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# System prepended metadata

title: Sequences by Rebekah

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# Sequences by Rebekah

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

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

**Example 1'.** $\{(-1)^n\}$ is not constant. 
Let $x_1=-1$, $x_2 =1$ and $x_1$ $\neq x_2$. ${\blacksquare}$ 
<!--Thus, it is an alternating sequence. ${\blacksquare}$ -->


.

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

**Example 2.** Prove that $\{(-1)^n\}$ is bounded above. 
*Proof*
Let $K=2$. Then $\forall$ n, $x_n = {(-1)^n}\leq2$ ${\blacksquare}$ 


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

**Example 2'.** $\{3n\}$  is NOT bounded above.
*Proof*
Given any $K \in{\mathbb R}$. Let $n^*=\lfloor |K|\rfloor+1$. 
Then $x_{n^*} =3n^*=3(\lfloor |K|\rfloor+1) > K$. ${\blacksquare}$
.
<!--
Improvisations
$K=2, n^*=3, x_3=9>2$ true
$K=13.1, n^*=14, x_{14}=52>13.1$ true
$K=0.5, n^*=1, x_1=3>0.5$ true
$K=-7, n^*=8, x_{8}=24>-7$ true
-->


**Definition 3.** $\{x_n\}$ is bounded below if $\exists ~K\in{\mathbb R}$ such that $\forall~ n$, $x_n\geq K$.


**Example 3.** $\{3\}$ is bounded below.
*Proof*
Let $K = 1$
Then $\forall~ n$, $~ x_n = 3 > K$. ${\blacksquare}$

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


**Example 3'.** $\{-n\}$ is NOT bounded below.
*Proof*
Given any K $\in{\mathbb R}$. Let $n^* = \lfloor|K|\rfloor + 1$.
Then, $x_{n^*} = -n^* = -\lfloor|K|\rfloor - 1 < K$ ${\blacksquare}$

<!--
Improvisations
$K=2, n^*=3, x_2=-3-1=-4<2$ true
$K=13.1, n^*=14, x_{14}=-15<13.1$ true
$K=0.5, n^*=1, x_1=-2<0.5$ true
$K=-7, n^*=8, x_{8}=-9<-7$ true
-->



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


**Example 4.** $\{\cos~n\}$ is bounded.
*Proof*
Let $K=1$.
Then $\forall~ n  \in{\mathbb N}$, $|\cos n|\leq 1=K$. ${\blacksquare}$



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

**Example 4'.** ${(-3)^n}$ is NOT bounded.
*Proof*
Given any $K \in{\mathbb R}$. Let ${n^*}=\lfloor|K|\rfloor+1$.
Then, $|x_{n^*}|=|(-3)^{n^*}| =|-3^{\lfloor|K|\rfloor+1}| > K$. ${\blacksquare}$


<!--
Improvisations
$K=2, n^*=3, x_3=9>2$ true
$K=13.1, n^*=14, x_{14}=|4782969|>13.1$ true
$K=0.5, n^*=1, x_1=|-3|>0.5$ true
$K=-7, n^*=8, x_{8}=6561>-7$ true
-->

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**Week 2**


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1. Prove that if $\{x_n\}$ is infinitely large then $\{x_n\}$ is unbounded
2. Prove that $\{(-1)^nn\}$ is inf. large
3. Counter the definition of infinitely large.
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
5. Counter the definition of $\lim x_n=L$.
6. Prove that $\lim(-1)^n\neq 0$
7. Prove that $\lim(-1)^n\neq 1$

<!--
*Notes:* 
Infinitely large is like a square due to $n \geq n{^*}$. Unbounded is like a rectangle. All squares are rectangles but not all rectangles are squares. Therefore, infinitely large can be unbounded but unbounded is not always infinitely large.-->
.

**Definition of infinitely large:**
$\{x_n\}$ is infinitely large if $\forall ~K \in{\mathbb R},~~ \exists ~m$ such that $\forall ~ n \geq m$, $|x_n|>K$.

***(1) Proof**.* (follows proof of unbounded)
$(-2)^n$ is infinitely large. 
Given any $K \in{\mathbb R}$. Let ${m}=\lfloor|K|\rfloor+1$. 
Then given any  $n\geq m$, we have
$$|x_n|=|(-2)^n| = 2^n\geq 2^m=2^{\lfloor|K|\rfloor+1}>K$$

<!--
Improvisations
$K=2, m=3, x_3=|-8|=8>|(-2)^2=4>2$ true
$K=13.1, n^*=, x_{}=||>$ true
$K=0.5, n^*=, x_=||>$ true
$K=-7, n^*=, x_{}=>$ true
-->



***(2) Proof***
$\{(-1)^nn\}$ is infinitely large
Given any $K \in{\mathbb R}$. Let ${m}=\lfloor|K|\rfloor+1$. 
Then given any  $n\geq m$, we have
$$|x_n| = n\geq m={\lfloor|K|\rfloor+1}>K$$


.


