# Sequences by Tim

## 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 $x_n=17=x_{n+1}$ $\forall n\in {\mathbb N}$.

**Counterdefinition 1.** $\{x_n\}$ is not constant if $\exists m\in{\mathbb N}$ such that $x_m\neq x_{m+1}$

**Example 1'.** $\{(-1)^n\}$ is not constant since
$$-1=x_1\neq x_2=1.$$

**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 $K=1$
Then $\forall n\in{\mathbb N},~~(-1)^n \leq 1$
$\blacksquare$

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

**Example 2'.** $\{2n\}$ is not bounded above.
*Proof*
Given any $K\in{\mathbb R}$, let $m=\lceil \lvert K \rvert \rceil + 1$
Then $x_m=2m=2\lceil \lvert K \rvert \rceil+2>K$ $\blacksquare$
[Link to Desmos Example 2'](https://www.desmos.com/calculator/noeav5nixt)

**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.** $\{2\}$ is bounded below.
*Proof*
let $K=0$
Then $\forall n\in{\mathbb N}$, $x_n=2 \geq 0$ $\blacksquare$

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

**Example 3'.** $\{-n\}$ is not bounded below.
*Proof*
Given any $K\in{\mathbb R}$, let $m=\lceil \lvert K \rvert \rceil +1$
Then
$$x_m=-m=-\lceil \lvert K \rvert \rceil -1<K$$ $\blacksquare$
[Link to Desmos for Example 3'](https://www.desmos.com/calculator/5dsr7ln4u7)

**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.** $\{\sin n\}$ is bounded.
*Proof*
Let $K=1$
Since $-1\leq \sin a\ \leq 1 ~~ \forall a\in {\mathbb R}$, it follows that
$\forall n\in{\mathbb N}$, $\lvert x_n \rvert = \lvert \sin n\ \rvert$ $\leq K$ $\blacksquare$

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

*The following lemma will be used in Example 4'*
**Lemma 1.** $2^a>a ~~\forall a\in {\mathbb N}$
*Proof*
$2^1 = 2 > 1$, so the statement is true for $a=1$
Now suppose $2^k > k$ for some $k\in {\mathbb N}$
Then
$$2^{k+1}=2 \cdot 2^k = 2^k + 2^k>k+k \geq k+1$$
So by the principle of induction, this proof is complete.
$\blacksquare$

**Example 4'.** $\{(-2)^n\},$ is not bounded.
*Proof*
Given any $K\in{\mathbb R}$, let $m=\lceil \lvert K \rvert \rceil +1$
Then since $2^a>a ~~\forall a\in {\mathbb N}$, and from the definition above,$~m\in {\mathbb N}$ , it follows that
$$\lvert x_m \rvert = \lvert (-2)^m \rvert = ~ \lvert(-2)^{\lceil \lvert K \rvert \rceil +1} \rvert = 2^{\lceil \lvert K \rvert \rceil +1}> K$$ $\blacksquare$
[Link to Desmos for Example 4'](https://www.desmos.com/calculator/2yl7z5r9bi)

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## **Week 2**

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**1. Prove that if $\{x_n\}$ is infinitely large then $\{x_n\}$ is unbounded**
*Proof*
Suppose $\{x_n\}$ is infinitely large but not unbounded.
Then $\{x_n\}$ is bounded.
This means $\exists K\in{\mathbb R}$ such that $\lvert x_n \rvert$ $\leq K$ $~\forall n\in {\mathbb N}$.
But since $\{x_n\}$ is infinitely large, there must be some $m\in {\mathbb N}$. such that $\lvert x_m \rvert>K$.
Thus we have a contradiction, and the proof is complete. $\blacksquare$

**2. Prove that $\{(-1)^nn\}$ is infinitely large**
[Link to Desmos Visual for #2](https://www.desmos.com/calculator/r6dg6pj7bh)
*Proof*
Given any $K\in{\mathbb R}$, let $m=\lceil \lvert K \rvert \rceil +1$.
Then
$$\lvert x_m \rvert=\lvert(\lceil \lvert K \rvert \rceil+1)(-1)^{\lceil \lvert K \rvert \rceil+1}\rvert=\lceil \lvert K \rvert \rceil +1>K$$
Now consider any $n\in{\mathbb N}$ such that $n\geq m$
Then it follows that
$$\lvert x_n \rvert=\lvert (-1)^nn \rvert = n \geq m>K$$
$\therefore \{(-1)^nn\}$ is infinitely large.$~~\blacksquare$

**3. Counter the definition of infinitely large.**
$\{x_n\}$ is not infinitely large if $\exists K\in{\mathbb R}$ such that $~\forall n\in {\mathbb N}, ~~\exists n^*>n$ where $\lvert x_{n^*} \rvert \leq K$

*Note that "infinitely large" is not equivalent to "unbounded". See link below.*
*This link shows a sequence that is unbounded but NOT infinitely large. Examples of $n^*$ are shown in orange. You can choose the n up to 100.*
[Desmos Link for Week 2 #3](https://www.desmos.com/calculator/mwca9hocam)

**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*
Part 1: $\{x_n\}$ is unbounded
$\{x_n\}$ is unbounded if, $\forall K\in {\mathbb R}$, $\exists m\in {\mathbb N}$ such that $\lvert x_m \rvert$ $> K$.
Given any $K$, let $m=2\lceil \lvert K \rvert \rceil +2$
Since $m$ is even, $\lvert x_m \rvert=\lvert m \rvert=2\lceil \lvert K \rvert \rceil +2 > K$
Therefore $\{x_n\}$ must be unbounded.
[Desmos Link to week 2 #4 Part 1](https://www.desmos.com/calculator/ghd8sf6rjw)

