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title: Sequences by Kourtney and Nange

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# Sequences by Kourtney and Nange

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**Week 6**
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**Lemma on Nested Intervasl:**
For sequence of nested intervals $\{[a_n,b_n]\}$ with $\lim(b_n-a_n)=0$, the intersection $\bigcap^{\infty}_{n=1}[a_n,b_n]\neq \emptyset$

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

Let $\lim a_n=:X$ and $\lim b_n=:Y$, suppose $\lim(b_n-a_n)=0$. By the limit properties this means that $\lim b_n-\lim a_n=0$ $\implies$ $Y-X=0$ $\implies$ $Y=X$. 
 
$\forall n \in {\mathbb N}, X \geq a_n$, and $Y \leq b_n$, so $a_n \leq X=Y \leq b_n$
Thus there is some element namely $Y=X$ that exists inbetween $a_n$ and $b_n$ for all n, therefore $\bigcap^{\infty}_{n=1}[a_n,b_n]\neq \emptyset$.

Now we want to prove there are no other points in $\bigcap^{\infty}_{n=1}[a_n,b_n]$ .

Assume there is any $a_k \neq X$ in {$a_n$} or any $b_k \neq Y$ in {$b_n$} that is also in $\bigcap^{\infty}_{n=1}[a_n,b_n]$, then we can make $\epsilon' = X-a_k$, and $\exists a_{k+1}$ s.t. $X-\epsilon'=a_k < a_{k+1}\leq X$. In the same way, we can make $\epsilon^* = b_k-Y$, and $\exists b_{k+1}$ s.t. $Y+\epsilon^*=b_k > b_{k+1}\geq Y$. We can see that $a_k \notin [a_{k+1}, b_{k+1}]$ and $b_k\notin [a_{k+1}, b_{k+1}]$, so $a_k \notin \bigcap^{\infty}_{n=1}[a_n,b_n]$ and $b_k\notin \bigcap^{\infty}_{n=1}[a_n,b_n]$.
This proves $X=Y$ is the only point in $\bigcap^{\infty}_{n=1}[a_n,b_n]$.
 
**2. What happens if we use open intervals in the Lemma on Nested Intervals instead of the closed ones?**
If open intervals are used then there is no guarantee that there is some element in the intersection of $a_n$ and $b_n$, it is possible that $\bigcap^{\infty}_{n=1}(a_n,b_n)=\emptyset$
For example, make {$a_n$}={0}, {$b_n$}={$\frac1n$}, $\{(0,\frac1n) \}$ is a sequence of nested open intervals  with $\lim(b_n-a_n)=0$, since $\lim0=0$ and $\lim\frac1n=0$, then $\bigcap^{\infty}_{n=1}(0,\frac1n)=\emptyset$
     
     
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**Week 5**
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**1.Exercise 2.1.15:** Let {$x_n$} be a sequence defined by $$x_n：=\begin{cases} n,&\text{if $n$ is odd}\\ 1/n,& \text{if $n$ is even}\end{cases}$$

a) Is the sequence bounded? (prove or disprove)
The sequence is not bounded.
For any $K~\in~{\mathbb R},$ make $n=2\lceil \mid K\mid\rceil+1,$ then $\mid x_n\mid=x_n = n=2\lceil \mid K\mid\rceil+1> K,$ That proves $\forall K~\in~{\mathbb R},$ $\exists~n\in~{\mathbb N}$ such that $\mid x_n\mid > K~$, so $\{x_n\}$ is unbounded. 

b) Is there a convergent subsequence? If so, find it.
Yes, there is a convergent subsequence $\{x_{2n}\}={\frac 1{2n}}$ . 

**Exercise 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 {$x_n$} be a sequence containing all rationals. Suppose there is a subsequence {$x_{n_i}$} with terms found as $x_{n_1}$, $x_{n_2}$, ...$x_{n_m}$. Let $\epsilon$ be the smallest distance between y and the finitely many terms of the subsequence chosen so far, then there are infinitely many rationals in the interval ($y-\frac\epsilon2, y+\frac\epsilon2$). This infinite collection of rationals exist somewhere among the sequence {$x_n$}, so there must be one of the $x_n$ with $n>n_m$ in the interval, and this gives the next term of the subsequence. Infinitely many terms can be added to the subsequence, repeating the same process. it is clear that the subsequence converges to y, since the choice of the first few terms is irrelevant for the limit.

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

**Lemma**: {$x_n$} is convergent, and $\exists$ b $\in~{\mathbb R}$ s.t. $\forall~n~\in~{\mathbb N}$, ${x_n}\leq b$ then $\lim x_n\leq b$.

