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# System prepended metadata

title: Sequences by Megan

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

**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 $\exists~n^*\in~{\mathbb N}$ such that $x_{n^*}=-1$ and $x_{n^*+1}=1$, where $n^*=1$.

**Definition 2.** $\{x_n\}$ is bounded above if $\exists K~\in~{\mathbb N}$ such that $x_n\leq K~\forall~n\in~{\mathbb N}$.

**Example 2.** Prove that $\{(-1)^n\}$ is bounded above.
Let $k=2$. Then $\forall~n$, $x_n=(-1)^n~\leq~2$. 

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

**Example 2'.** $\{(2)^n\}$ is not bounded above.
By contradiction, assume $\{(2)^n\}$ is bounded above. Let $K\in {\mathbb R}$ such that $\exists~x_{n^*} \in \{(2)^n\}$. Let $K=2^{n^*}$, using  $2^{{n^*}+1}$ with a base of 2, $2^{n} < 2^{{n^*}+1}$. Thus $K<2^{n}$, a contradiction. 

**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.
Let $k=-2$. Then $\forall n$, $x_n=(-1)^n \geq -2$.  

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

**Example 3'.** $\{2n\}$ is not bounded below.
By contradiction, assume $\{2n\}$ is bounded below. Let $K \in {\mathbb R}$ such that $\exists~n^* \in \{2n\}$. Let $K=2n^*$, using  $2{n^*}-1$, $2{n} > 2{n^*}-1$. Thus $K>2n$, a contradiction.

**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.** {$\sin n$} is bounded. 
Let $K = 1$. Then $\forall~K \in~ {\mathbb R}, \mid \sin(x_n) \mid \leq 1$. 

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

**Example 4'.** $\{4n\}$ is not bounded. 
Let $k=10$. Then $\exists~n^*$, such that $x_{n^*} = \mid 4n^* \mid \geq 10$. 
 

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**Week 2**
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**1. Prove that if $\{x_n\}$ is infinitely large then $\{x_n\}$ is unbounded.**

By contradiction, suppose $\{x_n\}$ is infinitely large. Then $\forall~K~\in~{\mathbb R}~\exists~n^*$ such that $\forall~n, ~n\geq n^*$, $\mid x_n \mid \gt K$. Let $\{x_n\}$ be bounded. Then $\exists~K_1 \in {\mathbb R}$ such that $x_n \leq K_1$, $\forall~n~\in~{\mathbb N}$. If $x_n \leq K_1$, then $\exists~K~\in~{\mathbb R}$ such that $K \geq x_n, \forall~n$. A contraction from $\forall~n, n \geq n^*$, $\mid x_n \mid \gt K$. Thus $\{x_n\}$ is unbounded. 



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

Suppose $\{(-1)^nn\}$ is bounded. Let $K_1~\in~{\mathbb R}$ such that $\mid K_1 \mid = \{(-1)^{n^*}{n^*}\}$. Then $\mid K_1 \mid \geq \{(-1)^nn\}, \forall~n$. Let $n^*$ be even. Then $\mid (-1)^{n^*+2}{n^*+2} \mid = \mid n^*+2 \mid \gt \mid n^* \mid = \mid (-1)^{n^*}{n^*} \mid = \mid K_1 \mid$. Thus a contraction. Then $\mid (-1)^{n^*+1}{n^*+1} \mid = \mid n^*+1 \mid \gt \mid n^* \mid = \mid (-1)^{n^*}{n^*} \mid = \mid K_1 \mid$. Thus a contraction. Showing $\{(-1)^nn\}$ is unbounded. Consider $\mid n^*+1 \mid$, then $\forall~n~\in~{\mathbb N}, n \geq n^*+1$ and $\{(-1)^nn\} > K$. Thus it is infinitely large. 



**3. Counter the definition of infinitely large.**

$\{x_n\}$ is not infinitely large if $\exists~K~\in~{\mathbb R}~\exists~n^*$ such that $\exists~n \geq n^*$, $\mid~x_n\mid \gt K$. 