Part 2: $\{x_n\}$ is not infinitely large.
$\{x_n\}$ is not infinitely large if $\exists K\in{\mathbb R}$ such that $~\forall n\in {\mathbb N}, ~~\exists n^*>n$ where $\lvert x_{n^*} \rvert \leq K$
Let $K=1$
Given any $n$, let $n^*=2n+1$
Since $n^*$ is odd, $\lvert x_{n^*} \rvert=0 \leq 1$
$\therefore \{x_n\}$ is not infinitely large $\blacksquare$
[Desmos Link to week 2 #4 Part 2](https://www.desmos.com/calculator/bp604c4i9o)

**5. Counter the definition of $\lim x_n=L$.**

$\lim x_n \neq L$ if $\exists~ \varepsilon > 0$ such that, $\forall n\in {\mathbb N}$, $\exists ~ n^* \geq n$
where $\lvert x_{n^*} - L \rvert \geq \varepsilon$

**6. Prove that $\lim(-1)^n\neq 0$**
*Proof*
Suppose $\lim(-1)^n = 0$
Then $\forall \varepsilon > 0, ~ \exists n^*$ such that $\forall n \geq n^*, ~\lvert x_n - 0 \rvert < \varepsilon$
Let $\varepsilon = \frac{1}{2}$
Since $\forall n \in {\mathbb N}, ~ \lvert x_n - 0 \rvert = \lvert x_n \rvert = 1 \geq \varepsilon$
It follows there is no $n^*$ that satisfies this condition, and this is a contradiction.
$\therefore \lim(-1)^n\neq 0 ~~~\blacksquare$

**7. Prove that $\lim(-1)^n\neq 1$**
Suppose $\lim(-1)^n = 1$
Then $\forall \varepsilon > 0, ~ \exists n^*$ such that $\forall n \geq n^*, ~\lvert x_n - 1 \rvert < \varepsilon$
Let $\varepsilon = \frac{1}{2}$
Choose any $n^*$ such that $\lvert x_{n^*} - 1 \rvert < \frac{1}{2}$
Then $x_{n^*}=(-1)^{n^*}=1$
and it follows that $x_{n^*+1}=(-1)^{n^*+1}=(-1)(-1)^{n^*}=(-1)(1)=-1$
but then $\lvert x_{n^*+1} - 1 \rvert = \lvert -1 - 1 \rvert = \lvert -2 \rvert = 2 \geq \varepsilon$
This is a contradiction.
$\therefore \lim(-1)^n\neq 1 ~~~\blacksquare$

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## **Week 3**

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**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 $\epsilon>0$.**

An equivalent way to say the original statement is:
"if $\epsilon > 0$ then $\epsilon$ is a bad number".

To logically negate the statement "if A then B", we need "A and not B".

So we need a case where the hypothesis ($\epsilon > 0$) is true, but the conclusion ($\epsilon$ is a bad number) is false.

The negation of this statement is:
$\exists ~ \epsilon > 0$ such that $\epsilon$ is not a bad number

The inequality does not change because if I found an $\epsilon \leq 0$ that is not a "bad number", it does not make the original statement false. The original statement only makes a claim about the "badness" of $\epsilon$ for $\epsilon > 0$.

**2. Let $A\subset \mathbb R$. State the definition of $m=\inf~A.$**

$m=\inf~A$ if the following two conditions are true:
$~~~~~$*i.*$~~~\forall a \in A,~ a \geq m$
$~~~~~$*ii.*$~~\forall \varepsilon > 0, ~\exists a^* \in A$ such that $m \leq a^* < m + \varepsilon$

**3. Prove that $\sup\{1-\frac{1}{n}\}=1$**

[Link to Desmos Week3 #3](https://www.desmos.com/calculator/r2vjigt8h0)

Since $\forall n \in {\mathbb N},~\frac{1}{n}>0$, it follows that $\forall n,~ 1-\frac{1}{n}\ < 1$.
So 1 is an upper bound of $\{1-\frac{1}{n}\}$.

Further, given any $\varepsilon > 0$, let $n^*=\lceil \frac{2}{\varepsilon} \rceil$.

Then

$$1-\frac{1}{\lceil\frac{2}{\varepsilon}\rceil} \geq 1-\frac{1}{(\frac{2}{\varepsilon})}=1-\frac{\varepsilon}{2} > 1 - \varepsilon$$

$\therefore \forall \varepsilon > 0, ~\exists n^* \in {\mathbb N}$ such that $1-\frac{1}{n^*} > 1-\varepsilon$
So, $\sup\{1-\frac{1}{n}\}=1~~\blacksquare$

**4. Prove that $\inf\{1-\frac{1}{n}\}=0$**
[Link to Desmos Week3 #4](https://www.desmos.com/calculator/hsq1dzumkc)

Since $\frac{1}{n} \leq 1 ~~\forall n \in {\mathbb N}$, it follows that $1-\frac{1}{n} \geq 0$.
So 0 is a lower bound of $\{1-\frac{1}{n}\}$.

Further, given any $\varepsilon > 0$, let $n^*=1$.
Then
$$1-\frac{1}{n^*}=1-\frac{1}{1}=0<\varepsilon$$
$\therefore \forall \varepsilon > 0, ~\exists n^* \in {\mathbb N}$ such that $1-\frac{1}{n^*} < 0 + \varepsilon$
So $\inf\{1-\frac{1}{n}\}=0~~\blacksquare$

**5. Prove that $\inf\{(0,1)\}=0$**
[Link to Desmos Week3 #5](https://www.desmos.com/calculator/fcrtfa39dt)

$\forall x \in \{(0,1)\}, ~~ x \geq 0$
So 0 is a lower bound of $\{(0,1)\}$

Also, given any $\varepsilon > 0, ~~\exists ~ x^* = \min\{\frac{1}{2}, \frac{\varepsilon}{2}\}$
such that
$$0 \leq \min\{\frac{1}{2}, \frac{\varepsilon}{2}\}<0+\varepsilon$$
$\blacksquare$