**Proof**: Let $\lim x_n=L$ and assume $L>b$. Let $\epsilon = \frac {L-b}2$, then $\exists~m\in~{\mathbb N}$ s.t. $\forall n \geq m$, $\mid x_n - L\mid < \frac {L-b}2$, then 

$\frac {b-L}2<x_n - L < \frac {L-b}2$, 

$\frac {b+L}2<x_n <\frac {3L-b}2$, 

$x_n>\frac {b+L}2>\frac {b+b}2=b$

This is contradiction to ${x_n}\leq b$, so $\lim x_n=L \leq b$ has to be true.

In the proof, we used the fact that the inequality cannot be strict by assumeing $L>b$, not $L \geq b$

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**Week 4**
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**1. 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.
 If {$x_n+y_n$} is convergent, while {$x_n$} is also convergent, we know that {$x_n+y_n-x_n$}={$y_n$} must be convergent. However this is against the existent statement.
 Therefore {$x_n+y_n$} must be divergent.

**2. If {$x_n$} is convergent, and {$y_n$} is divergent, what can you say about {$x_n*y_n$}?**
{$x_n*y_n$} could be convergent or divergent
{$1/n$} is convergent, {$n$} is divergent, {$(1/n)*n$}={1} is convergent.
{$2$} is convergent, {$n$} is divergent, {$2*n$}={2n} is divergent.

**3. If {$x_n$} is divergent, and {$y_n$} is divergent, what can you say about {$x_n+y_n$}?**
{$x_n+y_n$} could be convergent or divergent
{$n$} is divergent, {$1-n$} is divergent, {$n+(1-n)$}={1} is convergent.
{$n$} is divergent, {$2n$} is divergent, {$n+2n$}={3n} is divergent.

**4.  If {$x_n$} is divergent, and {$y_n$} is divergent, what can you say about {$x_n*y_n$}?** 
{$x_n*y_n$} could be convergent or divergent
{$(-1)^n$} is divergent, {$(-1)^{n+1}$} is divergent, {$(-1)^n*(-1)^{n+1}$}={-1} is convergent.
{$(-1)^n$} is divergent, {$n$} is divergent, {$(-1)^n*n$}={$n(-1)^n$} is divergent.


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**Week 3**
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**1. Prove that $\lim(-1)^n$ $DNE$.**

Pick any $a~\in~{\mathbb R}$. Prove that $\lim(-1)^n \neq a$.

If $a=1$, let $\epsilon^* =1$ and pick any $m~\in~{\mathbb N}$. Let $n^* = 2m+1 \geq m$. 
$\mid x_{n^*} - a\mid = \mid (-1)^{2m+1} - 1\mid = 2 > 1$, so $\lim(-1)^n \neq 1$.

If $a=-1$, let $\epsilon^* =1$ and pick any $m~\in~{\mathbb N}$. Let $n^* = 2m \geq m$. 
$\mid x_{n^*} - a\mid = \mid (-1)^{2m} - (-)1\mid = 2 > 1$, so $\lim(-1)^n \neq -1$.

For any $a$ that $\mid a \mid \neq 1$. Let $\epsilon^* =\frac {\mid \mid a \mid -1\mid}2$ and pick any $m~\in~{\mathbb N}$. Let $n^* = 2m \geq m$. 
$\mid x_{n^*} - a\mid = \mid (-1)^{2m} - a\mid =\mid 1-a \mid = \mid a -1 \mid \geq \mid \mid a \mid -1\mid > \frac {\mid \mid a \mid -1\mid}2 = \epsilon^*$ while $\mid a \mid \neq 1$. Thus, $\mid x_{n^*} - a\mid > \epsilon^*$. So $\lim(-1)^n \neq a$ when $\mid a \mid \neq 1$.

All of the above proves that $\lim(-1)^n$ $DNE$.

**2. Prove Proposition 2.1.6 - A convergent sequence has a unique limit.**

Assume that there is a $\lim \{x_n\} = x$, and $\lim \{x_n\} = y$, $x \neq y$. Let $\epsilon = \frac {\mid x-y \mid}3$, then $\exists n^*, n'$ such that $\forall n \geq n^*, \mid x_n - x \mid < \epsilon$ and $\forall n \geq n', \mid x_n - y \mid < \epsilon$. Let $m= max \{n^*, n'\}$, then $\forall n \geq m$ $\mid x_n - x \mid < \frac {\mid x-y \mid}3$, and $\mid x_n - y \mid < \frac {\mid x-y \mid}3$
(1) $\frac {-\mid x-y \mid}3 < x_n - x < \frac {\mid x-y \mid}3$
(2) $\frac {-\mid x-y \mid}3 < x - x_n < \frac {\mid x-y \mid}3$ (when line (1) multiplited by $-1$)
(3) $\frac {-\mid x-y \mid}3 < x_n - y < \frac {\mid x-y \mid}3$ 
(4) $\frac {-2\mid x-y \mid}3 < x - y < \frac {2\mid x-y \mid}3$ (result of adding lines (2) and (3))
(5) $\mid x - y \mid < \frac 23 \mid x - y \mid$
(6) $1< \frac 23$ this result of not possible

Therefore a convergent sequence has to have a unique limit. 

