# Sequences by Justin

## **Week 1**

**Definition 1.** $\{x_n\}$ is constant if $x_n=x_{n+1}$ $\forall~n\in {\mathbb N}$.

**Example 1.** $\{17\}$ is constant since $x_n=17=x_{n+1}$ $\forall~n\in {\mathbb N}$.
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**Counterdefinition 1.** $\{x_n\}$ is not constant if $\exists~m\in {\mathbb N}$ s.t. $x_m\neq x_{m+1}$.

**Example 1'.** $\{(-1)^n\}$ is not constant since $-1=x_1 \neq x_2=1$.
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**Definition 2.** $\{x_n\}$ is bounded above if $\exists~K\in{\mathbb R}$ such that $x_n\leq K$
$\forall~n\in {\mathbb N}$.

**Example 2.** Prove that $\{(-1)^n\}$ is bounded above.
*Proof*
Let $K=1$.
Further, $\forall~m \in{\mathbb N}$:
$(-1)^n=1$ when $n=2m$ and
$(-1)^n=-1$ when $n=2m-1$.
$\therefore$ $(-1)^n \leq K$ $\forall~n \in{\mathbb N}$. $\blacksquare$
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**Counterdefinition 2.** $\{x_n\}$ is not bounded above if $\exists~ m\in{\mathbb N}$ s.t. ${x_m}>K$
$\forall~K\in{\mathbb R}$.

**Example 2'.** Prove that $\{2^n\}$ is not bounded above.
*Proof*
$\forall~K \in{\mathbb R_{\leq 0}}$:
Let $m=1$.
Then, $2^m=2>K$.

$\forall~K \in{\mathbb R_{>0}}$:
Let $m=\lceil \lvert \log_{2} (K) \rvert \rceil+1$.
Then, $2^m=2^{\lceil \lvert \log_{2} (K) \rvert \rceil+1} \geq 2^{\log_{2} (K)+1}=2K>K$. $\blacksquare$
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**Definition 3.** $\{x_n\}$ is bounded below if $\exists~K\in{\mathbb R}$ such that $x_n\geq K$
$\forall~n\in {\mathbb N}$.

**Example 3.** Prove that $\{n^2\}$ is bounded below.
*Proof*
Let $K=1$.
Then, $n^2 \geq 1=K ~\forall~n \in{\mathbb N}$. $\blacksquare$
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**Counterdefinition 3.** $\{x_n\}$ is not bounded below if $\exists~ m\in{\mathbb N}$ s.t. ${x_m}<K$
$\forall~K\in{\mathbb R}$.

**Example 3'.** Prove that $\{-n\}$ is not bounded below.
*Proof*
Let $m=\lfloor \lvert K \rvert \rfloor +1$.
Then, $-m=-\lfloor \lvert K \rvert \rfloor -1<K$
$\forall~K \in{\mathbb R}$. $\blacksquare$
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**Definition 4.** $\{x_n\}$ is bounded if $\exists~K\in{\mathbb R}$ such that $\lvert x_n \rvert \leq K$
$\forall~n\in {\mathbb N}$.

**Example 4.** Prove that $\{\cos(n)\}$ is bounded.
*Proof*
Let $K=1$.
Since $\cos(n)=\pm \frac {Adjacent} {Hypotenuse}$, where $Adjacent \leq Hypotenuse$,
$\lvert \cos(n) \rvert \leq 1=K$ $\forall~n \in{\mathbb N}$. $\blacksquare$
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**Counterdefinition 4.** $\{x_n\}$ is not bounded if $\exists~ m\in{\mathbb N}$ s.t. $\lvert {x_m} \rvert >K$
$\forall~K\in{\mathbb R}$.

**Example 4'.** Prove that $\{\sqrt n\}$ is not bounded.
*Proof*
Let $m=\lceil K^2 \rceil +1$.
Then, $\lvert \sqrt m \rvert=\lvert \sqrt {\lceil K^2 \rceil +1} \rvert > \sqrt {\lceil K^2 \rceil} \geq \sqrt {K^2}=K$
$\forall~K \in{\mathbb R}$. $\blacksquare$
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## **Week 2**

1. Prove that if $\{x_n\}$ is infinitely large then $\{x_n\}$ is unbounded.

*Proof*
Let $\{x_n\}$ be an infinitely large sequence.
Then, $\forall~K \in{\mathbb R}$, $\exists~m \in{\mathbb N}$ s.t.
$\forall~n \geq m$, $\lvert x_n \rvert \geq K$.
Thus, $\forall~K \in{\mathbb R}$, $\exists~m \in{\mathbb N}$ s.t. $\lvert x_n \rvert \geq K$,
so $\{x_n\}$ is unbounded. $\blacksquare$
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2. Prove that $\{(-1)^nn\}$ is inf. large.

*Proof*
Let $m=\lceil \lvert K \rvert \rceil +1$.
Then, $\forall$ $n \geq m$,
$\lvert x_n \rvert =\lvert (-1)^nn \rvert =\lvert n \rvert \geq \lvert m \rvert \geq \lvert \lceil \lvert K \rvert \rceil +1 \rvert \geq \lvert \lceil \lvert K \rvert \rceil \rvert \geq K$
$\forall~K \in{\mathbb R}$, so $\{(-1)^nn\}$ is infinitely large. $\blacksquare$
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3. Counter the definition of infinitely large.
$\{x_n\}$ is not infinitely large if $\forall~m \in{\mathbb N}$, $\exists~n \geq m,~K \in{\mathbb R}$
s.t $\lvert x_n \rvert <K$.
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4. Prove that $$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$is not infinitely large, but unbounded.

*Proof*
Not infinitely large:
$\forall~m \in{\mathbb N}$, let $K=1$ and $n=2m+1$, so $n \geq m$.
Then, since n is odd, $x_n=0$.
$\therefore 0=\lvert x_n \rvert <K=1$, so $\{x_n\}$ is not infinitely large.