**6. Let $A=\{0\}\cup \{(1,2)\}$. Prove that $\inf A=0$.**
[Link to Desmos for Week3 #6](https://www.desmos.com/calculator/nwiz7x3vb8)

$\forall a \in A, ~~ a \geq 0$
So 0 is a lower bound of A.
Also, given any $\varepsilon > 0, ~~\exists ~ a^* = 0$
such that
$$0 \leq a^* < 0+\varepsilon$$
$\blacksquare$

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

***Exercise 2.1.3***
*Is the sequence $\{\frac{(-1)^n}{2n}\}$ convergent? If so, what is the limit?*

[Link to Desmos for Exercise 2.1.3](https://www.desmos.com/calculator/0t4fdmexap)

Yes, $\{\frac{(-1)^n}{2n}\}$ is convergent, and the limit is 0.
*Proof*
Given any $\varepsilon > 0$, let $n^* = \lceil \frac{1}{2\varepsilon} \rceil +1$.
Then $\forall n \geq n^*$,
$$\lvert\frac{(-1)^n}{2n}-0 \rvert=\frac{1^n}{2n}=\frac{1}{2n} \leq \frac{1}{2n^*}=\frac{1}{2\lceil\frac{1}{2\varepsilon}\rceil+1} \leq \frac{1}{2(\frac{1}{2\varepsilon})+1}=\frac{1}{\frac{1}{\varepsilon} +1}=\frac{\varepsilon}{\varepsilon + 1}<\varepsilon$$
$\blacksquare$

***Exercise 2.1.4***
*Is the sequence $\{2^{-n}\}$ convergent? If so, what is the limit?*

[Link to Desmos for Exercise 2.1.4](https://www.desmos.com/calculator/sy0ftlrpht)

Yes, $\{2^{-n}\}$ is convergent, and the limit is 0.
*Proof*

Given any $\varepsilon > 0$, let $n^* =\max\{1, \lceil \log_{2}(\frac{1}{\varepsilon}) \rceil +1\}$
Then for $\varepsilon ~\geq 1, n^* = 1$
and $\forall n \geq n^*$,
$$\rvert 2^{-n}-0 \lvert = 2^{-n} = \frac{1}{2^n} \leq \frac{1}{2^1} < \varepsilon$$
Further, $\forall \varepsilon$, $0<\varepsilon < 1,$ and $\forall n \geq n^*$
$$\lvert 2^{-n}-0 \rvert=2^{-n}=\frac{1}{2^n} \leq \frac{1}{2^{\lceil \log_2(\frac{1}{\varepsilon})\rceil+1}}<\frac{1}{(\frac{1}{\varepsilon})}=\varepsilon$$
$\blacksquare$
8. Redo Week 1 HW if you want to (for half of the remaining points)

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## **Week 4**

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

*Proof*

Let $\{a_n\}$ be a sequence that is infinitely large.
Then $\forall K \in {\mathbb R}$, $~\exists n^*$ such that $\forall n \geq n^*$,
$$|a_n| > K$$
Let $\{a_{n_i}\}$ be a subsequence of $\{a_n\}$.

Pick $n_{i^*} \geq n^*$

Then it follows that, $\forall n_i \geq n_{i^*} \geq n^*$,
$$\lvert a_{n_i} \rvert > K$$
$\blacksquare$

**2. p. 50 Ex. 2.1.7**
**Let $\{x_n\}$ be a sequence.
a) Show that $\lim{x_n}=0$ (that is, the limit exists and is zero) if and only if $\lim{|x_n|}=0$**

*Proof*

Suppose $\lim{x_n}=0$
Then by definition of a limit, we have:
$\forall \varepsilon > 0, ~\exists ~ n^*$ such that $\forall n \geq n^*$,
$$\lvert x_n - 0 \rvert < \varepsilon $$
It follows that, $\forall n \geq n^*$,
$$\vert \lvert x_n \rvert - 0 \rvert = \lvert x_n - 0 \rvert < \varepsilon $$
So $\lim{|x_n|}=0$.
$$ $$
Conversely, suppose $\lim{|x_n|}=0$
Then by definition of a limit, we have:
$\forall \varepsilon > 0, ~\exists ~ m^*$ such that $\forall n \geq m^*$,
$$\lvert \lvert x_n \rvert - 0 \rvert < \varepsilon$$
It follows that, $\forall n \geq m^*$,
$$\lvert x_n - 0 \rvert = \lvert \lvert x_n \rvert - 0 \rvert < \varepsilon$$
So $\lim{x_n}=0 ~~~\blacksquare$

**b) Find an example such that $\{|x_n|\}$ converges and $\{x_n\}$ diverges**

Consider $\{x_n\} = \{(-1)^n\}$.
$\forall n \in {\mathbb N}$,
$$|x_n|=|(-1)^n|=1$$
So $\{|x_n|\}$ converges, and $\lim{|x_n|}=1$. (Note that this is not a formal proof, as the exercise only asks for an example.)

Further, we have shown in class and in previous HW problems that $\{(-1)^n\}$ diverges.

**3. p. 50 Ex. 2.1.10
Show that the sequence $\{x_n\}=\{\frac{n+1}{n}\}$ is monotone, bounded, and use Theorem 2.1.10 to find the limit.**

[Desmos Link for Week 4 #3](https://www.desmos.com/calculator/peuzqn1cfy)