**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.**

Given $$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$. By contradiction, assume $x_n$ is infinitely large. Then $\forall ~K~\in~{\mathbb R}~\exists~n^*$ such that $\forall~n \geq n^*$, $\mid x_n \mid \gt K$. When $n^*$ is odd, $x_{n^*+2}$ is also odd. Thus $x_{n^*}$ and $x_{n^*+2}$ both equal zero. $x_{n^*+1}$ is even when $x_{n^*}$ is odd and $x_{n^*+1}=n+1$. Thus $n+1>0$ creating a contraction. Similarly, when $n^*$ is even, $x_{n^*+2}$ is also even. Thus $x_{n^*}=n$ and $x_{n^*+2}=n+2$. $x_{n^*+1}$ is odd when $x_{n^*}$ is even and $x_{n^*+1}=0$. Thus $0>n$ creating a contraction. Thus proving the sequence is not infinitely large. Suppose $x_n$  is bounded. Then $\exists~K~\in~{\mathbb R}$ such that $x_n~\leq~K, \forall~n$. Then $\exists~n^*$ which is even, such that $x_{n^*}=2K$. Then $x_{n^*} > K$ a contradition. Thus $x_n$  is unbounded.



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

$\lim x_n\neq L$ if $\exists~\epsilon \gt 0~\exists~n^*$ such that $\exists n\geq~n^*$, $\mid{x_n-L}\mid \gt\epsilon$.



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

By contradiction, suppose $\lim(-1^n)=0$ and let $\epsilon = \frac {1}{2}$. Then $\exists~n^*$ such that $\forall~n \geq n^*$, $\mid (-1)^n-L \mid \lt \epsilon$. Since $\lim(-1)^n=0$ and $\epsilon = \frac {1}{2}$, this implies $\mid(-1)^n-0 \mid \lt \frac {1}{2}$. Then $\forall$ even $n^*$, $\mid1-0 \mid \lt \frac {1}{2}$ and $\forall$ odd $n^*$, $\mid-1-0 \mid \lt \frac {1}{2}$. Then $\mid1 \mid \lt \frac {1}{2}$ and $\mid-1\mid \lt \frac {1}{2}$. Thus $2 \lt 1$ a contraction. 



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

By contradiction, suppose $\lim(-1^n)=1$ and let $\epsilon = \frac {1}{2}$. Then $\exists~n^*$ such that $\forall~n \geq n^*$, $\mid (-1)^n-L \mid \lt \epsilon$. Since $\lim(-1)^n=1$ and $\epsilon = \frac {1}{2}$, this implies $\mid(-1)^n-1 \mid \lt \frac {1}{2}$. Then $\forall$ even $n^*$, $\mid1-1 \mid \lt \frac {1}{2}$ and $\forall$ odd $n^*$, $\mid-1-1 \mid \lt \frac {1}{2}$. Then $\mid 0\mid \lt \frac {1}{2}$ and $\mid-2\mid \lt \frac {1}{2}$. Thus $2 \lt 1$ a contraction. 

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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 $\epsilon>0$.**

To negate, $\forall~\epsilon >0, \epsilon$ is a bad number becomes $\exists~\epsilon$ such that $\epsilon >0$, and $\epsilon$ is a good number. To negate a $\forall$ one needs to find one number or situation where the statement is not true. The inequality sign does not change becasue you negate $\forall$ and the second part of the statement.   

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

$m= \inf~A$ if $\forall~a \in A, a \geq M$. $\forall~\epsilon > 0, ~\exists~a^*\in A$ such that $M \leq a^* < M + \epsilon$. 

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

$1- \frac{1}{n} \leq 1, \forall~n$. By contraction, suppose $\exists~n^*$, such that $1-\frac{1}{n^*} >1$. Then $1< (-1)(0)(n^*) \implies 1 <0$. Given $\epsilon >0, \exists~n^*$ such that $n^*=\lceil \frac{1}{\epsilon} \rceil+1$. Since, $n^* > \frac{1}{\epsilon}$ then $1 -\epsilon < 1- \frac{1}{n^*} \leq 1$.

**4. Prove that $\inf\{1-\frac{1}{n}\}=0$**

$0 \leq 1 -\frac{1}{n}, \forall~n$. By contraduction, suppose $\exists~n^*$ such that $1- \frac{1}{n^*} < 0$. Then $1 < \frac{1}{n^*} \implies n^*<1$. Given $\epsilon >0$, $x_{1}=0$ so $0 \leq  x_1 < \epsilon.$ 

**5. Prove that $\inf\{(0,1)\}=0$**

$\forall~x \in (0,1)$, $x \geq 0$. Given $\epsilon > 0, \exists~x^*$ such that $x^*= \frac{\epsilon}{2}$, such that  $0 \leq \min\{\frac{1}{2}, \frac{\epsilon}{2} \}  < \epsilon$. 