Unbounded:
$\forall~K \in{\mathbb R}$, let $m=2 \lceil \lvert K \rvert \rceil$.
Then, since m is even,
$\lvert x_m \rvert=\lvert m \rvert= \lvert 2 \lceil \lvert K \rvert \rceil \rvert \geq \lvert 2 \lvert K \rvert \rvert \geq \lvert K \rvert \geq K$,
so $\{x_n\}$ is unbounded. $\blacksquare$
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5. Counter the definition of $\lim x_n=L$.
$\lim x_n \neq L$ if $\forall~m \in{\mathbb N}$, $\exists~\epsilon>0, n \geq m$
s.t. $\lvert x_n-L \rvert \geq \epsilon$.
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6. Prove that $\lim(-1)^n\neq 0$.

*Proof*
$\forall~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2m$, so $n \geq m$.
Then, $x_n=(-1)^{2m}=1$.
$\therefore \lvert x_n-L \rvert=\lvert 1-0 \rvert=1 \geq \epsilon$,
so $\lim (-1)^n \neq 0$. $\blacksquare$
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7. Prove that $\lim(-1)^n\neq 1$.

*Proof*
$\forall~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2m+1$, so $n \geq m$.
Then, $x_n=(-1)^{2m+1}=-1$.
$\therefore \lvert x_n-L \rvert=\lvert -1-1 \rvert =2 \geq \epsilon$,
so $\lim (-1)^n \neq 1$. $\blacksquare$
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## **Week 3**

1. Negate and justify how we negate the following statement
$\forall \epsilon>0, ~~~~\epsilon$ is a bad number.
In particular, why we do not change the inequality $\epsilon>0$.

*Negation*:
$\exists~\epsilon >0$ s.t. $\epsilon$ is a good number.

*Justification*:
Rewrite $\forall~\epsilon>0, ~\epsilon$ is a bad number as:
"$\forall~\epsilon$, if $\epsilon>0$, then $\epsilon$ is a bad number."
This is logically equivalent to:
"$\forall~\epsilon$, ($\epsilon \leq 0$) $\cup$ ($\epsilon$ is a bad number)."
The negation of the above statement is:
"$\exists~\epsilon$ s.t. ($\epsilon >0$) $\cap$ ($\epsilon$ is a good number)."
$\therefore$ The negation of $\forall~\epsilon>0, ~\epsilon$ is a bad number is:
"$\exists~\epsilon >0$ s.t. $\epsilon$ is a good number."
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2. Let $A\subset \mathbb R$. State the definition of $m=\inf~A.$

$m=\inf A$ if $\forall~\epsilon >0$, $\exists~a^* \in A$ s.t. $m \leq a^* <m+\epsilon$.
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3. Prove that $\sup\{1-\frac{1}{n}\}=1$.

$\forall~n \in \{1-\frac {1}{n}\}$, $n \leq 1$.

Then, $\forall~\epsilon>0$, let $m=\lfloor \frac {1}{\epsilon} \rfloor+1$ so that:
$$x_m=1-\frac {1}{\lfloor \frac {1}{\epsilon} \rfloor+1} > 1-\frac {1}{\frac {1}{\epsilon}}=1-\epsilon$$

$\therefore 1-\epsilon<x_m \leq 1$,
so $\sup\{1-\frac{1}{n}\}=1$.
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4. Prove that $\inf\{1-\frac{1}{n}\}=0$

If $\epsilon >0$, let $m=1$.
Thus, $x_m=1-\frac {1}{1}=0< \epsilon$.
$\therefore 0 \leq x_m<0+\epsilon$,
so $\inf\{1-\frac{1}{n}\}=0$.
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5. Prove that $\inf\{(0,1)\}=0$

$\forall~a \in (0,1)$, $a \geq 0$.
Then, if $\epsilon >0$, let $a^*=\min\{\frac{\epsilon}{2}, \frac{1} {2}\}$.
$\therefore 0 \leq a*<0+\epsilon$,
so $\inf\{(0,1)\}=0$. $\blacksquare$
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6. Let $A=\{0\}\cup \{(1,2)\}$. Prove that $\inf A=0$.

$\forall~a \in \{0\}\cup \{(1,2)\}$, $a \geq 0$.
Then, if $\epsilon >0$, let $a^*=0$.
$\therefore 0 \leq a*<0+\epsilon$,
so $\inf~\{0\}\cup \{(1,2)\}=0$. $\blacksquare$

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7. Exercise 2.1.3 and 2.1.4 in our text. You must prove your answers, not just provide them.

*2.1.3*
By definition of convergent, $\lim x_n=L$ if
$\forall~\epsilon>0, \exists~m$ s.t. $\forall~ n \geq m$, $\lvert x_n-L \rvert<\epsilon$.
Let $m=\lfloor \frac {1}{2\epsilon} \rfloor+1$
Then, $\forall~n \geq m$:
$$\lvert x_n-L \rvert=\lvert \frac {(-1)^n}{2n}-0 \rvert=\lvert \frac {(-1)^n}{2n} \rvert=\frac {1}{2n}\leq \frac {1}{2m}=\frac {1}{2(\lfloor \frac {1}{2\epsilon} \rfloor+1)}<\frac {1}{2(\frac {1}{2\epsilon})}=\epsilon$$
$\therefore \lim (\frac {(-1)^n}{2n})=0.$ $\blacksquare$

*2.1.4*
By definition of convergent, $\lim x_n=L$ if
$\forall~\epsilon>0, \exists~m$ s.t. $\forall~ n \geq m$, $\lvert x_n-L \rvert<\epsilon$.
Let $m=\lceil \log_2 {(\frac{1}{\epsilon}+1)} \rceil$
Then, $\forall~n \geq m$:
$$\lvert x_n-L \rvert=\lvert 2^{-n}-0 \rvert=\lvert 2^{-n} \rvert=\lvert \frac {1}{2^n} \rvert =\frac{1}{2^n} \leq \frac{1}{2^m}= \frac {1}{2^{\lceil \log_2 {(\frac{1}{\epsilon}+1)} \rceil}} \leq \frac {1}{\frac{1}{\epsilon}+1}<\frac{1}{\frac{1}{\epsilon}}= \epsilon$$
$\therefore \lim (2^{-n})=0.$ $\blacksquare$
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8. Redo Week 1 HW if you want to (for half of the remaning points)

**Counterdefinition 2.** $\{x_n\}$ is not bounded above if $\exists~ m\in{\mathbb N}$ s.t. ${x_m}>K$
$\forall~K\in{\mathbb R}$.