*$i.$ $\{x_n\}=\{\frac{n+1}{n}\}$ is monotone.*
*Proof*
$\forall n \in {\mathbb N}$,
$$x_n = \frac{n+1}{n} = \frac{n}{n}+\frac{1}{n} = 1 + \frac{1}{n} > 1 + \frac{1}{n+1} = \frac{(n+1)+1}{n+1} = x_{n+1}$$
So $\{x_n\}$ is monotone decreasing. $\blacksquare$

$$ $$

*ii. $\{x_n\}$ is bounded*
*Proof*
Choose $K=2$ (from our definition of bounded).
Then $\forall n \in {\mathbb N}$,
$$|x_n| = |\frac{n+1}{n}| \leq 2 = K$$
$\blacksquare$

iii. In order to use Theorem 2.1.10, I first need to show that $\inf{x_n}=1$
*Proof*
Given any $\varepsilon > 0$, let $n^* = \lceil \frac{1}{\varepsilon} \rceil + 1$
Then it follows that
$$x_{n^*} = \frac{(\lceil \frac{1}{\varepsilon} \rceil+1)+1}{\lceil \frac{1}{\varepsilon} \rceil+1} = 1 + \frac{1}{\lceil \frac{1}{\varepsilon} \rceil +1} < 1 + \frac{1}{\frac{1}{\varepsilon}}=1+\varepsilon$$
$\therefore \inf{x_n}=1$
Now by Theorem 2.1.10, since $\{x_n\}$ is monotone decreasing and bounded, we have
$$\lim{x_n}=\inf{x_n}=1$$
$\blacksquare$

**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 $x_n = x_k ~~\forall n \geq k$.**

*Proof*
By the definition of a limit, we have
$\forall \varepsilon > 0, ~\exists ~ n^*$ such that $\forall n \geq n^*$,
$$|x_n - x_k| < \varepsilon$$

Now suppose $\exists ~ m \geq k$ such that $x_m \neq x_k$
Then since $\{x_n\}$ is monotone, we have exactly one of the following:
$~~~~~$Case 1: $\{x_n\}$ is monotone increasing
$~~~~~~~~~~~~~~~~~~$so $\forall n \geq m, ~~~x_n \geq x_m > x_k$
$~~~~~$Case 2: $\{x_n\}$ is monotone decreasing
$~~~~~~~~~~~~~~~~~~$so $\forall n \geq m, ~~~x_n \leq x_m < x_k$

Both cases imply $\exists ~ \varepsilon = \frac{|x_m - x_k|}{2}$ such that $\forall n \geq m$,
$$|x_n - x_k| > \varepsilon$$
This contradicts our assumption that $\lim{x_n}=x_k$
$\therefore x_n = x_k ~~\forall n \geq k. ~~\blacksquare$

**5. p. 50 Ex. 2.1.16
Let $\{x_n\}$ be a sequence. Suppose there are two convergent subsequences $\{x_{n_i}\}$ and $\{x_{m_i}\}$. Suppose
$$ \lim{x_{n_i}} = a ~~~and~~~\lim{x_{m_i}} = b $$
where $a \neq b$. Prove that $\{x_n\}$ is not convergent, without using Proposition 2.1.17**

*Proof*
Suppose $\lim{x_n}=L$
Then $\forall \varepsilon > 0, \exists n^*$ such that $\forall n \geq n^*, |x_n - L|< \varepsilon$

By Proposition 2.1.6 (proven in class), we know that every convergent sequence has a unique limit. This means we have at least one of the following:
$~~~~~$*i.* $L \neq a$
$~~~~~$*ii.* $L \neq b$

Without loss of generality, suppose $L \neq a$.
Since $\lim{x_{n_i}} = a$, we have
$\forall \varepsilon > 0, \exists n_{i^*}$ such that $\forall n_i \geq n_{i^*}, |x_{n_i} - a|< \varepsilon$
But this means $\exists$ $\varepsilon = \frac{|L-a|}{3}$ where, for any given $n^*$, we can choose $n_{i^*} > n^*$ where, $\forall n_i \geq n_{i^*} > n^*$,
$$|x_{n_i}-L|>\frac{|L-a|}{3}=\varepsilon$$
This contradicts our assumption that $\lim{x_n}=L$
$\therefore \{x_n\}$ is divergent. $\blacksquare$

**6. p. 50 Ex. 2.1.17**
**Find a sequence $\{x_n\}$ such that for any y ∈ R, there exists a subsequence $\{x_{n_i}\}$ converging to y.**

We know that the set of rational numbers, ${\mathbb Q}$, is countable.
Consider the following sequence, which contains all rational numbers:
$$\{x_n\}=\{-1,0,1,-2,-\frac{3}{2}, -1, -\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2},2,-3,-\frac{8}{3},...\}$$
That is, the sequence that counts from $-m$ to $m$ in increments of $\frac{1}{m}, ~~ \forall m \in {\mathbb N}$
For any real number y, we can pick a subsequence $\{x_{n_i}\}$ that converges to y.
For example, pick $n_1$ such that $y+\frac{1}{2}<x_{n_1}<y+1$
Next, pick $n_2$ such that $y+\frac{1}{3}<x_{n_2}<y+\frac{1}{2}$
Continuing to construct the sequence choosing $n_k$ such that
$$y+\frac{1}{k+1} < x_{n_k} < y + \frac{1}{k}$$

we will have
$$\lim{x_{n_i}}=\lim{(y+\frac{1}{n})}=y$$

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## **Week 6**
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**2.2.4**
**$x_1=\frac{1}{2}$ and $x_{n+1}=(x_n)^2$**
**Show that $\{x_n\}$ converges and find $\lim{x_n}$**
First note that $\{x_n\}$ is bounded. In particular, we have
$$ 0<x_n\leq \frac{1}{2}$$
This can be shown with induction.
We have
$$0<\frac{1}{2}=x_1 \leq \frac{1}{2}$$
Now suppose for some $k$, we have
$$ 0<x_k\leq \frac{1}{2}$$
Then it follows that
$$ 0<x_k^2 = x_{k+1} \leq \frac{1}{4}$$

Next, it can be shown that $\{x_n\}$ is monotone decreasing. Since $\{x_n\}$ is bounded, we have

$$ 0<x_n\leq \frac{1}{2}$$
multiplying by $x_n$ yields
$$ 0<x_n^2=x_{n+1}\leq \frac{1}{2}x_n \leq x_n$$

Since the sequence is bounded and monotone, it must be convergent.

