N$, we have $$|a_n-a| < \e /|c|.$$Hence for all $n> N$ we have \begin{align*}|c\cdot a_n- c\cdot a|=|c||a_n-a|< |c|\cdot \e /|c|=\e.\end{align*}Hence $(c\cdot a_n)$ converges to $c\cdot a$.

#### Exercise 3.8

Solution:

(a) Since $(b_n)$ converges to $b$, by part (b) of Exercise 3.7, we have $(-b_n)$ converges to $-b$.

Note that $a_n-b_n= a_n+(-b_n)$, therefore it follows from part (a) of Exercise 3.7 that $(a_n-b_n)$ (the same as $(a_n+(-b_n))$) converges to $a+(-b)=a-b$.

(b) Due to part (b) of Exercise 3.7, we can assume $b > 0$. If $b < 0$, we can simply change $b_n$ to $-b_n$. We also assume that $a\ne 0$. The case of $a=0$ is similar and simpler.

Because $(b_n)$ converges to $b$, therefore there exists $N_1$ such that $|b_n-b| < b/2$ for all $n > N_1$. Hence we have $$\label{ex3-8-1} \frac{b}{ 2} < b_n < \frac{3b}{2}$$ for all $n > N_1$. There exists $N_2$ such that$$\label{ex3-8-2} |b_n-b|<\frac{b^2\e}{4|a|}$$ for all $n > N_1$.

Since $(a_n)$ converges to $a$, there exists $N_3$ such that $$\label{ex3-8-3}|a_n-a| < \dfrac{b \e}{4}$$ for all $n > N_3$.

Let $N=\max\{N_1,N_2,N_3\}$. Then for any $n > N$, we have \begin{align*}\left|\frac{a_n}{b_n}-\frac{a}{b}\right|&=\frac{|a_nb-ab_n|}{|b_nb|}=\frac{|(a_n-a)b-a(b_n-b)|}{|b_nb|}\\ &\leqslant \frac{|(a_n-a)b|}{|b_nb|}+\frac{|a(b_n-b)|}{|b_nb|}\leqslant \frac{|a_n-a|}{|b_n|}+\frac{|a|\left|\frac{b^2\e}{4a}\right|}{b^2/2}\\ & < \frac{b\e/4}{b/2}+\frac{\e}{2}=\frac{\e}{2}+\frac{\e}{2}= \e.\end{align*}Here, we used \eqref{ex3-8-1}, \eqref{ex3-8-2}, and \eqref{ex3-8-3}. Hence $(a_n/b_n)$ converges to $a/b$.

#### Exercise 3.9

Solution: Let $a_n=(-1)^n$ and $b_n=(-1)^{n+1}$. Then by Example 3.18, both $(a_n)$ and $(b_n)$ are divergent.

Note that $a_n+b_n=(-1)^n(1+(-1))=0$ for all $n$, therefore we have $(a_n+ b_n)$ converges.

#### Exercise 3.10

Solution: Let $a_n=b_n=(-1)^n$. Then by Example 3.18, both $(a_n)$ and $(b_n)$ are divergent.

Note that $a_n\cdot b_n=1$ for all $n$, we have $(a_n\cdot b_n)$ converges.

#### Exercise 3.11

Solution: Impossible. Suppose $(a_n+b_n)$ converges. Let $c_n=a_n+b_n$. Since $(a_n)$ is also convergent, by Theorem 3.21, we conclude that $(c_n-a_n)$ is convergent.

Note that $c_n-a_n=b_n$, we have $(b_n)$ is convergent. This is impossible because by assumption $(b_n)$ is divergent.

#### Exercise 3.12

Solution: Since $(a_n)$ is bounded, there exists $C>0$ such that $|a_n| < C$ for all $n$. In order to show that $(a_n\cdot b_n)\to 0$. We show that for any $\e > 0$, there exist $N$ such that for all $n> N$ we have $|a_nb_n-0| < \e$, namely $|a_nb_n| < \e$.

Since $(b_n)$ is convergent to $0$, there exist $N$ such that $|b_n| < \dfrac{\e}{C}$ for all $n> N$. Hence for $n> N$, we have$|a_n b_n|\leqslant C|b_n|\leqslant C \frac{\e}{C}=\e.$Hence $(a_nb_n)$ converges to zero.