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

$\forall~x \in \{0\}\cup \{(1,2)\}, x \geq 0$. Given $\epsilon > 0,~x_1=0$ so $0 \leq x_1 < \epsilon$. 

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

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

The sequence is convergent and the limit is $0$. $\forall~\epsilon > 0,\exists~n^* \in {\mathbb N}$ such that $\forall~n>n^*, \lim\frac{(-1)^n}{2n} = \mid \frac{(-1)^n}{2n} - L \mid = \mid \frac{(-1)^n}{2n} - 0 \mid = \mid \frac{(-1)^n}{2n} \mid < \epsilon$. Thus, $\mid \frac{(-1)^n}{2n} \mid = \frac{1}{2n} < \epsilon$. Thus, $1<2n\epsilon$ or $n> \frac{1}{2\epsilon}$. Let $n^*=\lfloor \frac{1}{2\epsilon}\rfloor+1$. Then $\forall~n\geq n^*, \mid \frac{(-1)^n}{2n} - L \mid = \mid \frac{(-1)^n}{2n} - 0 \mid = \frac{1}{2n} \leq \frac{1}{2n^*}=\frac{1}{2\lfloor \frac{1}{2\epsilon}\rfloor+1} < \epsilon.$



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

The sequence is convergent and the limit is $0$. Given any $\epsilon > 0,~\forall~n \in {\mathbb N}, \lim 2^{-n}=\mid 2^{-n} - L\mid = \mid 2^{-n} - 0 \mid = \mid 2^{-n}\mid=\frac{1}{2^n}<\epsilon$ or $2^n\epsilon >1$. Thus $\log_2(1)< \log_2(2^n\epsilon)= \log_2(2^n)+\log_2(\epsilon)=n+\log_2(\epsilon)$. Thus $\log_2(1) < n+\log_2(\epsilon) \implies \log_2(1)-\log_2(\epsilon) < n$ and $0-\log_2(\epsilon)= -\log_2(\epsilon)=\log_2(\frac{1}{\epsilon})<n$. Given $\epsilon > 0$, let $n^*=\lfloor \log_2(\frac{1}{\epsilon}) \rfloor +1$. Then $\forall n\geq n^*, \mid 2^{-n} - L\mid = \mid 2^{-n} - 0 \mid = \mid 2^{-n}\mid=\frac{1}{2^n} \leq \frac{1}{2^{n^*}}= \frac{1}{2^{\lfloor\log_2(\frac{1}{\epsilon})\rfloor+1}}< \epsilon$

**8. Redo Week 1 HW if you want to (for half of the remaning points)**

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**Week 4**
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**1. Prove that any subsequence of an infinitely large sequence is infinitely large.**
Let $\{x_n\}$ be a sequence thats infiniely large then $\forall ~K\in{\mathbb R}$
$\exists ~ n^*\in {\mathbb N}$ such that $\forall~ n\geq n^*$, $\mid x_n \mid>K$.

Let $\{x_{n_i}\}$ is a subsequence of $\{x_n\}$. Then $\exists ~ n_i^*$ such that $n_i^* > n^*$. So then $\forall~ n_i\geq n_i^*$, $\mid x_{n_i} \mid>K$. Therefore $\{x_{n_i}\}$ is infinitely large 


after some N^* for all n greater then nstar xn is greater than any K
the subset is also infinitly large for any K chose the same y>N^*
then for all y greater than this y the abs is greater than k


**2. p. 50 Ex. 2.1.7**

**2.1.7** Let $\{x_n\}$ be a sequence. Show that $\lim x_n =0$ if and only if $\lim\mid x_n \mid =0$. Find an example such that $\{\mid x_n\mid\}$ converges and $\{x_n\}$ diverges.