**(3) Counterdefinition of infinitely large**: 
$\{x_n\}$ is NOT infinitely large if $\exists ~ K \in{\mathbb R}$  $\forall ~m$, such that $\exists~ n \leq m, |x_n|< K$

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***(4) Proof*** 
$$x_n=\begin{cases} n,&\text{if $n$ is even}\\ 0,& \text{if $n$ is odd}\end{cases}$$ is not infinitely large, but unbounded.

Let $K=1$. Given any $m ~\exists ~ n~\leq m$ such that $|x_n|<K$
Let $n_1=2m$ when $\{x_{n}\}$ is even and let $n_2=0$ when $\{x_{n}\}$ is odd.
$x_n= x_{n_1}=2m \geq
$x_n$ is not infinitely large when $n$ is odd. $x_n$ is unbounded when $n$ is even. Thus $\{x_n\}$ is unbounded but not infinitely large.
 

<!--
In english, it is not infinitely large because it does not always have a point higher than the testing point. The $x_n=0$ shows that is is bounded below which limits the sequence from being infinietly large. It is still considered unbounded because it is only bounded below and not bounded above.-->


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**Definition of $\lim x_n=L$**
$\lim x_n=L$ if $\forall ~\epsilon >0 ~~\exists ~~n^*$ such that $\forall ~n\geq n^*, ~~|x_n - L|<\epsilon$

**(5) Counterdefinition of $\lim x_n\neq L$**
$\lim x_n\neq L$ if $\exists ~\epsilon >0 ~~\forall ~~n^*$  $\exists ~n\geq n^*$ such that $|x_n - L|>\epsilon$


<!-- 
no matter what you do, some of the points will be outside the bound -->


.

**(6)** ***Proof*** $\lim(-1)^n\neq 0$
Let's assume by contradiction that $\lim(-1)^n= 0$
$\{x_n\}=\{(-1)^n\}$
Let $\epsilon = 1/2$, $\exists ~ n^*$ such that $\forall ~ n \geq n^*,~~~ |x_n - 0|<1/2$
$x_n=x_{2n^*}=(-1)^{2n^*}=1$ Then $|1-0|=1>1/2$
Thus, we have a contradiction.${\blacksquare}$

.

**(7)** ***Proof*** $\lim(-1)^n\neq 1$
Let's assume by contradiction that $\lim(-1)^n= 1$
$\{x_n\}=\{(-1)^n\}$
Let $\epsilon = 1/2 ~~ \exists ~ n^*$ such that $\forall ~ n \geq n^*,~~~ |x_n - 1|<1/2$
$x_n=x_{2n^*+1}=(-1)^{2n^*+1}=1$ Then $|1-0|=2>1/2$
Thus, we have a contradiction.${\blacksquare}$

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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 $\epsilon>0$.

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

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

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

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

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

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

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



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**(1)** Does it deal with proof of the definition of a limit?
$\forall \epsilon>0, ~~~~\epsilon$ is a bad number
Using negation: $\exists ~ \epsilon >0$

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**Definition** $A\subset \mathbb R$, $M=\sup~A$ if:
$\bullet$ $\forall ~a \in A, a \leq M$
$\bullet$ $\bullet$ $\forall ~\epsilon >0, \exists ~a^* \in A$ such that $M- \epsilon < a^* \leq M$


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**(3)** ***Proof***.
$\sup\{1-\frac{1}{n}\}=1$
By the definition of the supA
$\bullet$ $\forall ~ n, 1-\frac{1}{n}\ \leq 1$, obviously true.
$\bullet$ $\bullet$ Given some $\epsilon >0$. Let $n^*= \lfloor \frac{1}{\epsilon} \rfloor+1$ such that 
$$1- \epsilon < 1- \frac{1}{n^*} \leq 1$$

Work
$1- \epsilon < 1 - \frac{1}{n^*}$
$- \epsilon < - \frac{1}{n^*}$
$- \epsilon n^* < -1$
$n^* > \frac{1}{\epsilon}$
$n^* = \lfloor \frac{1}{\epsilon} \rfloor+1$

.