*Correction*
$\{x_n\}$ is not bounded above if $\forall~K\in{\mathbb R}$, $\exists~ m\in{\mathbb N}$ s.t. ${x_m}>K$.
___

**Counterdefinition 3.** $\{x_n\}$ is not bounded below if $\exists~ m\in{\mathbb N}$ s.t. ${x_m}<K$
$\forall~K\in{\mathbb R}$.

*Correction*
$\{x_n\}$ is not bounded below if $\forall~K\in{\mathbb R}$ $\exists~ m\in{\mathbb N}$ s.t. ${x_m}<K$.
___

**Counterdefinition 4.** $\{x_n\}$ is not bounded if $\exists~ m\in{\mathbb N}$ s.t. $\lvert {x_m} \rvert >K$
$\forall~K\in{\mathbb R}$.

*Correction*
$\{x_n\}$ is not bounded if $\forall~K\in{\mathbb R}$, $\exists~m\in{\mathbb N}$ s.t. $\lvert {x_m} \rvert >K$.
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## **Week 4**

1. Prove that any subsequence of an infinitely large sequence is infinitely large.

Let $\{x_n\}$ be an infinitely large sequence.
Then, $\forall~K \in{\mathbb R}$, $\exists~m \in{\mathbb N}$ s.t.
$\forall~n \geq m$, $\lvert x_n \rvert \geq K$.
Let $\{x_{n_i}\}$ be a subsequence of $\{x_n\}$.
Let $m_i \geq m$.
Then, $\forall~n_i \geq m_i \geq m$,
$\lvert x_{n_i} \rvert \geq K$.

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2. p. 50 Ex. 2.1.7

Let $\{x_n\}$ be a sequence.

a. Show that $\lim {x_n}=0$ if and only if $\lim{|x_n|}=0$.

$\bullet~\lim {x_n}=0 \implies \lim{|x_n|}=0$
If $\lim{x_n}=0$ then $\forall~\epsilon>0$, $\exists~m$ s.t. $\forall~n \geq m$,
$$\lvert x_n-0 \rvert=\lvert \lvert x_n \rvert-0 \rvert<\epsilon$$
So $\lim{|x_n|}=0$. $\blacksquare$

$\bullet~\lim{|x_n|}=0 \implies \lim {x_n}=0$
If $\lim{\lvert x_n \rvert}=0$ then $\forall~\epsilon>0$, $\exists~m$ s.t. $\forall~n \geq m$,
$$\lvert \lvert x_n \rvert-0 \rvert=\lvert x_n-0 \rvert<\epsilon$$
So $\lim{x_n}=0$. $\blacksquare$

b. Find an example such that $\{\lvert x_n \rvert\}$ converges and $\{x_n\}$ diverges.

$\{\lvert x_n \rvert\}=\{\lvert (-1)^n \rvert \}$ is convergent.
*Proof*
Let $m=1$ and $L=1$.
Then, $\forall~n \geq m$ and $\forall~\epsilon>0$:
$\lvert x_n-L|=\lvert 1-1 \rvert=0<\epsilon$,
so $\lim{x_n}=1$. $\blacksquare$

$\{x_n\}=\{(-1)^n\}$ is divergent.
*Proof*
Suppose by contradiction that $\lim x_n=a$.

Case 1: $a \neq 1$ and $a \neq -1$
Let $\epsilon_1=\lvert a-1 \rvert$ and $\epsilon_2=\lvert a+1 \rvert$.
Then, $\exists~\epsilon=\frac{\min\{\epsilon_1,\epsilon_2\}}{2}$ s.t $\forall~m$, $\exists~n=m+1$
s.t. $\lvert x_n-a \rvert >\epsilon$.
So $\lim{x_n}=1$ or $\lim{x_n}=-1$.

Case 2: $a=1$
$\forall~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2m+1$, so $n \geq m$.
Then, $x_n=(-1)^{2m+1}=-1$.
$\therefore \lvert x_n-L \rvert=\lvert -1-1 \rvert =2 \geq \epsilon$,
so $\lim (-1)^n \neq 1$.

Case 3: $a=-1$
$\forall~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2m$, so $n \geq m$.
Then, $x_n=(-1)^{2m}=1$.
$\therefore \lvert x_n-L \rvert=\lvert 1+1 \rvert =2 \geq \epsilon$,
so $\lim (-1)^n \neq -1$.

$\therefore \{x_n\}$ is divergent by contradiction. $\blacksquare$
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3. p. 50 Ex. 2.1.10

Show that the sequence $\{\frac{n+1}{n}\}$ is monotone, bounded and use Theorem 2.1.10 to find the limit.