By definition of the limit, $\forall \epsilon > 0$, $\exists~n^*$ such that $\forall n \geq n^*$, $\mid x_n-L \mid < \epsilon$. Given $\lim x_n=0$, we know $\forall \epsilon > 0$, $\exists~n^*$ such that $\forall n \geq n^*$, $\mid x_n- 0 \mid < \epsilon  \implies \mid x_n \mid < \epsilon~\Leftrightarrow ~\mid \mid x_n \mid -~0\mid < \epsilon$. Thus by definition of the limit, $\lim \mid x_n \mid = 0$. Given $\lim \mid x_n \mid =0$, we know that $\forall \epsilon > 0, \exists~n'$ such that $\forall n \geq n', \mid \mid x_n \mid -~0~\mid < \epsilon \implies \mid \mid x_n \mid \mid < \epsilon \implies \mid x_n -~0~\mid < \epsilon$. Thus by definition of the limit, $\lim \mid x_n \mid=0$.An example of a sequence such that $\{\mid x_n\mid\}$ converges and $\{x_n\}$ diverges is $(-1)^n$. 

**3. p. 50 Ex. 2.1.10**

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

$\{x_n\}=\{\frac{n+1}{n} \}=\{1+\frac{1}{n}\}$. Let us show it is monotone decreasing. We will start with $n+1 \geq n$. Thus $\{1+\frac{1}{n+1}\} \leq \{1+\frac{1}{n}\}$. 
Applying Theorem 2.1.10, if a sequence is monotone decreasing and bounded, then $\lim_{n\to\infty} x_n=$ $\inf{\{x_n: n \in {\mathbb N}\}}$. $1 \leq 1 +\frac{1}{n},~ \forall~n$. By contradiction, suppose $\exists~n^*$ such that $1+\frac{1}{n^*} < 1$. Then $\frac{1}{n^*} < 0  \implies 1<0$, a contradiction. Given $\epsilon >0, \exists~n^*$ such that $n^*=\lceil \frac{1}{\epsilon} \rceil+1$. Since, $n^* > \frac{1}{\epsilon}$ then $1 +\epsilon > 1+\frac{1}{n^*} \geq 1$. Thus, $\inf{\{1 +\frac{1}{n}}\}=1$ and $\lim_{n\to\infty} x_n=1$. 



**4. p. 50 Ex. 2.1.13**

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

By contradiction, $\exists~k \in {\mathbb N}$ such that $\lim_{n\to\infty} x_n=x_k$. Suppose $\exists~n^*$ such that $n^* > k$ and $x_{n^*} < x_k$, without loss of generality. Let $\epsilon = x_k - x_{n^*}$. Given the monotone sequence converges to $x_k$, there $\exists~n'$ such that $n'>n^*>k$. $\mid x_{n'}-x_{k} \mid < \epsilon \implies -\epsilon < x_{n'}-x_{k} < \epsilon \implies x_{k} -\epsilon < x_{n'} < \epsilon + x_{k}$. By construction, $\epsilon = x_k - x_{n^*}$, $x_{k} -x_k + x_{n^*} < x_{n'} < x_k - x_{n^*} + x_{k} \implies x_{n^*} < x_{n'} < 2x_k - x_{n^*}$. We have $x_{n'} > x_{n^*}$ and by construction, $x_{n^*} < x_k$. By definition of a monotone sequence, the sequence can not decrease then increase.  


**5. p. 50 Ex. 2.1.16**
Let $\{x_n\}$ be a sequence. Suppose there are two convergent subsequences $\{x_{n_i}\}$ and $\{x_{m_i}\}$.

$\lim x_{n_i}=a$. and $\lim x_{m_i}=b$,
where $a\neq b$
Prove that $\{x_n\}$ is not convergent 

By contractiction suppose $\{x_n\}$was convergent and $\lim x_n=L$
Then let $p=max(\mid a - L\mid,\mid b - L\mid)$
Since it is possible that $L=a$ or $L=b$ this choise of p ensures that $L \neq a$ or $L \neq b$
So with out lose of generallity suppose $p=\mid a - L\mid$ then $L \neq a$
Let $\epsilon= p/2$ Then $\exists ~ n^*$ such that $\forall n>n^*, \mid x_n* - L\mid<\epsilon$ 

Also since $\lim x_{n_i}=a$. Then $\exists ~ n' 
\in \{x_{n_i}\}$ such that $\forall n_i>n', \mid x_n - a\mid<\epsilon$