**3. Desmos: Show that $\lim (2-\frac{(-1)^n}n)=2$**

https://www.desmos.com/calculator/dnxawgs0dc


**4. Prove that $\inf(0,1)=0$.**

Proof by definition:
(1) $\forall x \in (0,1)$, $x \geq 0$
(2) Take any $\epsilon > 0$, let $x^* = min\{\frac12,\ \frac \epsilon2\}$
When $x^* = \frac \epsilon2$
$\epsilon > \frac \epsilon2 \geq 0$
$0 + \epsilon > \frac \epsilon2 \geq 0$
$0 + \epsilon > x^* \geq 0$
When $x^* = \frac 12$
$\epsilon > \frac \epsilon2> \frac 12 \geq 0$
$0 + \epsilon > \frac 12 \geq 0$
$0 + \epsilon > x^* \geq 0$
In both cases, $0 + \epsilon > x^* \geq 0$ is a true statement, 
(3)Therefore $\inf(0,1)=0$.

**5. Let $A=(0,1) \cup {2}$. Prove that $\sup A=2$.**

Proof by definition:
(1) $\forall x \in A$, $x \leq 2$
(2) Pick any $\epsilon > 0$, let $x^* = 2$
$2 - \epsilon < x^* \leq 2$
$2 - \epsilon < 2 \leq 2$
This is a true statement, therefore $\sup A=2$.


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**Week 2**
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**5. Counter the definition of $\lim x_n=L$.**
For this definition, $n,m~\in~{\mathbb N}$ and $\epsilon~\in~{\mathbb R}$

$\lim x_n \neq L$ 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 $\epsilon = \frac 12$. Pick any $m~\in~{\mathbb N}$. Let $n=2m$, then $\mid x_n - L\mid = \mid (-1)^{2m} - 0\mid = \mid 1 - 0\mid = 1 \geq \epsilon = \frac 12$. Therefore the $\lim(-1)^n\neq 0$.

**7. Prove that $\lim\frac{1}{n^2}=0$.**
Take any $\epsilon > 0$. Let $n^*=$ $\lceil \sqrt {\frac1\epsilon}\rceil$. Then $\forall n \geq n^*$,
$\mid x_n - L\mid = \mid \frac1{n^2} - 0\mid = \frac1{n^2} \leq \frac1{n^{*2}} = \frac 1{\lceil \sqrt {\frac1\epsilon}\rceil^2} < \frac 1{\sqrt {\frac1\epsilon}^2} = \epsilon$. Therefore, $\mid x_n - L\mid  <  \epsilon$, it proves $\lim\frac{1}{n^2}=0$. 



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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 $x_n=17=x_{n+1}~\forall ~n~\in~{\mathbb N}$. All the points lie on the same horizontal line $y=17$. 


**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 $x_n\neq x_{n+1}$ for example, $x_1=-1\neq 1=x_2$. This is always true if $n$ is odd, then $x_n=-1$ and if n is even, then $x_n=1$. 

**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.
$\{(-1)^n\}$ is bounded above by any $K \geq 1$ because the "ceiling" of $(-1)^n$ is 1 when $n$ is even. When $n$ is odd, then $(-1)^n=-1$. Either way, $(-1)^n\leq 1$, so any $K$ value larger than or equal to $1$ will bound the sequence above.


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

**Example 2'.** Prove that $\{2^n\}$ is not bounded above.
$\{2^n\}$ is not bounded above because $\nexists K ~\in~{\mathbb R}$ such that $x_n\leq K~\forall~n\in~{\mathbb N}$. Since $n\in~{\mathbb N}$, then all values of $n$ must be positive, so the sequence would continue to get exponetially larger for each $n$. 
For any negative K,  $x_n > K$, $\forall~n\in~{\mathbb N}$. For any positive $K$, $x_n > K$ when $n=(\log_2 K)+1$ (rounded to the rearest natural number). Thus $\exists~n\in~{\mathbb N}$ such that $x_n > K~\forall K~\in~{\mathbb R}$, so $\{2^n\}$ is not bounded above. 


**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.
$\{(-1)^n\}$ is bounded below by any $K \leq -1$ because the "floor" of $(-1)^n$ is -1 when $n$ is any odd number. When $n$ is even, then $(-1)^n=1$. Either way, $(-1)^n\geq -1$. So any $K$ value less than or equal to $-1$ will bound the sequence below.