**(2)** **Definition** $A\subset \mathbb R$, $m=\inf~A$ if:
$\bullet$ $\forall ~a \in A, a \geq m$
$\bullet$ $\bullet$ $\forall ~\epsilon >0, \exists ~a^* \in A$ such that $m \leq a^* < m+ \epsilon$
.
**(4)** ***Proof***.
$\inf\{1-\frac{1}{n}\}=0$
$\bullet$ $\forall ~ n, 1-\frac{1}{n}\ \geq 0$, true.
$\bullet$ $\bullet$ Given some $\epsilon >0$. Let $n^*= 1$ such that 
$$0 \leq 1- \frac{1}{n^*} < \epsilon$$

<!--Work
$0 - \frac{1}{n^*} < 0+ \epsilon$
$- \frac{1}{n^*} < \epsilon$
$-1 < \epsilon n^*$
$\frac{-1}{\epsilon} < n^*$
$n^* > \frac{-1}{\epsilon}$ -->


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

$\bullet$ $\forall ~a \in (0,1),~ a \geq 0$.
$\bullet$ $\bullet$ Then, if $\epsilon >0$, let $a^* = min\{\frac{\epsilon}{2},\frac{1}{2}\}$. 
Therefore $0 \leq a^* < 0 + \epsilon$, so $\inf \{(0,1)\}=0$. $\blacksquare$


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

$\bullet$ $\forall ~a \in \{0\}\cup \{(1,2)\}, a \geq 0$.
$\bullet$ $\bullet$ Then, if $\epsilon >0$, let $a^* = 0$. 
Therefore $0 \leq a^* < 0 + \epsilon$, so $\inf \{0\}\cup \{(1,2)\}=0$. $\blacksquare$



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**(7) 2.1.3** Is the sequence $\{ \frac{(-1)^n}{2n}\}$ convergent? If so, what is the limit?

By definition of convergent: $\lim x_n=L$ if $\forall ~ \epsilon >0 ~ \exists ~n^*$ s.t. $\forall ~n \geq n^* ~|x_n - L|< \epsilon$
Let $n^* = \lfloor \frac{1}{2\epsilon} \rfloor+1$
Then, $\forall~ n \geq n^*$
$|x_n - L|= |\frac{(-1)^n}{2n}-0|=|\frac{(-1)^n}{2n}|=\frac{1}{2n} \leq \frac{1}{2n^*} = \frac{1}{2(\lfloor \frac {1}{2\epsilon} \rfloor+1)}<\frac {1}{2(\frac {1}{2\epsilon})} = \epsilon$.
$\therefore$ $\lim (\frac{(-1)^n}{2n})=0$. $\blacksquare$


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**(7) 2.1.4** Is the sequence $\{ 2^{-n}\}$ convergent? If so, what is the limit?

By definition of convergent: $\lim x_n=L$ if $\forall ~ \epsilon >0 ~ \exists ~n^*$ s.t. $\forall ~n \geq n^* ~|x_n - L|< \epsilon$
Let $n^*=\lceil \log_2 {(\frac{1}{\epsilon}+1) \rceil}$
Then, $\forall~ n \geq n^*$
$|x_n - L|= |2^{-n}-0|=|2^{-n^*}|=|\frac{1}{2^n}|=\frac{1}{2^n} \leq \frac{1}{2^{n^*}} = \frac{1}{2^{\lceil \log_2 {(\frac{1}{\epsilon}+1) \rceil}}} \leq \frac{1}{\frac{1}{\epsilon}+1} < \frac{1}{\frac{1}{\epsilon}}= \epsilon$
$\therefore$ $\lim (2^{-n})=0$. $\blacksquare$

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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 an infinitely large sequence.
Then $\forall ~K \in{\mathbb R} ~\exists ~n^* \in{\mathbb N}$ such that $\forall ~n \geq n^*, |x_n| \geq K$.

Let ${x_{n_i}}$ be a subsequence of ${x_n}$.
Let $n^*_i \geq n^*$, 
Then, $\forall ~n_i \geq n^*_i, |x_{n_i}| \geq K$.

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

Let $\{x_n\}$ be a sequence. 
**a.** Show that $\lim {x_n}=0$ if and only if $\lim{|x_n|}=0$.

$\bullet$ $\lim {x_n}=0 \implies \lim{|x_n|}=0$
If $\lim x_n=0$ then $\forall ~\epsilon>0, ~ \exists ~ n^*$ s.t. $\forall ~ n \geq n^*,$ 
$$|x_n -0| = ||x_n|-0|< \epsilon$$ So,  $\lim{|x_n|}=0$. 

$\bullet$ $\lim {|x_n|}=0 \implies \lim{x_n}=0$
If $\lim |x_n|=0$ then $\forall ~\epsilon>0, ~ \exists ~ n^*$ s.t. $\forall ~ n \geq n^*,$ 
$$||x_n| -0| = |x_n-0|< \epsilon$$ So,  $\lim{x_n}=0$. 