Let $n''=\max\{n^*,n'\}$
Then $\forall n>n''$, $\mid x_n - L\mid<\epsilon$ and $\forall n_i>n''$, $\mid x_{n_i} - a\mid<\epsilon$
This means  $\mid x_n - L\mid +  \mid x_{n_i} - a\mid < 2\epsilon$ 
Then  $\mid x_n - L\mid +  \mid x_{n_i} - a\mid < \mid a - L\mid$ 
In order for this to be true than $\mid x_n - L\mid=g\mid a - L\mid$ and $\mid x_{n_i} - a\mid=h\mid a - L\mid$ where $g$ and $h$ $\in$ $[0,1]$. However by construction $g<.5$ and $h<.5$ That is not possible 

**6. p. 50 Ex. 2.1.17**

Find a sequence $\{x_n\}$ such that $\forall ~y\in{\mathbb R}$ there exisits a subsquence $\{x_{n_i}\}$ converting to $y$.
$\{ \frac{1}{1},\frac{1}{2},\frac{2}{1},\frac{1}{3},\frac{3}{1},\frac{2}{3},\frac{3}{2},\frac{1}{4},\frac{4}{1},\frac{4}{3},\frac{3}{4},\frac{1}{5},\frac{5}{1},\frac{2}{5},\frac{5}{2},\frac{3}{5},\frac{5}{3},\frac{4}{5},\frac{5}{4},...  \}$


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**Week 6**
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**2.2.4** Suppose $x_1 := \frac{1}{2}$ and $x_{n+1} := x^2_n$. Show that $\{x_n\}$ converges and find $\lim x_n$.

Given $x_1 := \frac{1}{2}$ assume $x_n \geq 0,$ for some $n \in {\mathbb N}$. Then $x_{n+1} = x^2_n \geq 0^2 =0$. Thus by induction, $x_{n+1} := x^2_n$ is bounded below by $0$. Now we have $x_n \geq x_{n+1} \Leftrightarrow x_n \geq x^2_n \implies x_n \geq (x_n)(x_n)= \frac{x_n}{x_n} \geq x_n \implies 1 \geq x_n$. Hence $x_n$ is decreasing and bounded below, by the decreasing monotone convergence theorem, $\lim _{n\to\infty}(x_n)= \inf\{x_n: n\in {\mathbb N}\}$. Thus, $\lim _{n\to\infty}(x^2_n) = 0$.       


**2.2.5** Let $x_n := \frac{n-cos(n)}{n}$. Use the sequeeze lemma to show that $\{x_n\}$ converges and find the limit. 

Given $x_n := \frac{n-cos(n)}{n}$, we know that $x_n := \frac{n-cos(n)}{n} = 1- \frac{cos(n)}{n}$. Thus $\lim(\frac{n-cos(n)}{n})= \lim(1)-\lim(\frac{\cos(n)}{n}).$ By limit of a constant $\lim(1)=1.$ Since $-1 \leq \cos(n) \leq 1, \forall~n$ Then $-\frac{1}{n} \leq \frac{cos(n)}{n} \leq \frac{1}{n}$ and $\lim(-\frac{1}{n}) \leq \lim(\frac{\cos(n)}{n}) \leq \lim(\frac{1}{n})$. $0 \leq \lim(\frac{\cos(n)}{n}) \leq 0$ By the squeeze lemma, $\lim(\frac{\cos(n)}{n})=0$. Thus $\lim(1)-\lim(\frac{\cos(n)}{n})= 1-0 = 1.$

**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 $\{w_n\}$ Converage? What are the limits? Can you apply Proposition 2.2.5? Why or why not?

$z_n:=\frac{x_n}{y_n}$ , so $z_n:=\frac{\frac{1}{n^2}}{\frac{1}{n}}=\frac{n}{n^2}=\frac{1}{n}$  

The $lim\frac{1}{n}=0$
Given $\epsilon >0, \exists~n^*$ such that $n^*= \lceil \frac{1}{\epsilon} \rceil+1$. Then $\forall~~ n\geq n^*$, $n \geq n^* > \frac{1}{\epsilon}$ and $\mid \frac{1}{n}-0 \mid < \epsilon$.


$w_n:=\frac{y_n}{x_n}$ so $w_n:=\frac{\frac{1}{n}}{\frac{1}{n^2}}=\frac{n^2}{n}=n$
$\{{n}\}$ is not bounded above
Let $K\in{\mathbb R}$ let $\ n^*=$ $\mid \lceil K \rceil$ $\mid$ $+1$. Since $\mid\lceil K \rceil\mid$$\ +1$$\ >K$, then  $x_n*>K$ 
Since $\{n\}$ is not bounded it does not converage.

