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## Solution to Understanding Analysis Exercise 2.2

##### Exercise 2.2.1

Example: Take the sequence (1,0,1,0,1,0,1,0,…….). To check this sequence verconges $0$, take e.g. $\varepsilon=2$. The example defines a divergent sequence. It is not hard to see the sequence in our example also verconges $1$ if we take $\varepsilon=2$. “A sequence verconges” means exactly “the sequence is bounded”.

##### Exercise 2.2.2

(a) Let $\varepsilon>0$ be arbitrary. We would like to take an $N\in\mathbf N$ such that for all $n>N$ we have$\left|\frac{2n+1}{5n+4}-\frac{2}{5}\right|<\varepsilon.$Note that\begin{align*}&~\frac{2n+1}{5n+4}-\frac{2}{5}\\ =&~\frac{5(2n+1)-2(5n+4)}{5(5n+4)}\\=&~\frac{-3}{5(5n+4)}.\end{align*}Hence$\left|\frac{2n+1}{5n+4}-\frac{2}{5}\right|=\left|\frac{-3}{5(5n+4)}\right|<\frac{1}{5n}.$We want $\dfrac{1}{5n}<\varepsilon$, i.e. $n>\dfrac{1}{5\varepsilon}$. Now we only need to take $N>\dfrac{1}{5\varepsilon}$.Then$\left|\frac{2n+1}{5n+4}-\frac{2}{5}\right|<\frac{1}{5n}<\frac{1}{5N}<\varepsilon.$Therefore $\lim\dfrac{2n+1}{5n+4}=\dfrac{2}{5}$.

(b) Let $\varepsilon>0$ be arbitrary. Take $N>\dfrac{2}{\varepsilon}$, then $\dfrac{2}{N}<\varepsilon$. For all $n>N$, we have\begin{align*}\left|\frac{2n^2}{n^3+3}-0\right|= & ~\frac{2n^2}{n^3+3}\\ < & ~\frac{2n^2}{n^3}=\frac{2}{n}<\frac{2}{N}<\varepsilon.\end{align*}Therefore $\lim\dfrac{2n^2}{n^3+3}=0$.

(c) Let $\varepsilon>0$ be arbitrary. Take $N>\dfrac{1}{\varepsilon^3}$, then $\dfrac{1}{\sqrt{N}} < \varepsilon$.For all $n>N$, we have\begin{align*}\left|\frac{\sin n^2}{\sqrt{n}}-0\right|= & ~\left|\frac{\sin n^2}{\sqrt{n}}\right|\\ \leqslant & ~\frac{1}{\sqrt{n}} < \dfrac{1}{\sqrt{N}} < \varepsilon.\end{align*}Therefore $\lim\dfrac{\sin n^2}{\sqrt{n}}=0$.

##### Exercise 2.2.3

(a) There is a college in the United states, there are no students who are at least seven feet tall. (Alternatively, there is a college in the United states, every student is less than seven feet tall. )

(b) There is a college in the United states, there are no professors who give every student a grade of either A or B.

(c) For all colleges in the United states, there exists a student who is less than six feet tall.

##### Exercise 2.2.4

(a) Consider the sequence (1,2,1,2,1,2,1,2,……).

(b) Impossible. Let $\lim a_n=a$. Take a positive $\varepsilon$. Since there are infinitely many ones, therefore for any $N\in\mathbf N$ we find $n>N$ such that $a_n=1$. Hence $$|a_n-a|<\varepsilon\Longrightarrow |1-a|<\varepsilon.$$This implies $|1-a|<\varepsilon$ is true for any positive $\varepsilon$. This can only happen if $a=1$ (see Theorem 1.2.6), thus completing the proof.

(c) Consider the following sequence. We take a single one then take a zero, then take two ones followed by a zero, then three ones followed by a zero, ……, then $n$ zeros followed by a zero, then continue. Because of the appearance of infinitely many zeros, this sequence cannot be convergent. (Why? If the sequence is convergent, then, by part (b), we have the limit should be both one and zero. However, the limit of a sequence is unique, see Theorem 2.2.7.)

##### Exercise 2.2.5

(a) We have $\lim a_n=0$. For every positive $\varepsilon$, take $N=6$, then for all $n>N$, we have $$0<5/n<1\Longrightarrow [[5/n]]=0.$$ Therefore$\left|[[5/n]]-0\right|<\varepsilon,$for all $n>N=6$. Hence the limit is zero.