*Monotone*
$$x_{n+1}=\frac{(n+1)+1}{n+1}=\frac{n+2}{n+1}*\frac{n}{n}=\frac{n^2+2n}{n(n+1)} \leq \frac{n^2+2n+1}{n(n+1)}=\frac{n+1}{n}*\frac{n+1}{n+1}=x_n$$
So $\{x_n\}$ is monotone decreasing. $\blacksquare$

*Bounded*
Let $K=2$.
Since $x_n$ is monotone decreasing and positive $\forall~n \in{\mathbb N}$,
$0<x_n \leq x_1$.
Thus:
$$\lvert \frac{n+1}{n} \rvert =\frac{n+1}{n}=x_n \leq x_1=2=K$$
So $\{x_n\}$ is bounded. $\blacksquare$

*Limit*
$\forall~n \in{\mathbb N}$, $\frac{n+1}{n}>1$.
If $\epsilon >0$, let $m=\frac{2}{\epsilon}$.
Thus:
$$x_m=\frac{\frac{2}{\epsilon}+1}{\frac{2}{\epsilon}}=1+\frac{\epsilon}{2}<1+ \epsilon$$
$\therefore 1 \leq x_m<1+\epsilon \implies \inf\{\frac{n+1}{n}\}=1$.
So by Theorem 2.1.10, $\lim{x_n}=\inf\{x_n:n \in{\mathbb N}\}=1$. $\blacksquare$
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4. p. 50 Ex. 2.1.13

Let $\{x_n\}$ be a convergent monotone sequence. Suppose there exists a $k \in {\mathbb N}$ such that $\lim {x_n}={x_k}$. Show that ${x_n}={x_k}$ for all $n \geq k$.

Since $\{x_n\}$ is convergent s.t. $\lim {x_n}={x_k}$,
$\forall~\epsilon>0$, $\exists~m$ s.t. $\forall~n \geq m$, $\lvert x_n-x_k \rvert <\epsilon$.

When $n=k$, $x_n=x_k$.

When $n>k$, suppose by contradiction that $x_n \neq x_k$.
Since $\{x_n\}$ is monotone,
$\forall~n$, $x_n>x_k$ or $x_n<x_k$.
But then $\exists~\epsilon=\lvert \frac {x_n-x_k}{2} \rvert$ s.t $\lvert x_n-x_k \rvert \geq \epsilon$.
Thus, $x_n=x_k$ by contradiction.

$\therefore$ When $n \geq k$, $x_n=x_k$.

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5. p. 50 Ex. 2.1.16

Let $\{x_n\}$ be a sequence. Suppose there are two congruent subsequences $\{x_{n_i}\}$ and $\{x_{m_i}\}$, s.t. $\lim x_{n_i}=a$ and $\lim x_{m_i}=b$, where $a \neq b$. Prove that $\{x_n\}$ is not convergent, without using Proposition 2.1.17.

*Proof*
Since $\lim x_{n_i}=a$, then
$\forall~\epsilon>0$, $\exists~n^*_i$ s.t. $\forall~n_i \geq n^*_i$,
$\lvert x_{n_i}-a \rvert <\epsilon$.

Since $\lim x_{m_i}=b$, then
$\forall~\epsilon>0$, $\exists~m^*_i$ s.t. $\forall~m_i \geq m^*_i$,
$\lvert x_{m_i}-b \rvert <\epsilon$.

Suppose by contradiction that $\lim x_n=L$.

Case 1: $L \neq a$
Then, $\forall~m$, let $\epsilon=\frac{\lvert {L-a} \rvert}{2}$ and $n=n^*_i$, so:
$$\lvert L-a \rvert=\lvert L-x_n+x_n-a \rvert \leq \lvert x_n-a \rvert+\lvert x_n-L \rvert<\epsilon+\lvert x_n-L \rvert=\frac{\lvert {L-a} \rvert}{2}+\lvert x_n-L \rvert$$
$\therefore \lvert x_n-L \rvert \geq \frac{\lvert {L-a} \rvert}{2}=\epsilon$.
Thus, $\lim x_n=a$.

Case 2: $L \neq b$
Analogous to the above proof with the result that $\lvert x_n-L \rvert \geq \frac{\lvert {L-b} \rvert}{2}=\epsilon$.
Thus, $\lim x_n=b$, which is a contradiction.

$\therefore \{x_n\}$ is divergent. $\blacksquare$
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6. p. 50 Ex. 2.1.17

Find a sequence $\{x_n\}$ such that for any $y \in {\mathbb R}$, there exists a subsequence $\{x_{n_i}\}$ converging to y.

Let $\{x_n\}=\{-1,0,1,-2,-\frac{3}{2},\frac{2}{2},-\frac{1}{2},\frac{0}{2},\frac{1}{2},\frac{2}{2},\frac{3}{2},2,-3,-\frac{8}{3},-\frac{7}{3},-\frac{6}{3},...,\frac{8}{3},3,-4,...\}$
To create a subsequence $\{x_{n_i}\}$ converging to any $y \in{\mathbb R}$:

For the first instance of every $k \in{\mathbb N}$ in $\{x_{n}\}$,
let $x_{n_k}$ be the closest value to $y$ in the interval $[-k,k]$.

Since ${\mathbb Q}$ is dense in ${\mathbb R}$, then $\forall~\epsilon>0$,
$\exists~m_i$ s.t. $\forall~n_i \geq m_i$, $\lvert x_{n_i}-y \rvert < \epsilon$.
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## Week 5 ##

1. If $\{x_n\} \rightarrow a$ and $\{y_n\} \rightarrow b$ and $\exists~m$ s.t. $\forall~n \geq m$, $x_n<y_n$,
then $a \leq b$.