We can not use Proposition 2.2.5 because the limit of $\{y_n\}$ is zero.

**2.2.7**
If $\{x_n\}$ is a sequence such that  $\{x_n^2\}$ convergages, then $\{x_n\}$ converages is a false statment. 
Consider $\{(-1)^n\}$. $\{(-1)^n\}$ is divergent but $\{((-1)^n)^2\}=\{1\}$ converges.



**2.2.7**

$lim\frac{n^2}{2^n}=0$


$\forall ~n\geq10, 0<\frac{n^2}{2^n}<\frac{1}{n}$ 

$lim\{0\}=0$ and $lim\frac{1}{n}=0$ so by the squeeze theorem $lim\frac{n^2}{2^n}=0$


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**Week 7**
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**1. Without using Lemma on nested Intervals prove that
$$\bigcap_{n=1}^\infty[\frac{-1}{n}, \frac{-1}{n}]=\{0\}.$$**

$\forall ~n$, $-1/n < 0$ and $\forall ~n$, $1/n > 0$
so $-1/n < 0 <1/n$
Therefor 0 $\in$ $[-1/n, 1/n]$

By contradition suppose $\exists ~K\in{\mathbb R}$ that is also in $\bigcap_{n=1}^\infty[-1/n, 1/n].$

Then that means $\forall ~~n$, $k \in  [-1/n, 1/n]$

Let $n^*= \mid \lceil  \frac{1}{k} \rceil \mid +1$. Since $n^* > \mid \frac{1}{k} \mid$ then if $k>0, 1/n^*<k$ and if $k<0, k<-1/n^*$, In either case $k ~ \nexists [-1/n^*, 1/n^*]$ which is a contradiction. 

Therefor the interection only contains zero.


**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.**
Given $\epsilon >0, \exists~n^*$ such that $n^*=\lceil \frac{1}{\epsilon} \rceil+1$. Then $\forall~n\geq n^*, \mid \frac{(-1)^n}{n} - L \mid = \mid \frac{(-1)^n}{n} - 0 \mid = \frac{1}{n} \leq \frac{1}{n^*}=\frac{1}{\lfloor \frac{1}{\epsilon}\rfloor+1} < \epsilon.$ By proposition 2.3.5, since $\{x_n\}$ converges, then $\lim x_n= \lim\inf x_n=\lim\sup x_n$. $\lim x_n= 0$ thus $\lim\sup x_n = 0$ and $\lim\inf x_n = 0$. 


**B.**
Given $\frac {(n-1)(-1)^n}{n}$, 
$A_1 = \{0, \frac{1}{2}, -\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}, \frac{5}{6}, -\frac{6}{7}...\}$
$A_2 = \{\frac{1}{2}, -\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}, \frac{5}{6}, -\frac{6}{7}...\}$
$A_3 = \{-\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}, \frac{5}{6}, -\frac{6}{7}...\}$
$A_4 = \{\frac{3}{4}, -\frac{4}{5}, \frac{5}{6}, -\frac{6}{7}...\}$. 
Let $a_n= \sup A_n,~a_1= 1, a_2 = 1, a_3=1, a_4=1$ thus $a_n= \{1\}$.
Let $b_n= \inf A_n,~b_1= -1, b_2 = -1, b_3=-1, b_4=-1$ thus $b_n= \{-1\}$. By definition, $\lim a_n= \lim\sup x_n$ and $\lim b_n= \lim\inf x_n$.  Thus $\lim a_n= \lim 1= 1$ and $\lim\sup x_n=1$. Thus $\lim b_n= \lim -1= -1$ and $\lim\inf x_n= -1$.