To find the limit, use the fact that
$$\lim\{x_{n+1}\}=\lim{x_n}=x$$
Then $x=x^2$, so $x=\{0,1\}$
But we know that $x\neq 1$ because $x_n \leq \frac{1}{2}$
Thus $\lim{x_n}=0 ~~~\blacksquare$

**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.**
Since $-1 \leq \cos(n) \leq 1$, it follows that
$$1-\frac{1}{n} \leq 1 - \frac{\cos(n)}{n} \leq 1+\frac{1}{n}$$
We also know that
$$\lim{(1-\frac{1}{n})}=\lim{(1+\frac{1}{n})}=1$$
So by the squeeze lemma we have
$$\lim{(\frac{n-\cos(n)}{n})} = \lim{(1-\frac{\cos(n)}{n})}=1$$

**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 proposition 2.2.5? Why or why not?**

$z_n = \frac{x_n}{y_n} = \frac{\frac{1}{n^2}}{\frac{1}{n}}=\frac{1}{n}$, so $\{z_n\}$ converges to $\lim{\frac{1}{n}}=0$

$w_n = \frac{y_n}{x_n} = \frac{\frac{1}{n}}{\frac{1}{n^2}}=n$, so $\{w_n\}$ diverges to infinity.

We cannot use proposition 2.2.5 because $\lim{x_n}=\lim{y_n}=0$, and these sequences are the divisors of $\{w_n\}$ and $\{z_n\}$ respectively.

**2.2.7**
**True or false, prove or find a counter example. If $\{x_n\}$ is a sequence such that $\{x_n^2\}$ converges, then $\{x_n\}$ converges.**

False. Consider $x_n:=\{(-1)^n\}$
In this case, $\{x_n^2\}$ converges to 1, but $\{x_n\}$ diverges.

**2.2.8**
**Show that $\lim\limits_{n\to \infty}{\frac{n^2}{2^n}}=0$**

We can use Lemma 2.2.12 (ratio test for sequences).
$$L:=\lim\limits_{n\to \infty}{\frac{|\frac{(n+1)^2}{2^{n+1}}|}{|\frac{n^2}{2^n}|}}=\lim\limits_{n\to \infty}{|\frac{(n+1)^2}{2^{n+1}}||\frac{2^n}{n^2}|}=\lim\limits_{n\to \infty}{\frac{n^2+2n+1}{2n^2}}=\frac{1}{2}$$

Since $L<1$, by Lemma 2.2.12, the sequence must converge and the limit is 0.

**Theorem 2.3.4**
**If $\{x_n\}$ is a bounded sequence, then there exists a subsequence $\{x_{n_k}\}$ such that**
$$\lim\limits_{n\to \infty}{x_{n_k}}=\lim\limits_{n\to \infty}{\sup{x_n}}$$
**Similarly, there exists a (perhaps different) subsequence $\{x_{m_k}\}$ such that**
$$\lim\limits_{n\to \infty}{x_{m_k}}=\lim\limits_{n\to \infty}{\inf{x_n}}$$

*Proof (Part 1)*
Define $a_n:=\sup\{x_k ~|~ k \geq n\}$
Define $x:=\lim{a_n}$
We will construct the following sequence:
Let $x_{n_1}=x_1$
By the definition of supremum, $\exists~ x_{n_2}$ such that
$$a_2-\frac{1}{2} < x_{n_2} < a_2~~~~~or~~~~~0\leq a_2 - x_{n_2} < \frac{1}{2}$$
Now we cannot choose any $n \leq n_2$, so $n_3 \geq n_2 + 1$
Consider the tail of $\{a_n\}$, in particular $\{a_{{n_2}+1}\}$
Again, by the definition of supremum, $\exists ~ x_{n_3}$ such that
$$a_{{n_2}+1}-\frac{1}{3} < x_{n_3} < a_{{n_2}+1}~~~~~or~~~~~0\leq a_{{n_2}+1} - x_{n_3} < \frac{1}{3}$$
Now suppose we found $x_{n_{k-1}}$
We cannot pick any $n \leq n_{k-1}$, so $n_k \geq n_{k-1}+1$
Consider the tail of $\{a_n\}$, in particular $\{a_{n_{k-1}+1}\}$
By the definition of supremum, $\exists ~ x_{n_k}$ such that
$$a_{{n_{k-1}}+1}-\frac{1}{k} < x_{n_k} < a_{{n_{k-1}}+1}~~~~~or~~~~~0\leq a_{{n_{k-1}}+1} - x_{n_k} < \frac{1}{k}$$
Further, since $\{a_{n_k}\}$ is monotone decreasing, note that
$$|a_{n_k}-x_{n_k}|=a_{n_k}-x_{n_k} \leq a_{n_{k-1}+1}-x_{n_k} < \frac{1}{k}$$
Since $\lim{a_n}=x$, it follows that $\lim{a_{n_k}}=x$
So $\forall \varepsilon > 0, ~~\exists M_1$ such that $\forall k \geq M_1$
$$|a_{n_k}-x|<\frac{\varepsilon}{2}$$
Choose $M_2$ such that $\frac{1}{M_2} \leq \frac{\varepsilon}{2}$
Let $M=\max\{M_1,M_2\}$
Then $\forall ~ k \geq M$, we have
$$|x-x_{n_k}|=|x-a_{n_k}+a_{n_k}-x_{n_k}|\leq|x-a_{n_k}|+|a_{n_k}-x_{n_k}|<\frac{\varepsilon}{2} + \frac{1}{k} \leq \frac{\varepsilon}{2} + \frac{1}{M_2} \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2}=\varepsilon$$
$$ $$
*Proof (Part 2)*
Define $b_m:=\inf\{x_k ~|~ k \geq m\}$
Define $x:=\lim{b_m}$
We will construct the following sequence:
Let $x_{m_1}=x_1$
By the definition of infimum, $\exists~ x_{m_2}$ such that
$$b_2 \leq x_{m_2} < b_2+\frac{1}{2} ~~~~~or~~~~~0\leq x_{m_2}-b_2 < \frac{1}{2}$$
Now we cannot choose any $m \leq m_2$, so $m_3 \geq m_2 + 1$
Consider the tail of $\{b_m\}$, in particular $\{b_{{m_2}+1}\}$
Again, by the definition of infinum, $\exists ~ x_{m_3}$ such that
$$b_{{m_2}+1} \leq x_{m_3} < b_{{m_2}+1}+\frac{1}{3} ~~~~~or~~~~~0\leq x_{m_3}-b_{{m_2}+1} < \frac{1}{3}$$
Now suppose we found $x_{m_{k-1}}$
We cannot pick any $m \leq m_{k-1}$, so $m_k \geq m_{k-1}+1$
Consider the tail of $\{b_m\}$, in particular $\{b_{m_{k-1}+1}\}$
By the definition of infimum, $\exists ~ x_{m_k}$ such that
$$b_{{m_{k-1}}+1} \leq x_{m_k}< b_{{m_{k-1}}+1}+\frac{1}{k} ~~~~~or~~~~~0\leq x_{m_k} - b_{{m_{k-1}}+1} < \frac{1}{k}$$
Further, since $\{b_{m_k}\}$ is monotone increasing, note that
$$|x_{m_k}-b_{m_k}|=x_{m_k}-b_{m_k} \leq x_{m_k} - b_{m_{k-1}+1} < \frac{1}{k}$$