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

**Example 3'.** Prove that $\{-2n\}$ is not bounded below.
$\{-2n\}$ is not bounded below because $\nexists K ~\in~{\mathbb R}$ such that $x_n> K  ~\forall~n\in~{\mathbb N}$. Since $n\in~{\mathbb N}$, then all values of $n$ must be positive, so the sequence would continue to output larger and larger negative values. For any $K$, $x_n< K$ when $n=\frac K{-2} +1$ (rounded to the rearest natural number) thus $\exists~n\in~{\mathbb N}$ such that $x_n < K~\forall K~\in~{\mathbb R}$, so $\{-2n\}$ 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.
{$\sin n$} is bounded becasue $\exists$ K $\in~{\mathbb R}$ such that $\mid x_n\mid$ $\leq$ K $\forall~n~\in~{\mathbb N}$. The smallest possible value of $K$ that would bound {$\sin n$} is 1.


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

**Example 4'.** Prove that $\{4n\}$ is not bounded. 
$\{4n\}$ is not bounded because $\nexists K ~\in~{\mathbb R}$ such that $\mid x_n\mid$ $\leq K  ~\forall~n\in~{\mathbb N}$. Since $n \in~{\mathbb N}$, then all values of $n$ must be positive, so the sequence would continue to output larger and larger values as $n$ increases. For any $K$, $x_n > K$ when $n=\frac K{4} +1$ (rounded to the rearest natural number) thus $\exists~n\in~{\mathbb N}$ such that $x_n=\mid x_n\mid > K~\forall K~\in~{\mathbb R}$, so $\{4n\}$ is not bounded. 

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

Assume that $\{x_n\}$ is infinitely large. By definition, if $~\forall~K\in~{\mathbb R}$, $\exists~n^*\in~{\mathbb N}$  s.t. $~\forall~n\in~{\mathbb N}$ and $n\geq n^*$, then $\mid x_n\mid \geq K~$. In this case, $\{x_n\}$ must also unbounded because $\exists~n~\in~{\mathbb N}$ such that $\mid x_n\mid$ $>K$ $\forall K$ $\in~{\mathbb R}$. Taking out the quantifier of $n^*$ out of the definition of infinitely large makes the definition of the sequence more broad and when less specific it is the defintion of unbounded. For any sequence, unbounded is a larger umbrella term and infinitely large is a specific case of an unbound sequence. This means that if $\{x_n\}$ is infinitely larger than it is also unbounded.


2. Prove that $\{(-1)^nn\}$ is inf. large
 
$\forall K$ $\in~{\mathbb R}$ and $~\forall~n\in~{\mathbb N}$, $\exists~n^*\in~{\mathbb N}$ such that $n\geq n^*$ and $\mid x_n\mid \geq K~$. This statement is true as long as $n^*\geq K$. Since $n\geq n^*$, then $n\geq K$ because $\mid x_n\mid~=~\mid (-1)^nn\mid~ = n ~\geq K~$. This proves the statement.

3. Counter the definition of infinitely large.

$\{x_n\}$ is not infinitely large $~\forall~n^*\in~{\mathbb N}$, $\exists K$ $\in{\mathbb \ R}$ and $\exists~n\in~{\mathbb N}$, such that $n\geq n^*$ and  $\mid x_n\mid < K~$ 

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 

Proof by Contradiction for Not Infinitely Large:
Assume that $\{x_n\}$ is infinitely large. So by definition, $~\forall K$ $\in{\mathbb \ R}$ and $\exists~n^* and~ n\in~{\mathbb N}$ such that $\forall n\geq n^*$, s.t. $\mid x_n\mid \geq K~$. However this definition cannot be true because $\mid x_n\mid =0< K~$ when $n=n^* +1$ if $n^*$ is even, alternatively $n=n^*$ or $n=n^* +2$ if $n^*$ is odd. This is a contradiction to the definition, therefore $\{x_n\}$ is not infinitely large.
A specific case, let $K=2$. In this case, when $n^*=2$ all of the following terms should be greater than 2. However when $n=3$ then $\mid x_n\mid =0< K$ so the definition of infinitely large is contradiced and therefore $\{x_n\}$ is not infinitely large. 

Proof for Unbounded:
$\{x_n\}$ is unbounded because $\exists~n~\in~{\mathbb N}$ such that $\mid x_n\mid$ $>K$ $\forall K$ $\in~{\mathbb R}$. For any K, we can always find a $~n'~\in~{\mathbb N}$, and $n'>K$. If $n'$ is even, we can have $n=n'$, then $\mid x_n\mid=x_n=n=n'>K$. If $n'$ is odd, we can have $n=n'+1$, then $\mid x_n\mid=x_n=n=n'+1>n'>K$.

That proves the statement.