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


By construction let $\{x_{n_k}\}$ be a subsequence of an unbounded sequence $\{x_n\}$.
By definition if $\{x_n\}$ is unbounded then $\forall ~~ k~~ \exists ~~n^*$ such that $\mid n^*\mid > \mid k \mid$

For this construction it is important that $x_{n_1} \neq 0$ so with out loss of generallity lets assume that $x_{1} \neq 0$
Let $x_{n_1}=x_1$ 
Pick any $n_2$ such that $n_2 > 1$ and also $\mid x_{n_2} \mid > 2\mid x_1 \mid$
Pick any $n_3$ such that $n_3 > n_2$ and also $\mid x_{n_3} \mid >2 \mid x_2 \mid$
Pick any $n_3$ such that $n_3 > n_2$ and also $\mid x_{n_3} \mid > 2\mid x_2 \mid$

In general suppose you found an $x_{n_{k-1}}$
Pick any $n_k$ such that $n_k > n_{k-1}$ and also $\mid x_{n_k} \mid > 2\mid x_{k-1} \mid$

So by construction $\{x_{n_k}\}$ is an infinitly large subsequence 


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**Week 10**
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**2.5.1** Suppose the $k$th partial sum of $\sum_{n=1}^{\infty}x_n$ is $s_k=\frac{k}{k+1}$. Find the series, that is find $x_n$, prove that the series converges, and then find the limit. 

  $s_k=\frac{k}{k+1}$ has a limit of 1
  
 First, $\mid \frac{k}{k+1} -1 \mid =\mid \frac{-1}{k+1} \mid=\frac{1}{k+1}$  
 
 $\forall$ $\epsilon >0, \exists~k^*$ such that $k^*=\lceil \frac{1}{\epsilon} \rceil+1$. 
  Since, $k^* > \frac{1}{\epsilon}$ then $\frac{1}{k+1} < \epsilon$.
  So then for all $k \ge k^*$, $\mid \frac{k}{k+1} -1 \mid < \epsilon$
  
    
    
**2.5.2** Prove Proposition 2.5.5, that is for $-1<r<1$ prove $\sum^{\infty}_{n=0} r^n= \frac{1}{1-r}$. 

 Proposition 2.5.4 states $\sum^{\k}_{n=0} r^n= \frac{1-r^k}{1-r}$
 When $r=a/b$ such that $a<b$ and $b\neq 0$ then $lim\frac{1-r^k}{1-r}= \frac{lim 1-r^k}{lim 1-r}$
 
 $lim 1-r^k=1$
 $lim 1-r=1-r$ 
 
 so then $lim\frac{1-r^k}{1-r}=\frac{1}{1-r}$ when$r=a/b$ such that $a<b$ and $b\neq 0$


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**2.5.3** Decide the convergence or divergence of the following series. 

**A.**$\sum^{\infty}_{n=1}\frac{3}{9n+1}$

By the comparison test, given $y_n= \frac{3}{9n+1}$ and $x_n= \frac{1}{100n}, \forall n\geq 1, 0 \leq \frac{1}{100n} \leq \frac{3}{9n+1}$. $x_n= \frac{1}{100} \sum \frac{1}{n}$ is divergent thus $y_n$ is divergent. 


**B**$\sum^{\infty}_{n=1}\frac{1}{2n-1}$
By the comparison test, given $y_n= \frac{1}{2n-1}$ and $x_n= \frac{1}{2n}, \forall n\geq 1, 0 \leq \frac{1}{2n} \leq \frac{1}{2n-1}$. $x_n= \frac{1}{2} \sum \frac{1}{n}$ is divergent thus $y_n$ is divergent. 

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**Week 11**
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**2.5.3** Decide the convergence or divergence of the following series.

**C.** $\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$

By the alternating series, $x_n=\frac{1}{n^2}$. Thus $x_n$ is a monotone decreasing sequece of postive real numbers. $\lim_{n\to\infty} \frac{1}{n^2}=0$. Thus by the alternating series test, $\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$ converges. 

**D.** $\sum^{\infty}_{n=1}\frac{1}{n(n+1)}$

By the comparison test, given $y_n= \frac{1}{n^2}$ and $x_n= \frac{1}{n(n+1)}, \forall n\geq 1, 0 \leq \frac{1}{n(n+1)} \leq \frac{1}{n^2}$. $x_n= \sum \frac{1}{n^2}$ is convergent by the p-series test, $\sum \frac{1}{n^p}$ converges if $p >1$. Thus $y_n$ converges.