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

Sequence $\{|x_n|\} =\{|(-1)^n|\}$ is convergent,
*Proof*
Let $n^*=1$
Then $\forall n \geq n^*, and \forall ~ \epsilon>0$
$$|x_n-L|=|1-1| =0<\epsilon$$ So, $\lim x_n=1$. $\blacksquare$

But $\{x_n\} =\{(-1)^n\}$ is divergent
*Proof*
Case 1: $a \neq 1$ and $a \neq -1$
Let $\epsilon_1 = |a-1|$ and $\epsilon_2 = |a+1|$
Then, $\exists~~ \epsilon = \frac{min{\{\epsilon'},{\epsilon''}\}}{2}$ s.t. $\forall ~ n^* ~\exists ~ n=n^*+1$ s.t. $|x_n-a| \geq \epsilon$
So $\lim x_n =1, \lim x_n=-1,$ or $x_n$ is divergent.

Case 2: $a = 1$, 
$\forall ~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2n^* +1$, so $n \geq n^*$.
Then, $x_{n^*} = (-1)^{2n^*+1}=-1$.
$\therefore |x_n - L|=|-1-1|= 2 \geq \epsilon$
so $\lim (-1)^n \neq 1.$

Case 3: $a = -1$
$\forall ~m \in{\mathbb N}$, let $\epsilon =1$ and $n=2n^*$, so $n \geq n^*$.
Then, $x_{n^*} = (-1)^{2n^*}=1$.
$\therefore |x_n - L|=|1+1|= 2 \geq \epsilon$
so $\lim (-1)^n \neq 1.$

$\therefore \{x_n\}$ is divergent. $\blacksquare$

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

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

*Monotone*: $\{x_n\} \searrow$ if $\forall ~ n, ~ x_{n+1} \leq x_n$

Let $x_n = \frac{n+1}{n}$ and ${x_{n+1}}= \frac{(n+1)+1}{n+1}$
Then 
$${x_{n+1}}= \frac{(n+1)+1}{n+1}= \frac{n+2}{n+1}* \frac{n}{n}= \frac{n^2+2n}{n(n+1)} \leq \frac{n^2+2n+1}{n(n+1)} = \frac{n+1}{n}* \frac{n+1}{n+1} = x_n$$ So $\{x_n\}$ is monotone decreasing. $\blacksquare$

*Bounded*: $\{x_n\}$ is bounded if $\exists ~ K$ such that $|x_n| \leq ~ K ~ \forall ~ n$

Let $K=2$. 
Since $\{x_n\}$ is monotone decreasing, $x_n \geq x_1 \forall~ n  \in{\mathbb N}$
Then, since $x_n > 0$ 
$$|\frac{n+1}{n}| = \frac{n+1}{n} = x_n \leq x_1 =2=K.$$ So, $\{x_n\}$ is bounded. ${\blacksquare}$


*Limit: Theorem 2.1.10*: a monotone sequence is bounded if and only if it is convergent. $\{x_n\} \searrow \lim x_n= \inf\{x_n, n  \in{\mathbb N}\}.$

$\bullet$ $\forall ~ n, \frac{n+1}{n}\ \geq 1$, true.
$\bullet$ $\bullet$ Given some $\epsilon >0$. Let $n^*= \frac{2}{\epsilon}$ such that 
$$x_{n^*} = \frac{n^*+1}{n^*}= \frac{\frac{\epsilon}{2}+1}{\frac{\epsilon}{2}} = 1+\frac{\epsilon}{2} < 1+\epsilon$$ $\therefore 1 \leq x_{n^*} < 1+\epsilon$  so $\inf\{\frac{n+1}{n}\}=1$. 
Thus, by Theorem 2.1.10, $\lim x_n= \inf\{x_n, n  \in{\mathbb N}\}=1.$$\blacksquare$


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

*Proof*
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 ${x_n}={x_k}$ for all $n \geq k$.

Since ${x_n}$ is convergent s.t. $\lim{x_n} = x_k$.
$\forall ~ \epsilon >0, ~ \exists ~n$ s.t. $\forall ~ n \geq n^*, |x_n - x_k|<\epsilon$.
Thus, when $n=k, x_n=x_k$. 

Now, when $n>k$, suppose by contradiction $x_n \neq x_k$.
Further, since $\{x_n\}$ is monotone, $\forall ~n, x_n>x_k$ or $x_n<x_k$.
Then, $\exists ~\epsilon= |\frac{x_n-x_k}{2}|$ s.t. $|x_n - x_k|<\epsilon$
Thus, $x_n=x_k$ by contradiction.

$\therefore$ when $n \geq k, x_n=x_k$. $\blacksquare$



**5. p. 50 Ex. 2.1.16**

Let $\{x_n\}$ be a sequence. Suppose there are two congruent subsequences $\{x_{n_i}\}=a$ and $\{x_{m_i}\}=b$, where $a \neq b$. Prove that $\{x_n\}$ is not convergent, without using Proposition 2.1.17.