#### Exercise 3.13

Solution: Consider the  seqeunce $(a_n)$ given by$a_{2n-1}=6+\frac{1}{n},\quad a_{2n}=7-\frac{1}{n}.$Clearly, we have $6 < a_n < 7$.

Taking the subsequence $(a_{2n})$, we have $(a_{2n})$ converges to 7.

Taking the subsequence $(a_{2n-1})$, we have $(a_{2n-1})$ converges to 6.

#### Exercise 3.14

Solution: By Theorem 3.37, the bounded subsequence has a convergent subsequence (of the bounded subsequence).

But a subsequence of a subsequence is again a subsequence. Hence we are done!

#### Exercise 3.15

Solution: This is exactly Lemma 3.41.

#### Exercise 3.16

Solution: Suppose $(a_n)$ is decreasing, then $(-a_n)$ is increasing. By Theorem 3.27 for the increasing case, the sequence $(-a_n)$ converges iff it is bounded. Note that by Exercise 3.7 (b), $(a_n)$ converges iff $(-a_n)$ converges. It is also clear that $(a_n)$ is bounded iff $(-a_n)$ is bounded. Hence we conclude that $(a_n)$ converges iff $(-a_n)$ converges iff $(-a_n)$ is bounded iff $(a_n)$ is bounded.

Recall that $(-a_n)$ is increasing, we have from Theorem 3.27 (a) that either $(-a_n)$ diverges to $\infty$ or $\lim_{n\to \infty} -a_n=\sup \{-a_n:n\in \mathbb N\}$.

If $\lim_{n\to \infty} -a_n=\sup \{-a_n:n\in \mathbb N\}$. By Exercise 1.21, $\sup \{-a_n:n\in \mathbb N\}=-\inf\{a_n:n\in\mathbb N\}.$We also have $\lim_{n\to \infty} -a_n=-\lim_{n\to \infty}a_n$ by Exercise 3.7 (b). Hence $$\lim_{n\to \infty}a_n=\inf\{a_n:n\in\mathbb N\}.$$If $(-a_n)$ diverges to $\infty$, then for any $M<0$, we have $-M >0$, hence there exists $N$ such that $-a_n > -M$ for all $n > N$. That is $a_n < M$ for all $n > N$. Thus by Definition 3.15, $(a_n)$ diverges to $-\infty$.

#### Exercise 3.17

Solution: Recall the definition of $-S$ fro Exercise 1.21. If $S$ is bounded below, it means there exists $M$ such that we have $a > M$ for all $a\in S$. Thus $-a < -M$ for all $a\in S$. Therefore $-S$ is bounded above with $-M$ as an upper bound.

Therefore, by Proposition 3.29 for the increasing case, there exists a sequence $(a_n)$ where $a_n\in -S$ for each $n$ and$\lim_{n\to \infty} a_n=\sup(-S).$But $a_n\in -S$ means $a_n=-b_n$ where $b_n\in S$ for each $n$. Therefore, by Exercise 3.7 (b), we have$\lim_{n\to \infty} b_n=-\lim_{n\to \infty} a_n.$ Recall from Exercise 1.21 that $\sup(-S)=-\inf (S)$. Therefore$\lim_{n\to \infty} b_n=-\lim_{n\to \infty} a_n=-\sup(-S)=\inf (S).$

#### Exercise 3.18

Solution: A typical example is defined as follows. We consider the sequence $(a_n)$ given by$a_{2n}=n,\quad a_{2n-1}=1.$Since the subsequence $(a_{2n})$ diverges to $\infty$, this sequence $(a_n)$ is bounded. But the other subsequence $(a_{2n-1})$ converges to $1$. Since they have different limits, the limit of $(a_n)$ does not exist.

#### Exercise 3.19

Remark, this exercise follows immediately from the Stolz–Cesàro theorem.