(b) We have $\lim a_n=1$. For every positive $\varepsilon$, take $N=12$, then for all $n>N$, we have $$1< \frac{12+4n}{3n} =\frac{4}{n}+\frac{4}{3}<\frac{4}{12}+\frac{4}{3}<2.$$Thus $a_n=1$ for $n>12$. Therefore$\left|[[(12+4n)/(3n)]]-1\right|<\varepsilon,$for all $n>N=12$. Hence the limit is one.
The notation for (b) is a little confusing. Whether we consider it as $\dfrac{12+4n}{3n}$ or $\dfrac{12+4n}{3}n$. If the latter, then the limit should be infinity.

##### Exercise 2.2.6

Solution 1. For any $\varepsilon>0$, since $(a_n)\to a$, by definition there exists $N_1\in \mathbf{N}$ such that$|a_n-a|<\varepsilon$fro all $n>N_1$.
Similarly, since $(a_n)\to b$, by definition there exists $N_2\in \mathbf{N}$ such that$|a_n-b|<\varepsilon$fro all $n>N_2$.

Let $N=\max\{N_1,N_2\}$, then for any $n> N$, we have\begin{align*}|a-b|\leqslant |a-a_n|+|a_n-b|<\varepsilon+\varepsilon=2\varepsilon.\end{align*}
Therefore, we have $|a-b|<2\varepsilon$ for any $\varepsilon>0$. This may only happen if $a=b$ (see Theorem 1.2.6). (A nonnegative number which is less than any positive number can only be zero).

Remark: Here we used the famous inequality $$|\alpha-\beta|+|\beta-\gamma|\geqslant |\alpha-\gamma|.$$

Solution 2. Suppose $a\ne b$. Since $(a_n)\to a$, by definition there exists $N_1\in \mathbf{N}$ such that$|a_n-a|<\frac{|a-b|}{2}$fro all $n>N_1$.

Similarly, since $(a_n)\to b$, by definition there exists $N_2\in \mathbf{N}$ such that$|a_n-b|<\frac{|a-b|}{2}$fro all $n>N_2$.
Let $N=\max\{N_1,N_2\}$, then for any $n> N$, we have\begin{align*}|a-b|\leqslant |a-a_n|+|a_n-b|<\frac{|a-b|}{2}+\frac{|a-b|}{2}=|a-b|.\end{align*}Namely, we get $|a-b|<|a-b|$ which is impossible. Hence our assumption does not hold. Therefore $a=b$.

##### Exercise 2.2.7

A sequence $(a_n)$ is eventually in a set $A\subset \mathbf R$ means from some point ($n$) on, all $a_n\in A$.
A sequence $(a_n)$ is frequently in a set $A\subset \mathbf R$ means there infinitely many $n$ such that $a_n\in A$.
If we can figure out this, this exercise is easy.

(a) It is frequently.

(b) “Eventually” is stronger. “Eventually” implies “frequently”.

(c) A sequence $(a_n)$ converges to a real number $a$ if, for every positive number $\varepsilon$, the sequence is eventually in the $\varepsilon$-neighborhood of $a$. “Eventually” is used.

(d) Not necessarily “Eventually”. For example, consider the sequence $$(1,2,1,2,1,2,\cdots).$$It is frequently in$(1.9,2.1)$.

##### Exercise 2.2.8

(a) Yes. We can take $M=2$ ($M=1$ is also fine). Then for any $N\in\mathbf{N}$ there exists an odd integer $n$ satisfying $N\leqslant n\leqslant N+2$. Note that for odd $n$, $x_n=0$. Therefore, this sequence is zero-heavy.

(b) Yes. If a sequence is zero-heavy, then, by definition, for any $k\in\mathbf N$ there exists $n$ satisfying $kM+k+1\leqslant n\leqslant (k+1)M+k+1$ where $x_n=0$. Note that for different $k\in\mathbf{N}$, these numbers $kM+k+1,\cdots,(k+1)M+k+1$ are disjoint. Hence we must have infinitely many zeros.

(c) No. Consider the example $x_n=0$ if $n$ is a perfect square and $x_n=1$ otherwise. For any $M\in\mathbf N$, let $k>M$ be an integer. Take $N=k^2+1$. Then $$k^2+1\leqslant N+M=k^2+1+M < k^2+k+1\leqslant (k+1)^2.$$Therefore, by definition of the sequence, for any $N\leqslant n\leqslant N+M$, we have $x_n=1$. Hence this sequence is not zero-heavy.

(d) A sequence is not zero-heavy if for any $M\in\mathbf N$ there exists $N\in \mathbf N$ such that $x_n\ne 0$ for all $N\leqslant n\leqslant N+M$.