# Sequences by Catherine and Liz
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**Week 1**
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**Definition 1.** $\{x_n\}$ is constant if $x_n=x_{n+1}~\forall ~n~\in~{\mathbb N}$.

**Example 1.** $\{17\}$ is constant since $17=x_n=x_{n+1}~\forall ~n~\in~{\mathbb N}$.

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

**Example 1'.** $\{(-1)^n\}$ is not constant since $(-1)^1= -1$ and $(-1)^2 = 1$.

**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.
The only values that $(-1)^n$ can be are 1 and -1. Therefore, $~\forall ~K~\geq1,\,x_n\leq K~\forall~n\in~{\mathbb N}$.

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

**Example 2'.** Prove that $\{2^n\}$ is not bounded above.
Pick any K. Let n*=$\lceil log_2 K\rceil +1$
$2^{n*}=2^{\lceil log_2 K\rceil +1}$ > K

**Example 2'.** Prove that $\{2^n\}$ is not bounded above.
Pick any $K\in~{\mathbb R}$. Let n*=$\lceil \log_2 |K|\rceil +1$
$2^{n*}=2^{\lceil \log_2 |K|\rceil +1}$ > K.

**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 $\{(-1)^n\}$ is bounded below. The only values that $(-1)^n$ can be are 1 and -1. Therefore, $~\forall ~K~\leq-1,\,x_n\geq K~\forall~n\in~{\mathbb N}$.

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

**Example 3'.** Prove that $\{-2n\}$ is not bounded below.
Pick any $K\in~{\mathbb R}$. Let n*= $\mid\lceil\frac{-K}{2}+1\rceil\mid$
-2n*$=-2(\mid\lceil\frac{-K}{2}+1\rceil \mid$) $\leq -\mid K\mid<K$

**Definition 4.** $\{x_n\}$ is bounded if $\exists$ K $\in~{\mathbb R}$ such that $\mid x_n\mid$ $\leq$ K $\forall~n~\in~{\mathbb N}$.

**Example 4.** Prove that {$\sin n$} is bounded.
$\forall~n\in~{\mathbb N}$, $-1\leq$ $\sin n$ $\leq 1$
Therefore, $~\forall ~K~\geq1,\,x_n\leq K,~\forall~n\in~{\mathbb N}$. Thus, {$\sin n$} is bounded above.
Therefore, $~\forall ~K~\leq-1,\,x_n\geq K,~\forall~n\in~{\mathbb N}$. Thus, {$\sin n$} is bounded below.

**Counterdefinition 4.** $\{x_n\}$ is not bounded if $\forall~K\in~{\mathbb R}$, $\exists \mid x_n\mid$ >K.

**Example 4'.** Prove that $\{4n\}$ is not bounded.
Pick any $K\in~{\mathbb R}$. Let n*= $\mid\lceil\frac{K}{4}+1\rceil\mid$.
4n*$=4(\mid\lceil\frac{K}{4}+1\rceil\mid$)$\geq \mid K\mid>K$
Therefore $\{4n\}$ is not bounded above, and thus is not bounded.

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

Infinitely large means that $\forall K \in~{\mathbb R}$, $\exists$ $n^* \in~{\mathbb N}$ such that $\forall n\geq n^* ,\mid x_n\mid$$\geq K$.

Pick any $K\in~{\mathbb R}$. Let $n^* = n$. By the definition of infinitely large, $\mid x_n\mid$$\geq K$, so $\exists$ $n \in~{\mathbb N}$ such that $\mid x_n\mid$$\geq K$. Thus $\{x_n\}$ is unbounded.

2. Prove that $\{(-1)^nn\}$ is inf. large.

Pick any $K\in~{\mathbb R}$. Let n*= $|\lceil K \rceil|$ $+1$.
$\mid x_n\mid$ = $\mid(-1)^nn\mid$ $= n$
So $\mid x_n*\mid$ $= n^* > K$

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}$$,$ $\exists$ $n^*\geq n$ such that $\mid x_n\mid$$\leq K$.

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.