*Proof*
Since $\lim x_{n_i}=a$, then
$\forall~\epsilon>0$, $\exists~n^*_i$ s.t. $\forall~n_i \geq n^*_i$, 
$\lvert x_{n_i}-a \rvert <\epsilon$.
    
Since $\lim x_{m_i}=b$, then
$\forall~\epsilon>0$, $\exists~m^*_i$ s.t. $\forall~m_i \geq m^*_i$, 
$\lvert x_{m_i}-b \rvert <\epsilon$
    
Suppose by contradiction that $\lim x_n=L$.

Case 1: $L \neq a$
Then, $\forall~m$, let $\epsilon=\frac{\lvert {L-a} \rvert}{2}$ and $n=n^*_i$, so:
$$\lvert L-a \rvert=\lvert L-x_n+x_n-a \rvert \leq \lvert x_n-a \rvert+\lvert x_n-L \rvert<\epsilon+\lvert x_n-L \rvert=\frac{\lvert {L-a} \rvert}{2}+\lvert x_n-L \rvert$$ $\therefore \lvert x_n-L \rvert \geq \frac{\lvert {L-a} \rvert}{2}=\epsilon$.

Thus, $\lim x_n=a$

Case 2: $L \neq b$
Analogous to the above proof with the result that $\lvert x_n-L \rvert \geq \frac{\lvert {L-b} \rvert}{2}=\epsilon$.
Thus, $\lim x_n=b$, which is a contradiction.

$\therefore \{x_n\}$ is divergent. $\blacksquare$


**6. p. 50 Ex. 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 $\{x_n\}=\{-1,0,1,-2,-\frac{3}{2},-\frac{2}{2},-\frac{1}{2},\frac{0}{2},\frac{1}{2},\frac{2}{2},\frac{3}{2},2,-3,-\frac{8}{3},-\frac{7}{3},-\frac{6}{3},...,\frac{8}{3},3,-4,...\}$
To create a subsequence $x_{n_i}$ converging to any $y \in{\mathbb R}$, 

For the first instance of every $k \in{\mathbb N}$ in $\{x_{n_i}\}$, 
let $x_{n_i}$ be the closest value to $y$ in the interval $[-k,k]$.
    
Since ${\mathbb Q}$ is dense in ${\mathbb R}$, then $\forall~\epsilon>0$,
$\exists~m_i$ s.t. $\forall~n_i \geq m_i$, $\lvert x_{n_i}-y \rvert < \epsilon$. $\blacksquare$

.
<!--
If meant $y \in {\mathbb N}$
Let $\{x_n\}=\{1,1,2,1,2,3,1,2,3,4,...\}$
Note that for any constant sequence $\{x_m\}=\{k\}$, $\lim x_m=k$ since 
$\forall~\epsilon >0$, $\exists~m^*$ s.t. $\forall~m \geq m^*$,
$\lvert x_m-k \rvert =\lvert k-k \rvert=0<\epsilon$.
Since $$ -->
    
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**Week 5 Quiz Prep**

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Lemma 2.2.3 (Modified)
$\{x_n\} \rightarrow a$ and $\{y_n\} \rightarrow b$ and $\exists ~ m$ s.t. $\forall ~n \geq m$ and $\forall ~n ~x_n \leq y_n \implies a \leq b$

Proof.
Let $a= \lim x_n$ and $b= \lim y_n$. 
Let $\epsilon >0$
Find an $M_1$ s.t. $\forall ~ n \geq M_1$ then $|x_n - a|< \frac{\epsilon}{2}$ 
Find an $M_2$ s.t. $\forall ~ n \geq M_2$ $|y_n - b|< \frac{\epsilon}{2}$

Then, $n \geq \max\{M_1, M_2\}$, we have $a-x_n < \frac{\epsilon}{2}$ and $y_n-b < \frac{\epsilon}{2}$

If we add these, we get
$y_n-x_n+a-b< \epsilon$ or $y_n-x_n<b-a+ \epsilon$

$\forall~\epsilon >0$, $\exists~m_1$ s.t $\forall~n \geq m_1$,
$\lvert x_n-a \rvert < \frac{\epsilon}{2}$.
$\forall~\epsilon >0$, $\exists~m_2$ s.t $\forall~n \geq m_2$,
$\lvert y_n-b \rvert < \frac{\epsilon}{2}$.
Let $m= \max\{m_1, m_2\}$, then
$a-x_m < \frac{\epsilon}{2}$ and $y_m-b < \frac{\epsilon}{2}$.
By addition of inequalities, $y_m-x_m+a-b<\epsilon$,
or $x_m-y_m<b-a+ \epsilon$.
Since $x_m<y_m$,
$0<b-a+ \epsilon$
or $a-b<\epsilon$.
Since $\epsilon>0$ is arbitrary, $a-b \leq 0$,
so $a \leq b$.