Solution: Since $(a_n)$ converges to $a$, for any $\e >0$, there exists a positive integer $k$ such that $$|a_n-a| < \frac{\e}{2}$$ for all $n> k$. We also set $$m=\max\{|a_1-a|,|a_2-a|,\dots,|a_{k}-a|\}.$$ Let $N=\max\left\{\dfrac{2km}{\e},k\right\}$. Then for any $n > N$, we have $\dfrac{km}{n}< \dfrac{\e}{2}$. Therefore
\begin{align*}|b_n-a|&=\left|\frac{a_1+\cdots+a_n}{n}-a\right|=\left|\frac{(a_1-a)+\cdots+(a_n-a)}{n}\right|\\ &\leqslant \frac{1}{n}\left(|a_1-a|+\cdots+|a_{k}-a|+|a_{k+1}-a|+\cdots+|a_n-a|\right)\\ & < \frac{1}{n}\left(km+(n-k)\frac{\epsilon}{2}\right)=\frac{km}{n}+\left(1-\frac{k}{n}\right)\frac{\e}{2}\\ & < \frac{\e}{2}+\frac{\e}{2}=\e.
\end{align*}Hence $(b_n)$ converges to $a$ as well.

#### Exercise 3.20

Solution: Recall from Exercise 1.14 that if $A\subset B$, then $\sup A\leqslant \sup B$. Since $\{a_{n+1},a_{n+2},\dots\}\subset \{a_n,a_{n+1},\dots\},$we have $b_{n+1} \leqslant b_n$. Therefore, the sequence $(b_n)$ is monotone. By assumption, the sequence $(a_n)$ is bounded. Hence there exist $\alpha$ and $\beta$ such that $\alpha \leqslant a_n \leqslant \beta$ for all $n$. In particular, we see that $\alpha \leqslant b_n\leqslant \beta$ as $\beta$ is an upper bound of the set $\{a_n,a_{n+1},\dots\}$. Hence the sequence $(b_n)$ is bounded.

Therefore, by Theorem 3.27, the sequence $(b_n)$ converges.

#### Exercise 3.21

Solution: Consider the sequence $(a_n)$ given by$a_{3n-2}=1+\frac{1}{n},~ a_{3n-1}=17+\frac{1}{n},~a_{3n}=-\pi+\frac{1}{n}.$Clearly, the subsequence $(a_{3n-2})$ converges to 1; the subsequence $(a_{3n-1})$ converges to 17; the subsequence $(a_{3n})$ converges to $-\pi$.

#### Exercise 3.22

Solution: By Proposition 3.32, it suffices to show that every subsequence $(a_{n_k})$ converges to the same limit.

Take any two subsequences $(a_{n_k})$ and $(a_{m_k})$. Suppose $(a_{n_k})$ converges to $\alpha$ and $(a_{m_k})$ converges to $\beta$. We show that $\alpha=\beta$. Consider the subsequence $a_{l_k}$ obtained from the union of $(a_{n_k})$ and $(a_{m_k})$. The sequence $(a_{l_k})$ converges by assumption. As subsequences of $(a_{l_k})$, it follows from Proposition 3.32 that both $(a_{n_k})$ and $(a_{m_k})$ converge to the same limit. Therefore $\alpha=\beta$.

#### Exercise 3.23

Solution: Let $(a_{n_k})$ be a subsequence of $(a_n)$. For any $M>0$, since $(a_n)$ diverges to $\infty$, there exists $N$ such that $a_k > M$ for all $k> N$.

Note that $1\leqslant n_1< n_2 < n_3< \cdots$, we have $k \leqslant n_k$. Therefore, for all $k> N$, we have $n_k\geqslant k > N$ as well. Hence $a_{n_k} > M$ for all $k> N$. This shows that $(a_{n_k})$ diverges to $\infty$.

#### Exercise 3.24

Solution:

(a) We show $0 < a_n \leqslant 2$ by induction on $n$. It is clear that $0 < a_1\leqslant \sqrt 2$. Suppose $0 < a_n\leqslant 2$, then$a_{n+1}=\sqrt{2+a_n} > \sqrt{2} > 0,$$a_{n+1}=\sqrt{2+a_n} \leqslant \sqrt{2+2}=2.$Therefore, $0 < a_n \leqslant 2$ for all $n$.