$\forall~\epsilon >0$, $\exists~m_1$ s.t $\forall~n \geq m_1$,
$\lvert x_n-a \rvert < \frac{\epsilon}{2}$.
$\forall~\epsilon >0$, $\exists~m_2$ s.t $\forall~n \geq m_2$,
$\lvert y_n-b \rvert < \frac{\epsilon}{2}$.
Let $m= \max\{m_1, m_2\}$, then
$a-x_m < \frac{\epsilon}{2}$ and $y_m-b < \frac{\epsilon}{2}$.
By addition of inequalities, $y_m-x_m+a-b<\epsilon$,
or $x_m-y_m<b-a+ \epsilon$.
Since $x_m<y_m$,
$0<b-a+ \epsilon$
or $a-b<\epsilon$.
Since $\epsilon>0$ is arbitrary, $a-b \leq 0$,
so $a \leq b$. $\blacksquare$

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## Week 6 ##

1. *Exercise 2.2.4*: Suppose $x_1 := \frac{1}{2}$ and $x_{n+1}=x_n^2$. Show that $\{x_n\}$ converges and find $\lim{x_n}$.

We can define $\{y_n\}_{n=1}^\infty=\{x_{k+n}\}_{n=1}^\infty={\frac{1}{(k+1)^2}}$.
Let $m=\sqrt{\frac{1}{\epsilon}}$.
Then, $\forall~\epsilon >0; ~k \geq m$,
$\lvert y_n-0 \rvert = \lvert \frac{1}{(k+1)^2} \rvert= \frac{1}{(k+1)^2} <\frac{1}{k^2} \leq \frac{1}{m^2} = \frac{1}{\sqrt{\frac{1}{\epsilon}}^2} = \frac{1}{\frac{1}{\epsilon}} =\epsilon$.
Notice $\{y_n\} \rightarrow 0$ is the first tail of $\{x_n\}$.
$\therefore \{x_n\}$ converges and $\lim{x_n}=0$. $\blacksquare$
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2. *Exercise 2.2.5*: Let $x_n := \frac{n-\cos{n}}{n}$. Use the squeeze lemma to show that $\{x_n\}$ converges and find the limit.

First, rewrite $x_n := 1+\frac{\cos{n}}{n}$ and note that $\forall~n \in{\mathbb N}$,
$-1 \leq \cos{n} \leq 1$.
So,
$$1+ \frac{-1}{n} \leq 1+ \frac{\cos{n}}{n} \leq 1+ \frac{1}{n}$$
$\forall~\epsilon>0$, let $m=\frac{2}{\epsilon}$.
Then, $\forall~n \geq m,$
$$\lvert (1+ \frac{-1}{n})-1 \rvert=\lvert \frac{-1}{n} \rvert =\frac{1}{n} \leq \frac{1}{m}=\frac{1}{\frac{2}{\epsilon}}=\frac{\epsilon}{2}<\epsilon$$
and
$$\lvert (1+ \frac{1}{n})-1 \rvert=\lvert \frac{1}{n} \rvert =\frac{1}{n} \leq \frac{1}{m}=\frac{1}{\frac{2}{\epsilon}}=\frac{\epsilon}{2}<\epsilon$$
Since $\{1+\frac{-1}{n}\} \rightarrow 1$ and $\{1+\frac{1}{n}\} \rightarrow 1$,
by the Squeeze Lemma, $\{x_n\}$ converges and $\lim{x_n}=1$. $\blacksquare$
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3. *Exercise 2.2.6*: Let $x_n := \frac{1}{n^2}$ and $y_n := \frac{1}{n}$. Define $z_n := \frac{x_n}{y_n}$ and $w_n := \frac{y_n}{x_n}$. Do $\{z_n\}$ and $\{w_n\}$ converge? What are the limits? Can you apply Prop. 2.2.5? Why or why not?

We can NOT use Prop. 2.2.5 because for the division of two sequences, $\{x_n\}$ and $\{y_n\}$, the requirement is $\lim{x_n} \neq 0$ and $\lim{y_n} \neq 0$. However, in this problem, $\lim{x_n}=\lim{\frac{1}{n^2}}=0$ and $\lim{y_n}=\lim{\frac{1}{n}}=0$.

Instead, rewrite $z_n$ and $w_n$ as
$$z_n := \frac{\frac{1}{n^2}}{\frac{1}{n}}=\frac{n}{n^2}=\frac{1}{n}$$
and
$$w_n := \frac{\frac{1}{n}}{\frac{1}{n^2}}=\frac{n^2}{n}=n$$
We have previously shown that $\{z_n\}=\{\frac{1}{n}\}$ converges and $\lim{z_n}=0$.
We know $\{w_n\}=\{n\}$ diverges and $\lim{w_n}=\infty$. $\blacksquare$
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4. *Exercise 2.2.7*: True or false, prove or provide a counterexample. If $\{x_n\}$ is a sequence such that $\{x_n^2\}$ converges then $\{x_n\}$ converges.

False.
*Counterexample*:
Let $\{x_n\}=\{(-1)^n\}$.
Then, it is trivial to show that $\{x_n^2\}=\{(-1)^{2n}\}=\{1\} \rightarrow 1$,
but $\{x_n\}$ diverges. $\blacksquare$
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5. *Exercise 2.2.8*: Show that $\lim{\frac{n^2}{2^n}}=0$.

Using the ratio test for sequences (Lemma 2.2.12), we look at:
$$\frac{\lvert x_{n+1} \rvert}{\lvert x_n \rvert}=\frac{(n+1)^2/2^{n+1}}{n^2/2^n}=\frac{2^n}{2^{n+1}}\frac{(n+1)^2}{n^2}=\frac{n^2+2n+1}{2n^2}$$
Then, $\forall~\epsilon >0$ let $m=\frac{2}{\sqrt{\epsilon}}$, s.t. $\forall~n \geq m$,
$$\lvert \frac{n^2+2n+1}{2n^2}-\frac{1}{2} \rvert=\lvert \frac{n^2+2n+1}{2n^2}-\frac{n^2}{2n^2} \rvert=\lvert \frac{2n+1}{2n^2} \rvert=\frac{1}{n}+\frac{1}{2n^2}$$
$$\leq \frac{1}{m}+\frac{1}{2m^2}=\frac{1}{\frac{2}{\sqrt{\epsilon}}}+\frac{1}{2(\frac{2}{\sqrt{\epsilon}})^2}=\frac{\sqrt{\epsilon}}{2}+\frac{\epsilon}{8} < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