Problem 1.  Consider the converse of Proposition 2.2.7. If it is true, prove it. If it is false provide a counterexample.



Problem 2.  Finish the proof of Proposition 2.2.6 after " We leave the rest of the proof to the reader". Do not copy the beginning of the proof. 

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**Week 6**

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Solve Exercises 2.2.4 --- 2.2.8 (5 problems).

**2.2.4**
Suppose $x_1:=\frac{1}{2}$ and $x_{n+1}:={x_n^2}$. Show that $\{x_n\}$ converges and find $\lim x_n$. Hint: you cannot divide by 0.

*Proof*
We can define $\{y_n\}_{n=1}^\infty=\{x_{k+n}\}_{n_1}^\infty={\frac{1}{(k+1)^2}}$.
Let $m= \sqrt{\frac{1}{\epsilon}}$.
Since $\forall ~ n$, ${\frac{1}{(k+1)^2}}>0$, 
$\forall~\epsilon >0; n \geq m$,
$|y_n-0| = |\frac{1}{(k+1)^2}|= \frac{1}{(k+1)^2} <\frac{1}{k^2} \leq \frac{1}{m^2} = \frac{1}{\sqrt{\frac{1}{\epsilon}}^2} = \frac{1}{\frac{1}{\epsilon}} =\epsilon$.
Notice, $\{y_n\} \rightarrow 0$ is the tail of $\{x_n\}$.
$\therefore \{x_n\}$ converges s.t. $\lim{x_n}=0$. $\blacksquare$

.

**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.

*Proof*
First, rewrite $x_n := 1+\frac{\cos{n}}{n}$ and note that $\forall~n \in{\mathbb N}$, $-1 \leq \cos{n} \leq 1$.
So, $$1+ \frac{-1}{n} \leq 1+ \frac{\cos{n}}{n} \leq 1+ \frac{1}{n}$$
$\forall \epsilon >0$ let $m=\frac{2}{\epsilon}$, 
Then $\forall ~ n \geq m$
$$|(1+ \frac{-1}{n})-1| = |\frac{-1}{n}| = \frac{1}{n} \leq \frac{1}{m} = \frac{1}{\frac{2}{\epsilon}} = \frac{\epsilon}{2} < \epsilon$$
and
$$|(1+ \frac{1}{n})-1| = |\frac{1}{n}| = \frac{1}{n} \leq \frac{1}{m} = \frac{1}{\frac{2}{\epsilon}} = \frac{\epsilon}{2} < \epsilon$$
Since $\{1+\frac{-1}{n}\} \rightarrow 1$ and $\{1+\frac{1}{n}\} \rightarrow 1$, 
by the Squeeze Lemma, $\{x_n\}$ converges and $\lim{x_n}=1$. $\blacksquare$

.

**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 Prop 2.2.5? Why or why not?

We can NOT use Prop 2.2.5 because for the division of two sequences, $\{x_n\}$ and $\{y_n\}$, the requirement is $\lim x_n \neq 0$ and $\lim y_n \neq 0$
However, in this problem, $\lim x_n = \lim \frac{1}{n^2} = 0$ and $\lim y_n = \lim \frac{1}{n} = 0$. 

Instead, rewrite $z_n$ and $w_n$ as:
$$z_n := \frac {\frac{1}{n^2}}{\frac{1}{n}} = \frac{n}{n^2} = \frac{1}{n}$$
and
$$w_n := \frac {\frac{1}{n}}{\frac{1}{n^2}} = \frac{n^2}{n} = {n}$$
We have shown that $\{z_n\} = \{\frac{1}{n}\}$ converges s.t. $\lim z_n = 0$  
We know that $\{w_n\} = \{n\}$ diverges s.t. $\lim w_n = \infty$



.

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

*False.
Counterexample*
Let $\{x_n\} = \{(-1)^n\}$.
Then, it is trival to show that $\{x_n^2\} = \{(-1)^2n\} = \{1\} \rightarrow 1$
But $\{x_n\}$ diverges.


.