Since $\lim{b_m}=x$, it follows that $\lim{b_{m_k}}=x$
So $\forall \varepsilon > 0, ~~\exists N_1$ such that $\forall k \geq N_1$
$$|b_{m_k}-x|<\frac{\varepsilon}{2}$$
Choose $N_2$ such that $\frac{1}{N_2} \leq \frac{\varepsilon}{2}$
Let $N=\max\{N_1,N_2\}$
Then $\forall ~ k \geq N$, we have
$$|x-x_{m_k}|=|x-b_{m_k}+b_{m_k}-x_{m_k}|\leq|x-b_{m_k}|+|b_{m_k}-x_{m_k}|<\frac{\varepsilon}{2} + \frac{1}{k} \leq \frac{\varepsilon}{2} + \frac{1}{N_2} \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2}=\varepsilon$$
$\blacksquare$

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## **Week 7**

---
**1. Without using Lemma on nested Intervals prove that
$$\bigcap_{n=1}^\infty[-\frac{1}{n}, \frac{1}{n}]=\{0\}.$$**
*Proof*
First note that $\forall ~ n \in {\mathbb N}$
$$-\frac{1}{n} < 0 < \frac{1}{n}$$
It follows that $\{0\} \in \bigcap_{n=1}^\infty[-\frac{1}{n}, \frac{1}{n}]$
Now suppose $\exists ~ [a,b] \in \bigcap_{n=1}^\infty[-\frac{1}{n}, \frac{1}{n}]$ such that $a \neq 0$ or $b \neq 0$
Without loss of generality, suppose $b \neq 0$
Then $b > 0$. But, $\forall n > \frac{1}{b}$ we have $\frac{1}{n} < b$.
This implies that $[a,b] \not\in \bigcap_{n=1}^\infty[-\frac{1}{n}, \frac{1}{n}]$, which contradicts our assumption. $\blacksquare$

*Alternate Proof (With Subsets)*
Note that $[-\frac{1}{n}, \frac{1}{n}]=[-\frac{1}{n},0] \cup [0,\frac{1}{n}].$
Suppose $\exists x \in \bigcap_{n=1}^\infty[-\frac{1}{n}, 0]$.
Then it follows that $\forall n \in {\mathbb N}$
$$-\frac{1}{n} \leq x \leq 0$$
Since $\lim\limits_{n\to \infty}{-\frac{1}{n}}=0$, we know that $x=0$, so $x \in [0,\frac{1}{n}]$
Therefore
$$\bigcap_{n=1}^\infty[-\frac{1}{n}, 0] \subseteq \bigcap_{n=1}^\infty[0,\frac{1}{n}]$$
Conversely, suppose $\exists y \in \bigcap_{n=1}^\infty[0,\frac{1}{n}]$
Then it follows that $\forall n \in {\mathbb N}$
$$0 \leq y \leq \frac{1}{n}$$
Since $\lim\limits_{n\to \infty}{\frac{1}{n}}=0$, we know that $y=0$, so $y \in [0,\frac{1}{n}]$
Therefore
$$\bigcap_{n=1}^\infty[0,\frac{1}{n}] \subseteq \bigcap_{n=1}^\infty[-\frac{1}{n}, 0]$$
We can concluded that
$$\bigcap_{n=1}^\infty[0,\frac{1}{n}] = \{0\} = \bigcap_{n=1}^\infty[-\frac{1}{n}, 0]$$
And so
$$\bigcap_{n=1}^\infty[-\frac{1}{n}, \frac{1}{n}]= \bigcap_{n=1}^\infty[-\frac{1}{n}, 0] \cup \bigcap_{n=1}^\infty[0,\frac{1}{n}]=\{0\}$$
$\blacksquare$

**2. Exercise 2.3.5**
**Let $x_n:=\frac{(-1)^n}{n}$. Find $\lim{\sup{x_n}}$ and $\lim{\inf{x_n}}$**

Since $\lim{x_n}=0$ (shown previously in class), it follows by Theorem 2.3.5 that
$$\lim{\sup{x_n}}=\lim{\inf{x_n}}=\lim{x_n}=0$$