Not bounded:
Pick any $K\in~{\mathbb R}$. Let $n^* = 2|\lceil K \rceil| > K$.

Not infinitely large:
Let $n^*=|\lceil K \rceil|$.
If $n$ is odd, then $x_n=0$, $x_{n+1}=n + 1$, and $x_{n+2}=0$.
If $n$ is even, then $x_n=n$, $x_{n+1}=0$.
Therefore, $\forall n \in~{\mathbb N}$$,$ $\exists$ $n^*\geq n$ such that $\mid x_n\mid$$\leq K$.

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**Week 2**
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**5. Counter the definition of $\lim x_n=L$.**

$\lim x_n\neq L$ iff $\exists \varepsilon^* > 0$ such that $\forall m$ $\exists n^*\geq m$ such that $\mid x_{n^*} - L\mid \geq \varepsilon$

**6. Prove that $\lim(-1)^n\neq 0$.**

Let $\varepsilon ^*= \frac{1}{2}$. Pick any $m \in~{\mathbb N}$.
Let $n^* = 2m$.
Then $\mid x_{n^*} - L\mid = \mid (-1)^{2m} - 0\mid = \mid 1 - 0\mid = 1\geq \frac{1}{2}$ .

**7. Prove that $\lim\frac{1}{n^2}$=0.**
Take any $\varepsilon > 0$.
Let $n^* = \lceil \frac{1}{\sqrt\varepsilon} \rceil$ .
Then $\forall n > n^*$,
$\mid \frac{1}{n^2} - 0\mid$ $= \frac{1}{n^2} \leq\frac{1}{({n^*})^2}$ $= \frac{1}{(\lceil \frac{1}{\sqrt\varepsilon} \rceil)^2}$ $\leq \frac{1}{( \frac{1}{\sqrt\varepsilon} )^2}$ $=\varepsilon$

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**Week 3**
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**1. Prove that $\lim (-1)^n$ does not exist.**
Pick any $a\in R$. Prove $\lim (-1)^n \neq a.$
Case 1: $\mid a\mid = 1$
Let $\varepsilon^* = \frac{1}{2}$. Let $n^* = 2n + 1$.
$\mid (-1)^{2n+1} - 1 \mid =\mid -1 - 1\mid = 2 \geq \frac{1}{2}$
Therefore $\lim (-1)^n \neq \mid 1\mid.$

Case 2: $\mid a\mid \neq 1$
Let $\varepsilon^* = \frac{\mid\mid a \mid - 1\mid}{2}$. Let $n^* = 2n$.
$\mid (-1)^{2n} - \mid a\mid \mid$ $= \mid 1- \mid a \mid\mid$ $=\mid \mid a \mid -1\mid$$>\frac{\mid\mid a \mid - 1\mid}{2} = \varepsilon^*$
Therefore $\lim (-1)^n \neq \mid a\mid$, when $a \neq 1$.

Due to Case 1 and Case 2, $\lim (-1)^n$ does not exist.

**2. Proof of Prop 2.1.6**
A convergent sequence has a unique limit.

Proof by contradiction.

Let $x, y \in R$.
Assume tht $\lim x_n = y$ , $\lim x_n = y$ , and $x \neq y$.

Let $\varepsilon = \frac {\mid x-y \mid}{3}$.
Then $\exists$ $n^*$ such that $\forall$ $n \geq n^*$, $\mid x_n - x \mid < \varepsilon$
and $\exists$ $n'$ such that $\forall$ $n \geq n'$, $\mid x_n - y \mid < \varepsilon$

Let $m =$ max{$n^*, n'$}

$\mid x - y\mid = \mid x -x_m + x_m - y \mid$
$\leq \mid x - x_m \mid + \mid x_m - y \mid = \mid x_m - x\mid + \mid x_m - y \mid$
$<\frac {\mid x-y \mid}{3} + \frac {\mid x-y \mid}{3} = \frac{2}{3}\mid x-y \mid$

So $\mid x-y\mid <\frac{2}{3}\mid x-y \mid$, which is a contradiction.
Therefore, x = y and the limit is unique.