Thus, $\lim{\frac{\lvert x_{n+1} \rvert}{\lvert x_n \rvert}}=\frac{1}{2}<1$.
$\therefore$ By Lemma 2.2.12, $\{x_n\}$ converges and $\lim{x_n}=0$. $\blacksquare$

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## Week 7 ##

1. Without using Lemma on Nested Intervals prove that
$$\bigcap_{n=1}^\infty[-1/n, 1/n]=\{0\}.$$

Let $x \in \bigcap_{n=1}^\infty[-1/n, 1/n]$.
Then $\forall~n$,
$$-\frac{1}{n} \leq x \leq \frac{1}{n}$$
$$\implies \lim_{n\to\infty}(-\frac{1}{n}) \leq \lim_{n\to\infty}(x) \leq \lim_{n\to\infty}(\frac{1}{n})$$
$$\implies 0 \leq x \leq 0$$
$\therefore~\bigcap_{n=1}^\infty[-1/n, 1/n]=\{x\}=\{0\}$.
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2. *Exercise 2.3.5*:
a) Let $x_n :=\frac{(-1)^n}{n}$. Find $\lim \sup x_n$ and $\lim \inf x_n$.
b) Let $x_n :=\frac{(n-1)(-1)^n}{n}$. Find $\lim \sup x_n$ and $\lim \inf x_n$.

a) By Theorem 2.3.5, if $\{x_n\} \rightarrow L$, then $\lim \sup x_n=\lim \inf x_n=\lim x_n=L$.
Rewrite $x_n :=\frac{(-1)^n}{n}=\frac{1}{n}(-1)^n$.
Let $\{a_n\}=\{\frac{1}{n}\}$ and $\{b_n\}=\{(-1)^n\}$.
Then, $\{a_n\} \rightarrow 0$ and $\{b_n\}$ is bounded.
Thus, $\{x_n\}=\{a_nb_n\} \rightarrow 0$.
$\therefore \lim x_n=\lim \sup x_n=\lim \inf x_n=0$

b) Rewrite $x_n:=\frac{(n-1)(-1)^n}{n}=(1-\frac{1}{n})(-1)^n$.
Let $a_n:=\sup\{x_k:k \geq n\}$ and $b_n:=\inf\{x_k:k \geq n\}$.
Notice that $\{1-\frac{1}{n}\} \rightarrow 1$ and $\{(-1)^n\}=\{-1,1\}$.
Thus, there are two convergent subsequences:
$\{x_{2m}\} \rightarrow 1$ and is monotone increasing.
$\{x_{2m-1}\} \rightarrow -1$ and is monotone decreasing.
So $\{a_n\}=\{1\}$ and $\{b_n\}=\{-1\}$.
By Proposition 2.3.2,
$\lim \sup x_n=\inf\{a_n:n\in{\mathbb N}\}=\inf\{1:n\in{\mathbb N}\}=1$
$\lim \inf x_n=\sup\{b_n:n\in{\mathbb N}\}=\sup\{-1:n\in{\mathbb N}\}=-1$.

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3. Prove that every unbounded sequence, contains an infinitely large subsequence. The proof goes by construction.

Let $\{x_n\}$ be an unbounded sequence.
Then, $\forall~B \in{\mathbb R}$, $\exists~m$ s.t.
$\lvert x_m \rvert >B$.

We can construct an infinitely large subsequence of $\{x_n\}$ as follows:
Let $x_{n_1}:=x_1$.

Then, choose any value of $B$.
Let $x_{n_2}:=x_{m_1}$ s.t. $m_1>1$ and $\lvert x_{m_1} \rvert>B$
(which is always possible, since $\{x_n\}$ is unbounded).

In general, $\forall~k\in{\mathbb N}$ s.t $k \geq 2$,
let $x_{n_{k+1}}:=x_{m_k}$ s.t. $m_{k}>m_{k-1}+1$ and $\lvert x_{m_k} \rvert>\lvert x_{n_{k}} \rvert$
(which is always possible, since if $\{x_n\}$ is unbounded, its tail is unbounded).

Then for $\{x_{n_k}\}$, since $B$ was arbitrary in our construction,
$\forall~B \in{\mathbb R}$, $\exists~n_k^*=n_2$ s.t. $\forall~n_k \geq n_k^*$
$\lvert x_{n_k} \rvert >B$.
$\therefore$ $\{x_{n_k}\}$ is infinitely large.
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## Week 8 ##

Tim's Desmos Problems

1. $m=6$
2. $m=7$
3. $15 < k \leq 31$
4. $m=17$
5. $m=26$
6. $n^*=31$
7. $x_n^*=0.95$
8. $x^*=0.5$
9. $n^*=18$
10. $\frac{1}{3} \leq \epsilon <\frac{1}{2}$
11. $n_1^*=4$
$n_2^*=9$
$\implies n^*=9$
12. Since $a_4=a_5$, we cannot simultaneously have $\lvert a_4-1 \rvert>\epsilon$ and $\lvert a_5-1 \rvert <\epsilon$.
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## Week 10 ##

1. Exercise 2.5.1: For $r \neq 0$ prove:
$$\sum_{k=0}^{n-1}r^k=\frac{1-r^n}{1-r}$$

Let $s=\sum_{k=0}^{n-1} r^k$
Then, $$=s(1-r) = s-rs$$
$$=\sum_{k=0}^{n-1} r^k - r\sum_{k=0}^{n-1} r^k$$
$$=\sum_{k=0}^{n-1} r^k - \sum_{k=0}^{n-1} r^{k+1}$$
$$=(1+r+r^2+...+r^{n-1}) - (r+r^2+...+r^n)$$
$$=(1+r-r+r^2-r^2+...+r^{n-1}-r^{n-1}-r^n)$$
$$=(1-r^n)$$
This implies that $s(1-r)=(1-r^n)$.
Thus, $s=\frac{1-r^n}{1-r}$ $\blacksquare$.
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2. Exercise 2.5.2: Prove that for $-1<r<1$ we have $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$