**Part b**
**Let $x_n:=\frac{(n-1)(-1)^n}{n}$**
**Find $\limsup{x_n}$ and $\liminf{x_n}$**
First note that this sequence is bounded (let $K=2$ from our definition of bounded.)
By Theorem 2.3.4, we can construct a subsequence that converges to $\limsup{x_n}$ and a (perhaps different) subsequence that convergest to $\liminf{x_n}$
Consider the subsequences:
$\{x_{2k}: k \geq 1\}$
$\{x_{2k-1}: k \geq 1\}$
Note that $\forall n \in {\mathbb N}$, $x_n$ belongs to exactly one of the two subsequences defined above. Further, note that $\forall k \in {\mathbb N}$
$$x_{2k-1} \leq 0 < x_{2k}$$
It follows that $\limsup{x_{2k}}=\limsup{x_n}$
and
$\liminf{x_{2k-1}}=\liminf{x_n}$
Since $\{x_{2k}\}$ is monotone increasing and converges to 1, by Theorem 2.3.5,
$$\limsup{x_{2k}}=1=\limsup{x_n}$$
Similarly, since $\{x_{2k-1}\}$ is monotone decreasing and converges to -1, by Theorem 2.3.5,
$$\liminf{x_{2k-1}}=-1=\liminf{x_n}$$

**3. Prove that every unbounded sequence, contains an infinitely large subsequence. The proof goes by construction.**
*Proof*
Let $\{x_n\}$ be an unbounded sequence. Then $\forall K \in {\mathbb R}, \exists n^*$ such that $|x_{n^*}|>K$.
We can construct an infinitely large subsequence, $\{x_{n_i}\}$ as follows:
Let $x_{n_1}=x_1$
From the definition of unbounded, let $K=|x_{n_1}|+1$. Then $\exists ~ n_2>n_1$ such that
$$|x_{n_2}| > |x_{n_1}|+1$$

Since the sequence is unbounded, and there are a finite number of points in the head of the sequence $\{x_n: n \leq n_2\}$, the tail of the sequence $\{x_n: n \geq n_2+1\}$ must be unbounded. From the definition of unbounded, let $K=|x_{n_2}|+1$. Then we can choose $n_3 \geq n_2+1$ such that
$$|x_{n_3}| > |x_{n_2}|+1$$

We continue constructing the sequence in this way. Now suppose we have found $x_{i-1}$

From the definition of unbounded, let $K=|x_{n_{i-1}}|+1$. Then we can choose $n_i \geq n_{i-1}+1$ such that
$$|x_{n_i}| > |x_{n_{i-1}}|+1$$.

Now that our sequence $\{x_{n_i}\}$ has been constructed, consider any $M \in {\mathbb R}$.

Since the sequence is unbounded, $\exists$ $n_{i^*}$ such that $x_{n_{i^*}} > M$
And further, $\forall i \geq i^*$
$$|x_{n_i}| \geq |x_{n_{i^*}}| > M $$
$\therefore$ the subsequence $\{x_{n_i}\}$ is infinitely large. $\blacksquare$

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## **Week 10**
___

**Exercise 2.5.1**

For $r\neq1$, $\sum_{k=0}^{n-1}r^k=\frac{1-r^n}{1-r}$

*Proof*

Let $s:=\sum_{k=0}^{n-1}r^k$
So $s=1+r+r^2+...+r^{n-1}$
Note also that $rs=r+r^2+...r^{n-1}+r^n$
It follows that
$$s(1-r)=s-rs=1-r^n$$
And solving for s, we get
$$s=\frac{1-r^n}{1-r}$$
$\blacksquare$

**Exercise 2.5.2**
For $-1<r<1$,
$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r}$$
*Proof*
From exercise 2.5.1, we know that
$$\sum_{n=0}^{x-1}r^n=\frac{1-r^x}{1-r}$$
It follows that, for $-1<r<1$
$$\sum_{n=0}^{\infty}r^n=\lim_{x\to\infty}\frac{1-r^x}{1-r}=\frac{1}{1-r}$$
$\blacksquare$
**Exercise 2.5.3**
**Decide the convergence or divergence of the following series**
a)
$$\sum_{n=1}^{\infty}{\frac{3}{9n+1}}$$
This series diverges, as can be shown by the comparison test. $\forall n\geq1$:
$$\sum_{n=1}^{\infty}{\frac{3}{9n+1}}=3\sum_{n=1}^{\infty}{\frac{1}{9n+1}}\geq3\sum_{n=1}^{\infty}{\frac{1}{10n}}=\frac{3}{10}\sum_{n=1}^{\infty}{\frac{1}{n}}$$
Since $\sum{\frac{1}{n}}$ diverges to infinity, so too must $\sum{\frac{3}{9n+1}}$

b)
$$\sum_{n=1}^{\infty}{\frac{1}{2n-1}}$$
This series diverges, as can be shown by the comparison test. $\forall n\geq1$:
$$\sum_{n=1}^{\infty}{\frac{1}{2n-1}}\geq\sum_{n=1}^{\infty}{\frac{1}{2n}}=\frac{1}{2}\sum_{n=1}^{\infty}{\frac{1}{n}}$$
Since $\sum{\frac{1}{n}}$ diverges to infinity, so too must $\sum{\frac{1}{2n-1}}$

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## **Week 11**
___

**Exercise 2.5.3 c,d,e**

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

First, we can show that this series converges absolutely:

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

So $\sum_{n=1}^{\infty}{|\frac{(-1)^n}{n^2}|}$ is a p-series with p=2, and by **Proposition 2.5.15** (p-series test), since p>1, it converges.