(b) We have\begin{align*}a_{n+1}-a_n &=\sqrt{2+a_n}-a_n=\frac{2+a_n-a_n^2}{\sqrt{2+a_n}+a_n}\\\text{use (a)}\quad &=\frac{(2-a_n)(1+a_n)}{\sqrt{2+a_n}+a_n}\geqslant 0.\end{align*}Hence $(a_n)$ is an monotone increasing sequence. By Theorem 3.27, the sequence $(a_n)$ is convergent.

(c) Let $x$ be the limit of $(a_n)$, then it is clear that $x > \sqrt{2}$. First we show that $\sqrt{2+a_n}\to \sqrt{2+x}$. For any $\e > 0$, there exists $N$ such that $|a_n-x| < 2\sqrt{2}\e$. Therefore for all $n> N$,\begin{align*}|\sqrt{2+a_n}-\sqrt{2+x}|&=\frac{|(2+a_n)-(2+x)|}{\sqrt{2+a_n}+\sqrt{2+x}}\\ <\frac{2\sqrt 2 \e}{\sqrt 2+\sqrt 2}=\e.\end{align*}. Thus $\lim \sqrt{2+a_n}=\sqrt{2+x}$. Then we have$\lim a_{n+1}=\lim \sqrt{2+a_n}\Longrightarrow x=\sqrt{2+x}.$Hence $x^2=2+x$, solving it we get that $x=2$ or $x=-1$. But $x > \sqrt{2}$, therefore we conclude that $x=2$, i.e. $(a_n)$ converges to $2$.

#### Exercise 3.25

Solution: Yes. Indeed we can find the formula of $a_n$ in the sequence, see here. Using this method, solve the corresponding Characteristic polynomial, we get solutions $x=1$ and $x=-1/2$. Therefore, we have $$a_n=A\left(-\frac{1}{2}\right)^n+B(1)^n=A\left(-\frac{1}{2}\right)^n+B.$$Here $A$ and $B$ depend on $a_1$ and $a_2$ (and you can solve them directly). Since $(-1/2)^n$ converges to zero, we conclude that $(a_n)$ converges to $B$.

#### Exercise 3.26

Solution: Let $a_n=n$. This sequence is monotone bu not Cauchy as it is not bounded, see Lemma 3.41.

#### Exercise 3.27

Solution: For any $\e > 0$. Since $(a_n)$ is Cauchy, there exists $N_1$ such that $|a_n-a_m| < \frac{\e}{2}$for all $m,n > N_1$. Similarly, since $(b_n)$ is Cauchy, there exists $N_2$ such that $|b_n-b_m| < \frac{\e}{2}$for all $m,n > N_2$.

Let $N=\max \{N_1,N_2\}$. Then for all $m, n > N$, we have$|a_n-a_m| < \frac{\e}{2},\quad |b_n-b_m| < \frac{\e}{2}.$Hence  by Corollary 1.14, \begin{align*}||a_{n}-b_n|-|a_m-b_m|| & \leqslant |(a_{n}-b_n)-(a_m-b_m)|\\ & =|(a_n-a_m)-(b_n-b_m)|\\ \text{triangle inequality}\quad & \leqslant |(a_n-a_m)|+|(b_n-b_m)| \\ & < \frac{\e}{2}+\frac{\e}{2}=\e.\end{align*}Hence $(|a_n-b_n|)$ is Cauchy.

#### Exercise 3.28

Solution: Note that $(a_n)$ is convergent, and hence Cauchy by Theorem 3.42. Therefore, for any $\e >0$, there exists $N$ such that for all $m> n >N$ we have $$\label{ex3-28-1}|a_m-a_n| < \e.$$Note that $|a_m-a_n|=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{m^2}.$Similarly, $|b_m-b_n|=\frac{1}{(n+1)^3}+\frac{1}{(n+2)^3}+\cdots+\frac{1}{m^3}.$Thus using \eqref{ex3-28-1}$|b_m-b_n| \leqslant |a_m-a_n| < \e$ for all $m> n >N$. It follows from that $(b_n)$ is Cauchy and hence converges by Theorem 3.42.