*Proof*
Since we have
$$\sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}$$
Then take the limit, s.t.
$$\lim \sum_{k=0}^{n-1} r^k = \lim \frac{1-r^n}{1-r}=\frac{\lim 1-\lim r^n}{\lim 1-\lim r}$$
Since $-1<r<1$, $\lim r^n=0$.
$\therefore \sum_{k=0}^{n-1} r^k = \frac{1}{1-r}$ $\blacksquare$.
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3. Exercise 2.5.3a: Decide the convergence or divergence of
$$\sum_{n=1}^{\infty}\frac{3}{9n+1}$$

Let $x_n := \frac{1}{10n}$ and $y_n:=\frac{3}{9n+1}$.
Then $\forall~n$,
$$0 \leq \frac{1}{10n} \leq \frac{3}{9n+1} \implies 0 \leq x_n \leq y_n$$
But,
$$\sum_{n=1}^{\infty}x_n=\sum_{n=1}^{\infty}\frac{1}{10n}=\frac{1}{10}\sum_{n=1}^{\infty}\frac{1}{n}$$ diverges.

$\therefore$ By the Comparison Test, $\sum_{n=1}^{\infty}\frac{3}{9n+1}$ also diverges.
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4. Exercise 2.5.3b: Decide the convergence or divergence of
$$\sum_{n=1}^{\infty}\frac{1}{2n-1}$$

Let $x_n := \frac{1}{2n}$ and $y_n:=\frac{1}{2n-1}$.
Then $\forall~n$,
$$0 \leq \frac{1}{2n} \leq \frac{1}{2n-1} \implies 0 \leq x_n \leq y_n$$
But,
$$\sum_{n=1}^{\infty}x_n=\sum_{n=1}^{\infty}\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}$$ diverges.

$\therefore$ By the Comparison Test, $\sum_{n=1}^{\infty}\frac{1}{2n-1}$ also diverges.
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## Week 11 ##

1. Exercise 2.5.3c: Decide the convergence or divergence of
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$$

Notice:
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^2}$$

Let $\{x_n\}:=\{\frac{1}{n^2}\}$.
Note that, $\{x_n\} \searrow$ and only contains positive real numbers.
Further, $\lim {x_n}=0.$
Thus, by Prop. 2.6.2 (Alternating Series), $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$ converges. $\blacksquare$
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2. Exercise 2.5.3d: Decide the convergence or divergence of
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$

Notice:
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}$$
Let $\{x_n\}:=\{\frac{1}{n^2}\}$ and let $\{y_n\}=\{\frac{1}{n^2+n}\}$.
Then $\forall~n$,
$$0 \leq \frac{1}{n^2+n} \leq \frac{1}{n^2} \implies 0 \leq x_n \leq y_n$$

By the convergence of the p-series, we know that $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$.

$\therefore$ By the Comparison Test, $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ also converges. $\blacksquare$
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3. Exercise 2.5.3e: Decide the convergence or divergence of
$$\sum_{n=1}^{\infty}ne^{-n^2}$$

Let $\{x_n\}:=\{ne^{-n^2}\}$. Note that $\forall~n$, $x_n>0$.
Then,
$$L:=\lim \sup \lvert x_n \rvert^{\frac{1}{n}}=\lim \sup (x_n)^{\frac{1}{n}}=\lim \sup (ne^{-n^2})^\frac{1}{n}=\lim \sup (n^{\frac{1}{n}}e^{-n})>0$$

By the Ratio Test for monotonicity we see,
$$\frac{n^{\frac{1}{n+1}}e^{-n-1}}{n^{\frac{1}{n}}e^{-n}}=n^{\frac{-1}{n(n+1)}}e^{-1}<1$$

Thus, $\{(x_n)^{\frac{1}{n}}\} \searrow$.

Then, since the first term $(x_1)^{\frac{1}{1}}=e^{-1}<1$,
$$0<\lim \sup \lvert x_n \rvert^{\frac{1}{n}}=L<1$$

$\therefore$ By the Root Test, $\sum_{n=1}^{\infty}ne^{-n^2}$ converges. $\blacksquare$
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4. Exercise 2.5.4a: Prove that if $\sum_{n=1}^{\infty}x_n$ converges, then $\sum_{n=1}^{\infty}(x_{2n}+x_{2n+1})$ also converges.

$$\sum_{n=1}^{\infty}(x_{2n}+x_{2n+1})=(x_2+x_4+x_6+...)+(x_3+x_5+x_7+...)=x_2+x_3+x_4+...=\sum_{n=2}^{\infty}x_n$$

Notice that $\sum_{n=1}^{\infty}x_n$ converges $\implies$ $\sum_{n=2}^{\infty}x_n$ converges, since removing finite terms in the head does not affect the convergence of the series.

$\therefore \sum_{n=1}^{\infty}(x_{2n}+x_{2n+1})$ also converges. $\blacksquare$
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5. Exercise 2.5.4b: Find an explicit example where the converse does not hold.

Let $\{x_n\}:=\{(-1)^n\}$.
Then, $x_{2n}=(-1)^{2n}=1$ and $x_{2n+1}=(-1)^{2n+1}=-1$.

Thus,
$$\sum_{n=1}^{\infty}(x_{2n}+x_{2n+1})=\sum_{n=1}^{\infty}(1+-1)=\sum_{n=1}^{\infty}0=0$$

so $\sum_{n=1}^{\infty}(x_{2n}+x_{2n+1})$ converges.