Therefore the series $\sum_{n=1}^{\infty}{\frac{(-1)^n}{n^2}}$ converges absolutely, so by **Proposition 2.5.13**, the series converges. $\blacksquare$

$$ $$

**d)** Decide the convergence or divergence of:
$$\sum_{n=1}^{\infty}{\frac{1}{n(n+1)}}$$

Since $\forall n\geq 1, \frac{1}{n^2+n} < \frac{1}{n^2}$
it follows that
$$\sum_{n=1}^{\infty}{\frac{1}{n(n+1)}}=\sum_{n=1}^{\infty}{\frac{1}{n^2+n}}<\sum_{n=1}^{\infty}{\frac{1}{n^2}}$$

We know that $\sum{\frac{1}{n^2}}$ converges by the p-series test. Therefore by the comparison test, $\sum{\frac{1}{n(n+1)}}$ converges. $\blacksquare$

$$ $$

**e)** Decide the convergence or divergence of:
$$\sum_{n=1}^{\infty}{ne^{-n^2}}$$

We can apply the ratio test

$$\lim_{n \to \infty}{\frac{|(n+1)e^{-(n+1)^2}|}{|ne^{-n^2}|}}=\lim_{n \to \infty}{\frac{(n+1)e^{n^2}}{ne^{(n+1)^2}}}=\lim_{n \to \infty}{\frac{(n+1)e^{n^2}}{ne^{n^2+2n+1}}}=\lim_{n \to \infty}{\frac{(n+1)}{ne^{2n+1}}}$$

$$=\lim_{n \to \infty}{\frac{n}{ne^{2n+1}}}+\lim_{n \to \infty}{\frac{1}{ne^{2n+1}}}=\lim_{n \to \infty}{\frac{1}{e^{2n+1}}}+\lim_{n \to \infty}{\frac{1}{ne^{2n+1}}}=0+0=0$$

So by **Proposition 2.5.17** (Ratio test), the series converges. $\blacksquare$

**2.5.4**
**a)**
**Prove that if $\sum_{n=1}^{\infty}{x_n}$ converges, then $\sum_{n=1}^{\infty}({x_{2n}}+{x_{2n+1}})$ also converges**

Note that the union of the sequences $x_{2n}$ and $x_{2n+1}$ is a subsequence of $x_n$.
Specifically, this union includes all terms in $x_n$ except for $x_1$.

It follows that

$$\sum_{n=1}^{\infty}({x_{2n}}+{x_{2n+1}})=\sum_{n=1}^{\infty}{x_{2n}}+\sum_{n=1}^{\infty}{x_{2n+1}}=-x_1+\sum_{n=1}^{\infty}{x_n}$$

We know $\sum{x_n}$ converges, and $x_1$ is clearly finite.
Therefore $\sum_{n=1}^{\infty}({x_{2n}}+{x_{2n+1}})$ also converges. $\blacksquare$

**b)**
**Find an explicit example where the converse does not hold.**

Note that $\sum{((-1)^{2n}+(-1)^{2n+1})}=\sum{0}$ and therefore converges.
But, the series $\sum{(-1)^n}$ does not converge.

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## **Week 12**
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**Lemma 3.1.7**
Let $S=(a,b)$ where $a,b \in {\mathbb{R}}$ and let c be a cluster point on the interior of $S$ ($c \neq a$ and $c \neq b$). Let $f:S \to {\mathbb{R}}$ be a function. Then $f(x) \to L$ as $x \to c$ if and only if for every sequence {$x_n$} of numbers such that $x_n \in S$ and $x_n \neq c ~~\forall n$, and such that $\lim{x_n}=c$, we have that the sequence {$f(x_n)$} converges to $L$.

*Proof*

*Calc definition $\to$ Sequential definition*

Suppose $f(x) \to L$ as $x \to c$, and {$x_n$} is a sequence such that $x_n \in S$, $x_n \neq c$, and $\lim{x_n}=c$

We wish to show that {$f(x_n)$} converges to $L$.
Let $\varepsilon > 0$ be given. Find $\delta > 0$ such that if $x \in S, x \neq c$ and $|x-c|<\delta$ then $|f(x)-L|<\varepsilon$

From the definition of a convergent sequence, we can chooose $M$ such that $\forall n \geq M$, $|x_n - c| < \delta$
$\therefore ~|f(x_n)-L| < \varepsilon$
So {$f(x_n)$} converges to $L$.

*Sequential definition $\to$ Calc definition*

Suppose the calculus definition fails.
That is, $\exists \varepsilon > 0$ such that $\forall \delta > 0$, $\exists x \in S, x \neq c$ such that $|x-c|<\delta$ and $|f(x)-L| \geq \varepsilon$.
Choose $\delta = \frac{1}{n}$ to construct a sequence {$x_n$}.

We know that $\exists \varepsilon > 0$ such that $\forall n$, $\exists x_n \in S, x_n \neq c$ where $|x_n - c| < \frac{1}{n}$ and $|f(x_n)-L| \geq \varepsilon$
The sequence {$x_n$} converges to c, but the sequence {f$(x_n)$} does not converge to $L$. So the sequential definition fails as well, and this direction is proven via contrapositive.
$\blacksquare$

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## **Week 13**
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**Exercise 6.6.1**
Let $f$ and $g$ be bounded functions on $[a,b]$. Prove
$$||f+g||_u \leq ||f||_u + ||g||_u$$

*Proof*
By **definition 6.1.9**, for all $x$ in the domains of $f$ and $g$, we have
$$||f+g||_u = \sup\{|f(x)+g(x)|\}$$
We know by the triangle inequality that $|f(x)+g(x)| \leq |f(x)|+|g(x)|$
It follows that

$$\sup\{|f(x)+g(x)|\} \leq \sup\{|f(x)|\}+\sup\{|g(x)|\}=||f||_u + ||g||_u$$
$\blacksquare$

**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 ${f_{n}g_{n}}$ (the multiple) does not converge uniformly to $fg$ on $A$.*

Let $A=[0,1]$
Let ${f_n}={x+\frac{1}{n}}$
Let ${g_n}={x+\frac{2}{n}}$
So $f_{n}g_{n} = x^2+\frac{3x}{n}+\frac{2}{n^2}$

${f_n}$ and ${g_n}$ both converge uniformly to $x$, but $f_{n}g_{n}$ does not converge uniformly to $x^2$, as shown in the Desmos link below.

[Link to Desmos for 6.1.6](https://www.desmos.com/calculator/glsjic9uey)

$\gamma$