**E.** $\sum^{\infty}_{n=1} ne^{-n^2}$

By the root test, $L=\lim_{n \to \infty} \mid ne^{-n^2}\mid^{\frac{1}{n}}= \lim_{n \to \infty} \mid n^{\frac{1}{n}}e^{-n}\mid =$
$\lim_{n \to \infty} \mid n^{\frac{1}{n}}\mid*\lim_{n \to \infty}\mid e^{-n}\mid= \lim_{n \to \infty} \mid n^{\frac{1}{n}}\mid*\lim_{n \to \infty}\mid \frac{1}{e^n} \mid = 1*0= 0$.
Thus $L=0$ and by the root test, $\sum^{\infty}_{n=1} ne^{-n^2}$ converges. 

**2.5.4**
**A.** Prove that if $\sum^{\infty}_{n=1} x_n$ converges, then $\sum^{\infty}_{n=1} (x_{2n}+x_{2n+1})$ also converges. 
Let $\sum^{\infty}_{n=1} x_n=L$
$\sum^{\infty}_{n=1} (x_{2n}+x_{2n+1})=x_2+x_3+x_4+x_5...=L-x_1$
Therefor $\sum^{\infty}_{n=1} (x_{2n}+x_{2n+1})$ converges 

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

$\sum^{\infty}_{n=1} [(-1)^{2n}+(-1)^{2n+1}]$ converges but $\sum^{\infty}_{n=1} (-1)^n$ diverges
$s_1=1+(-1)=0$
$s_2=0+[1+(-1)]=0$
$s_3=0+0+[1+(-1)]=0$
$s_4=0+0+0+0+[1+(-1)]=0$
$s_n=\{0,0,0,0,...\}$ which converges. 


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**Week 12**
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Let $\ S$ be a subset of $\ R$ and $\ S=(a,b)$ and let $\ c$ be a a cluster point of $\ S$ in the interior of $\ S$.
Then $\ f(x) \to L$ as $\ x \to c$ if and only if for every sequence $\ {x_n}$ of numbers such that $\ x_n \in S$ but not$\ c$ for all $\ n$ and such that $\ lim x_n=c$,we have that the sequence ${f(x)} \to L$

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**Week 13**
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**6.1.1** Let $f$ and $g$ be bounded functions on $[a,b]$. Prove $\mid\mid f+g \mid\mid _u~\leq~\mid\mid f \mid\mid_u + \mid\mid g \mid\mid_u$.

By definition of the uniform norm, $\mid\mid f+g \mid\mid _u=\sup\{\mid f(x)+g(x) \mid, x \in [a,b]\}$. By the triangle inequality, $\sup\{\mid f(x)+g(x) \mid, x \in [a,b]\} \leq \sup\{\mid f(x)\mid, x \in [a,b]\} + \sup\{\mid g(x) \mid, x \in [a,b]\} = mid\mid f \mid\mid_u + \mid\mid g \mid\mid_u$. 


**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 $\{f_ng_n\}$ does not converge uniformly to $fg$ on $A$.

Let $A={\mathbb N}$ and $f_n(x)=g_n(x)= x+\frac{1}{n}$. Given $\epsilon > 0,~\exists~N \in {\mathbb N}$ such that, $N=\lceil \frac{1}{\epsilon} \rceil+1$. Since, $N > \frac{1}{\epsilon}$ then  $n \geq N$ and $\forall x \in A$, $\mid f_n(x)-f(x)\mid = \mid x+\frac{1}{n}- (x)\mid =\mid \frac{1}{n}\mid = \frac{1}{n} \leq \frac{1}{N}< \epsilon$. Thus $f_n(x)=g_n(x)$ converges uniformly to some $f$ and $g$ on $A$.

However, suppose that $\{f_ng_n\}$ converges uniformly to $fg$ on $A$ if $f_n(x)=g_n(x)= x+\frac{1}{n}$. Then given $\epsilon> 0,~\exists~N \in {\mathbb N}$ such that,  $\forall ~~n \geq N$ and $\forall x \in A$, $\mid f_n(x)g_n(x)-f(x)g(x)\mid = \mid (x+\frac{1}{n})^2 -x^2 \mid = \mid x^2+\frac{2x}{n}+\frac{1}{n^2} -x^2\mid = \mid\frac{2x}{n}+\frac{1}{n^2} \mid< \epsilon$. 
Sine $n>0$ and $x>0$ then $\frac{2x}{n}<\frac{2x}{n}+\frac{1}{n^2} < \epsilon$ so $\frac{2x}{n} < \epsilon$ $\implies$ $n>\frac{2x}{\epsilon}$
Therefor this does not coverge uniformly since $n$ depends on $x$