**4. Prove that inf(0,1) = 0**
1) $\forall$ $x \in (0,1)$, $x \geq 0$.
2) Take any $\varepsilon > 0$.
$x^* = 0 + \frac{\varepsilon}{2}$
$0 + \varepsilon > 0 + \frac{\varepsilon}{2} > 0$

**5. Let A = (0,1) $\cup$ {2}. Prove that sup A = 2.**
1) $\forall$ $x \in (0,1)$ $\cup$ {2}, $x \leq 2$.
2) Take any $\varepsilon > 0$.
$x^* =$ max {$2 - \frac{\varepsilon}{2}, \frac{1}{2}$}
$2 - \varepsilon < 2 - \frac{\varepsilon}{2} <2$

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**Week 4**
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**2. If $\{x_n\}$ is convergent, and $\{y_n\}$ is divergent, what can you say about $\{x_n+y_n\}$?**

$\{x_n+y_n\}$ is divergent.

$\{x_n\}$ is convergent so lim$\{x_n\}$ = x.
$\{x_n\} + \{y_n\} := \{z_n\}$
Assume $\{z_n\}$ converges to z.

Let $\varepsilon > 0$. $\exists$ $n^*$ such that $\forall n>n^*$, $\mid x_n - x \mid < \varepsilon$ and $\mid z_n - z \mid < \varepsilon$
So $\mid (x_n\ +y_n)-z\mid <\varepsilon$.
$\mid y_n -(z-x)\mid$ = $\mid y_n -z+x\mid$ = $\mid x - x_n +x_n +y_n - z \mid$
$\leq \mid x - x_n \mid + \mid x_n +y_n - z \mid$ by the triangle inequality
$< \varepsilon + \varepsilon = 2\varepsilon$
So $\{y_n\}$ converges to z - x.
However, this is a contradiction, since $\{y_n\}$ is divergent.
Therefore, if $\{x_n\}$ is convergent, and $\{y_n\}$ is divergent, then $\{x_n+y_n\}$ is divergent.

**3. If $\{x_n\}$ is convergent, and $\{y_n\}$ is divergent, what can you say about $\{x_n \bullet y_n\}$?**

The product may be convergent or divergent.

Let $\{x_n\} = \{\frac{1}{2^n}\}$ and $\{y_n\} = {(\frac{3}{2})}^n\}$. $\{x_n \bullet y_n\}$ converges to $0$.

Let $\{x_n\} = \{\frac{1}{n}\}$ and $\{y_n\} = \{{2}^n\}$. $\{x_n \bullet y_n\}$ diverges.

**4. If $\{x_n\}$ is divergent, and $\{y_n\}$ is divergent, what can you say about $\{x_n+y_n\}$?**

The sum may be convergent or divergent.

Let $\{x_n\} = \{(-1)^n\}$ and $\{y_n\} = \{(-1)^{n+1}\}$. $\{x_n + y_n\}$ converges to $0$.

Let $\{x_n\} = \{2^n\}$ and $\{y_n\} = {(\frac{3}{2})}^n\}$. $\{x_n + y_n\}$ diverges.

**5. If $\{x_n\}$ is divergent, and $\{y_n\}$ is divergent, what can you say about $\{x_n \bullet y_n\}$?**

The product may be convergent or divergent.

Let $\{x_n\} = \{(-1)^n\}$ and $\{y_n\} = \{(-1)^{n+1}\}$. $\{x_n \bullet y_n\}$ converges to $-1$.

Let $\{x_n\} = \{2^n\}$ and $\{y_n\} = {(\frac{3}{2})}^n\}$. $\{x_n \bullet y_n\}$ diverges.