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

*Proof*
Using the ratio test for sequences (Lemma 2.2.12), we look at: 
$$\frac{|x_{n+1}|}{|x_n|} = \frac{(n+1)^2/2^{n+1}}{n^2 / 2^n} = \frac{2^n}{2^{n+1}} \frac{(n+1)^2}{n^2} = \frac{1}{2} \frac{n^2+2n+1}{n^2} = \frac{n^2+2n+1}{2n^2}$$

Then, $\forall ~ \epsilon > 0$. Let $m = \frac{2}{\sqrt\epsilon}$ s.t. $\forall ~ n \geq m$ we have
$$|\frac{n^2+2n+1}{2n^2}- \frac{1}{2}| = |\frac{n^2+2n+1}{2n^2}- \frac{n^2}{2n^2}| = |\frac{2n+1}{2n^2}| = \frac{1}{n} + \frac{1}{2n^2}$$ 
$$\leq$$
$$\frac{1}{m} + \frac{1}{2m^2} = \frac{1}{\frac{2}{\sqrt\epsilon}} + \frac{1}{2(\frac{2}{\sqrt\epsilon})^2} = \frac{\sqrt\epsilon}{2} + \frac{\epsilon}{8} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

Thus, $\lim \frac{|x_{n+1}|}{|x_n|} = \frac{1}{2} <1$.
$\therefore$ by Lemma 2.2.12, $\{x_n\}$ converges and $\lim x_n =0$

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**Week 7**


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1. Without using Lemma on nested Intervals prove that
$$\bigcap_{n=1}^\infty[\frac{-1}{n}, \frac{1}{n}]=\{0\}.$$
This is the Lemma we cannot use: $\{[a_n,b_n]\}$ s.t. $\forall ~ n ~ a_n \leq b_n$ and $a_n \leq a_{n+1}, b_{n+1} \leq b_n$ then $\bigcap_{n=1}^\infty [a_n, b_n] \neq 0$

    *Proof:*
    Let $x \in \bigcap_{n=1}^\infty[-1/n, 1/n]$.
Then $\forall~n$, 
$$-\frac{1}{n} \leq x \leq \frac{1}{n}$$
$$\implies \lim_{n\to\infty}(-\frac{1}{n}) \leq \lim_{n\to\infty}(x) \leq \lim_{n\to\infty}(\frac{1}{n})$$ 
$$\implies 0 \leq x \leq 0$$
$\therefore~\bigcap_{n=1}^\infty[-1/n, 1/n]=\{x\}=\{0\}$
.

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

    By Theorem 2.3.5, if $\{x_n\} \rightarrow L$, then $\lim \sup x_n=\lim \inf x_n=\lim x_n=L$.
    Notice $x_n :=\frac{(-1)^n}{n}=\frac{1}{n}(-1)^n$.
    Let $\{a_n\}=\{\frac{1}{n}\}$ and $\{b_n\}=\{(-1)^n\}$.
    Then, $\{a_n\} \rightarrow 0$ and $\{b_n\}$ is bounded.
    So $\{a_nb_n\} \rightarrow 0$
    Thus, $x_n =\frac{(-1)^n}{n} \rightarrow 0$
    $\therefore$ $x_n$ is a convergent sequence and $\lim \sup x_n=\lim \inf x_n = 0$

    .
    b) Let $x_n := \frac{(n-1)(-1)^n}{n}$, find $\lim \sup x_n$ and $\lim \inf x_n$


    Rewrite $x_n:=\frac{(n-1)(-1)^n}{n}=(1-\frac{1}{n})(-1)^n$.
    Let $a_n:=\sup\{x_k:k \geq n\}$ and $b_n:=\inf\{x_k:k \geq n\}$.
    Notice that $\{1-\frac{1}{n}\} \rightarrow 1$ and $\{(-1)^n\}=\{-1,1\}$.
    Thus, there are two convergent subsequences: 
    $\{x_{2m}\} \rightarrow 1$ and is monotone increasing.
    $\{x_{2m-1}\} \rightarrow -1$ and is monotone decreasing.
    So $\{a_n\}=\{1\}$ and $\{b_n\}=\{-1\}$.
    By Proposition 2.3.2, 
    $\lim \sup x_n=\inf\{a_n:n\in{\mathbb N}\}=\inf\{1:n\in{\mathbb N}\}=1$
    $\lim \inf x_n=\sup\{b_n:n\in{\mathbb N}\}=\sup\{-1:n\in{\mathbb N}\}=-1$.

.

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

    *Proof:*
Let $\{x_n\}$ be an unbounded sequence.
Then, $\forall~B \in{\mathbb R}$, $\exists~m$ s.t.
$\lvert x_m \rvert >B$.
    
    We can construct an infinitely large subsequence of $\{x_n\}$ as follows:
Let $x_{n_1}:=x_1$.
    
    Then, choose any value of $B$.
Let $x_{n_2}:=x_{m_1}$ s.t. $m_1>1$ and $\lvert x_{m_1} \rvert>B$
(which is always possible, since $\{x_n\}$ is unbounded).
    