However, $\sum_{n=1}^{\infty}x_n=\sum_{n=1}^{\infty}(-1)^n$ diverges. $\blacksquare$
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## Week 12 ##

Lemma 3.1.7:
Let $S⊂{\mathbb R}$ and $c$ be a cluster point of $S$. Let $f:S → {\mathbb R}$ be a function.
Then $f(x) \rightarrow L$ as $x \rightarrow c$, iff for every sequence ${x_n}$ of numbers s.t. $x_n ∈ S~\backslash \{c\}$ for all $n$,
and s.t. $\lim x_n = c$, we have that the sequence $\{f(x_n)\}$ converges to $L$.
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Lemma 3.1.7 (Open Interval):
Let $(a,b)⊂{\mathbb R}$ and $c$ be a in $(a,b)$. Let $f:(a,b) \rightarrow {\mathbb R}$ be a function.
Then $f(x) → L$ as $x \rightarrow c$, iff for every sequence ${x_n}$ of numbers s.t. $x_n ∈ (a,b)~\backslash \{c\}$ for all $n$,
and s.t. $\lim x_n = c$, we have that the sequence $\{f(x_n)\}$ converges to $L$.
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*Proof*
$(\implies)$
Suppose $f(x) \rightarrow L$ as $x \rightarrow c$ and $\{x_n\}$ is a sequence s.t. $x_n \in (a,b)~\backslash \{c\}$ and $\lim x_n=c$.
Let $\epsilon>0$.
Note before the next step that since all points in $[a,b]$ are cluster points of $(a,b)$,
$c$ is a cluster point of $(a,b)$.
Thus, we can choose a $\delta>0$ s.t. if $x \in (a,b)~\backslash \{c\}$ and $\lvert x-c \rvert<\delta$, then $\lvert f(x)-L \rvert<\epsilon$.
Since $\{x_n\}$ converges to $c$, we can find an $m$ s.t. $\forall~n \geq m$,
$\lvert x_n-c \rvert<\delta \implies \lvert f(x_n)-L \rvert<\epsilon$.
$\therefore \{f(x_n)\} \rightarrow L$. $\blacksquare$

$(\impliedby)$
Suppose $f(x) \nrightarrow L$ as $x \rightarrow c$.
Then, $\forall~\delta>0$, $\exists~x∈(a,b)~\backslash \{c\}$,
where $\lvert x−c \rvert<\delta$ and $\lvert f(x)−L \rvert \geq \epsilon$.
As seen in the proof of the other direction, note that $c$ is a cluster point of $(a,b)$.
Let $\delta= \frac{1}{n}$ to construct a sequence $\{x_n\}$.
We have that $\exists~\epsilon > 0$ s.t. $\forall ~ n$, $\exists~x_n \in (a,b)\backslash \{c\}$,
where $\lvert x_n−c\rvert < \frac{1}{n}$ and $|f(x_n)−L| \geq \epsilon$.
Then $\{x_n\} \rightarrow c$, but $\{f(x_n)\} \nrightarrow L$. $\blacksquare$
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## Week 13 ##
1. *Exercise 6.1.1* Let $f$ and $g$ be bounded funtions on $[a,b]$. Prove
$$\|f+g\|_u \leq \|f\|_u+\|g\|_u$$

*Proof*
Given that $f, g$ are bounded functions on $[a,b]$, $\exists ~ A, B >0$ s.t.
$$|f(x)|<A, ~ \forall ~ x \in [a,b]$$
$$|g(x)|<B, ~ \forall ~ x \in [a,b]$$
So, $\sup |f(x)|$ exists and $x \in [a,b]$ and $\sup |g(x)|$ exists ($\in {\mathbb R}) ~ x \in [a,b]$
Then, $$||f||_u := \sup_{x\in[a,b]}|f(x)|, ||g||_u := \sup_{x\in[a,b]}|g(x)|$$
Now $\forall ~ \in [a,b]$
$$|(f+g)(x)|= |f(x)+g(x)| \leq |f(x)|+|g(x)| \leq ||f||_u + ||g||_u$$
Thus, $|(f+g)(x)| \leq ||f||_u + ||g||_u ~\forall ~ x\in[a,b]$
So, $$\sup_{x\in[a,b]}|(f+g)(x)| \leq ||f||_u + ||g||_u$$
$\implies ||f+g||_u \leq ||f||_u + ||g||_u$. $\blacksquare$

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2. *Exercise 6.1.6* Find an example of a sequence of functions $\{f_n\}$ and $\{g_n\}$ that converge uniformly to some $f$ and $g$ on some set $A$, but such that $\{f_ng_n\}$ (the multiple) does not converge uniformly to $fg$ on $A$. Hint: Let $A:={\mathbb R}$, let $f(x):=g(x):=x$. You can even pick $f_n=g_n$.

*Proof*
Let $A:={\mathbb R}$, $f(x)=g(x)=x$ and $f_n(x)=g_n(x)=x+\frac{1}{n}$
Then for $\epsilon > 0$, choose $m > \frac{1}{\epsilon}$ so that $\forall ~n\geq m$ and $\forall~x,$
$$|f_n(x)-f(x)|=|x+\frac{1}{n}-x|=\frac{1}{n} \leq \frac{1}{m} < \epsilon$$
Thus, $\{f_n(x)\}$ converges uniformly to $f$.
Similarly, we can prove $\{g_n(x)\}$ converges uniformly to $g$.

However,

$$|f_n(x)g_n(x) - f(x)g(x)| = |(x+\frac{1}{n})^2 - x^2| = |x^2 + \frac{2x}{n} + \frac{1}{n^2} - x^2| = |\frac{2x}{n} + \frac{1}{n^2}| = \frac{2}{n}|x| + \frac{1}{n^2} > \epsilon$$
Hence, $\{f_n(x)g_n(x)\}$ does not converge uniformly to $f(x)g(x)$. $\blacksquare$