# Sequences by Alyssa Eller and Samantha Rangos
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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 $x_n=17$ and $x_{n+1}=17$.

**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 $(-1)^1$ $\neq$ $(-1)^2$.

**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.

Let $k=1$. $\{(-1)^n\} \leq 1$ $\forall~n\in~{\mathbb N}$. Therefore $\{(-1)^n\}$ is bounded above.

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

**Example 2'.** Prove that $\{2^n\}$ is not bounded above.

First, assume that $\{2^n\}$ is bounded above. This means $\exists K~\in~{\mathbb R}$ such that $2^n\leq K~\forall~n\in~{\mathbb N}$. However, $2^{n+1}>2^n$, so $2^n$ has no upper limit and therefore is not bounded above.

Proof by counterexample: Let $k=5$. For $n=3$, $2^3>5$.

**Definition 3.** $\{x_n\}$ is bounded below if $\exists K~\in~{\mathbb R}$ such that $x_n\geq K~\forall~n\in~{\mathbb N}$; $K$ is called the lower bound.

**Example 3.** Prove that $\{(-1)^n\}$ is bounded below.

Let $k=-1$. $\{(-1)^n\} \geq -1$ $\forall~n\in~{\mathbb N}$. Therefore $\{(-1)^n\}$ is bounded below.

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

**Example 3'.** Prove that $\{-2n\}$ is not bounded below.

First, assume that $\{-2n\}$ is bounded below. This means $\exists K~\in~{\mathbb R}$ such that $-2n\geq K~\forall~n\in~{\mathbb N}$. However, $-2n>-2(n+1)$, so $-2n$ has no lower limit and therefore is not bounded below.

**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.

The values of {$\sin n$} lie between $[-1, 1]$ $\forall~n~\in~{\mathbb N}$. Suppose $K=2$. $\mid \sin n\mid$ $=1$ which is $\leq2$. Thus, {$\sin n$} is bounded.

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

**Example 4'.** Prove that $\{4n\}$ is not bounded.

First, assume that $\{4n\}$ is bounded. This means $\exists K~\in~{\mathbb R}$ such that $\mid4n\mid\leq K~\forall~n\in~{\mathbb N}$. However, $4n<4(n+1)$, so $4n$ has no upper limit and therefore is not bounded.

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

Suppose $\{x_n\}$ is infinitely large. Then, we know $\forall~K~\in~{\mathbb R}$, $\exists~n^*~\in~{\mathbb N}$ such that $\forall~n~\geq n^*$ $\mid x_n\mid > K$. Since $\forall~n~$ $\mid x_n\mid > K$, by the definition of unbounded $\{x_n\}$ is also unbounded.

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

$\forall~K~\in~{\mathbb R}$, $\exists~n^*~\in~{\mathbb N}$, namely $n^*=5$, such that $\forall~n~\geq 5$ $\mid (-1)^nn\mid > K$.

3. Counter the definition of infinitely large.

$\{x_n\}$ is not infinitely large if $\exists~K~\in~{\mathbb R}$ where $\forall~n^*$ $\exists~n \in~{\mathbb N}<n^*$, $\mid x_n\mid < K$.

Geometrically, for any pair of horizontal lines, there is a tail not between the lines; the sequence is getting further from the x-axis.

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

Suppose $x_n$ is infinitely large. Then, $\forall~K~\in~{\mathbb R}$ there exists an $n^*$ such that for all $n\geq n^*$, $\mid x_n\mid \geq K$. However, if $n$ is odd, $x_n =0$, so there will always be an $n \geq n^*$ such that $x_n< K$. Therefore $x_n$ is not infinitely large. In addition, $\forall~K~\in~{\mathbb R}$, $\exists~n~\in~{\mathbb N}$ such that $\mid x_n\mid$ > $K$ because $2(n+1)>2n$. Therefore, $\{x_n\}$ is unbounded.

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

$\lim x_n \neq L$ if and only if $\exists \epsilon>0$ $\forall m$ $\exists n^* \geq m$ such that $\mid x_{n^*}-L\mid \geq \epsilon$.

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

Let $\epsilon^*=1/2.$ Then, pick any $m\in~{\mathbb N}$. Let $n^*=m+1 \geq m$. Then, $\mid x_{n^*}-L\mid = \mid(-1)^{m+1}-0\mid =1 \geq1/2.$

**7. Prove that $\lim\frac{1}{n^2}$=0.**

Take any $\epsilon >0$. Let $n^*=\lceil\frac{1}{\sqrt\epsilon} \rceil$. Then, $\forall n \geq \lceil\frac{1}{\sqrt\epsilon} \rceil, \mid \frac{1}{n^2} - 0\mid = \frac{1}{n^2} \leq \lceil\frac{1}{\sqrt\epsilon} \rceil < \epsilon$.