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**Week 5**
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**1. Ex. 2.1.15**
Let ${x_n}$ be a sequence defined by
$x_n :=$ \{$n$ if $n$ is odd,
$\space$ $\space$ $\space$ $\space$ $\space$ \{$\frac{1}{n}$ if $n$ is even

*a) Is the sequence bounded?*
No, ${x_n}$ is not bounded.
For any $m \in R$, $\exists n\in N$ such that $n>m$.
If $n$ is odd, then $x_n = n >m.$
If $n$ is even, then $n + 1$ is odd, and $x_{n+1} = n + 1 >m$

*b) Is there a convergent subsequence?*
Yes, the subsequence of all of the elements where $n$ is even converges to 0.
{$x_{n_k}$} = {$x_{2n}$} = {$\frac{1}{2n}$} converges to 0.

**Ex. 2.1.17**

Within [0, 1], define a sequence $S_0$ := {0, 1, 0, $\frac {1}{2}$, 1, 0, $\frac {1}{4}$, $\frac {2}{4}$, $\frac {3}{4}$, 1, 0, $\frac {1}{8}$, $\frac {2}{8}$,$\frac {3}{8}$, $\frac {4}{8}$, $\frac {5}{8}$, $\frac {6}{8}$, $\frac {7}{8}$, 1, ...}
To get the elements in the rest of the unit intervals:
$S_{-1}$ := -1 + $S_0$
$S_{1}$ := 1 + $S_0$
$S_{-2}$ := -2 + $S_0$
$S_{2}$ := 2 + $S_0$
$\space$ $\space$ $\vdots$
$S_{-n}$ := $-n$ + $S_0$
$S_{n}$ := $n$ + $S_0$
So $S$ = $S_k$ where $\mid k \mid \in N \cup \{0\}$.

**2. Find where in the proof of the lemma we used the fact that the inequality cannot be strict.**

Lemma
{$x_n$}$\to L$ and $\exists$ b $\in R$ such that $\forall$ $n$ $\in N$, $x_n \leq b \Rightarrow$ lim $x_n \leq b$.

In our proof by contradiction, we assumed that L > b.
Let $\epsilon = \frac {L-b}{2}$.
Then $\exists$ m such that $\forall n \geq m$,$\mid x_n - L \mid< \frac {L-b}{2}$
then, $\frac {b-L}{2} <x_n - L < \frac {L-b}{2}$
$\frac {b+L}{2}< x_n <\frac {3L-b}{2}$
$x_n > \frac {b+L}{2}>\frac {b+b}{2}=b$
$x_n > b$ contradicts $x_n \leq b$.

We used the fact that the inequalities in the lemma cannot be strict in our assumption that L > b.

If the lemma said "$\forall n$, $x_n < b \Rightarrow$lim $x_n < b$", in our proof by contradiction we would need to assume that L $\geq$ b. This allows the possibility that L = b.
In that case, $\frac {b-L}{2} = \frac {L-b}{2} = 0$.
$\Rightarrow0 < x_n - L< 0$
$\Rightarrow L < x_n < L$
$\Rightarrow b < x_n$

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**Week 6**
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**1. Prove that if $lim(b_n - a_n)=0$, then the intersection in the Lemma on Nested Intervals contains exactly one point.**

Lemma on Nested Intervals: $\forall~i~\in~{\mathbb N}$, [$a_i, b_i$] $\supset$ [$a_{i+1}, b_{i+1}$]

$lim(b_n - a_n)=0$
then $lim$ $b_n -$ $lim$ $a_n=0$
$lim$ $b_n =$ $lim$ $a_n$
inf $b_n$ = sup $a_n$
B = A
So $\forall n$, $a_n \leq A = B\leq b_n$
So the point A = B $\in$ [$a_n, b_n$] is unique.

**2. What happens if we use open intervals in the Lemma on Nested Intervals instead of the closed ones?**
If the intervals are open, there could be no intersection, a single point of intersection, or an interval.

No intersection:
$\overset{\infty}{\underset{i=1}{\bigcap}}$ (0, $\frac{1}{n}$)= $\varnothing$

Single point of intersection:
$\overset{\infty}{\underset{i=1}{\bigcap}}$ ($\frac{-1}{n}$, $\frac{1}{n}$)= {0}

An interval:
$\overset{\infty}{\underset{i=1}{\bigcap}}$ (1 $-\frac{1}{n}$, 2 $+\frac{1}{n}$) = (1, 2)