    In general, $\forall~k \in{\mathbb N}$ s.t $k>2$,
    let $x_{n_{k+1}}:=x_{m_k}$ s.t. $m_{k}>m_{k-1}+1$ and $\lvert x_{m_k} \rvert>\lvert x_{n_{k}} \rvert$
(which is always possible, since if $\{x_n\}$ is unbounded, its tail is unbounded).
        
    Then for $\{x_{n_k}\}$, since $B$ was arbitrary in our construction,
$\forall~B \in{\mathbb R}$, $\exists~n_k^*=n_2$ s.t. $\forall~n_k \geq n_k^*$
$\lvert x_{n_k} \rvert >B$.
$\therefore$ $\{x_{n_k}\}$ is infinitely large.


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Week 8 - Tim's Desmos

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1. $m = 6$
2. $m = -7$
3. $15 < k \leq 31$
4. $m = 17$
5. $m = 26$
6. $n^* = 31$
7. $x_{n^*} = 0.95$
8. $x^* = 0.5$
9. $n^* = 18$
10. $0.34 \leq \epsilon < 0.5$
11. $n^*_1 = 4, n^*_2 = n^* = 9$
12. not possible, $a_4 = a_5$


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**Week 10**

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**2.5.1**
For $r\neq1$ prove, $\sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}$
Hint: Let $s:= \sum_{k=0}^{n-1} r^k$, then compute $s(1-r)=s-rs$, and solve for $s$.

*Proof*
Let $s=\sum_{k=0}^{n-1} r^k$
Then $$=s(1-r) = s-rs$$ 
$$=\sum_{k=0}^{n-1} r^k - r\sum_{k=0}^{n-1} r^k$$
$$=\sum_{k=0}^{n-1} r^k - \sum_{k=0}^{n-1} r^{k+1}$$
$$=(1+r+r^2+...+r^{n-1}) - (r+r^2+...+r^n)$$
$$=(1+r-r+r^2-r^2+...+r^{n-1}-r^{n-1}-r^n)$$
$$=(1-r^n)$$
This implies that $s(1-r)=(1-r^n)$
Then $s=\frac{1-r^n}{1-r}$ $\blacksquare$.

.

**2.5.2**
Prove that for $-1<r<1$ we have $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$
Hint: Use 2.5.1

*Proof*
Since we have
$$\sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}, -1<r<1$$
Then take the limit.
$$\therefore \lim \sum_{k=0}^{n-1} r^k = \lim \frac{1-r^n}{1-r} = \frac{{\lim{1}}-{\lim{r^n}}}{{\lim{1}}-{\lim{r}}}$$
This implies that $\sum_{k=0}^{n-1} r^k = \frac{1}{1-r}$ $\blacksquare$. 

.

**2.5.3 (a,b), use the comparison test**
Decide the convergence or divergence of the following series.
**a) $\sum_{n=1}^{\infty} \frac{3}{9n+1}$**

The series diverges.

*Proof*
Let $x_n := \frac{1}{10n}$ and $y_n:=\frac{3}{9n+1}$.
Then $\forall ~n$, 
$$0 \leq \frac{1}{10n} \leq \frac{3}{9n+1} \implies 0 \leq x_n \leq y_n$$
But, 
$$\sum_{n=1}^{\infty}x_n=\sum_{n=1}^{\infty}\frac{1}{10n}=\frac{1}{10}\sum_{n=1}^{\infty}\frac{1}{n}$$ 
diverges. $\therefore$ By the Comparison Test, 
$$\sum_{n=1}^{\infty}\frac{3}{9n+1}$$ 
also diverges. $\blacksquare$

.

**b) $\sum_{n=1}^{\infty} \frac{1}{2n-1}$**

The series diverges.

*Proof*
Let $x_n := \frac{1}{2n}$ and $y_n:=\frac{1}{2n-1}$.
Then $\forall ~n$, 
$$0 \leq \frac{1}{2n} \leq \frac{1}{2n-1} \implies 0 \leq x_n \leq y_n$$
But, 
$$\sum_{n=1}^{\infty}x_n=\sum_{n=1}^{\infty}\frac{1}{2n}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}$$ 
diverges. $\therefore$ By the Comparison Test, 
$$\sum_{n=1}^{\infty}\frac{1}{2n-1}$$ 
also diverges. $\blacksquare$

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**Week 11**

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**2.5.3 (c,d,e) use any convergence tests you like, even the ones we have not proved**
Decide the convergence or divergence of the following series.

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

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$$ 
    
Notice:
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^2}$$
    
Let $\{x_n\}:=\{\frac{1}{n^2}\}$. 
Note that, $\{x_n\} \searrow$ and only contains positive real numbers.
Further, $\lim {x_n}=0.$
Thus, by Prop. 2.6.2 (Alternating Series), $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$ converges.

.

**d) $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$**

$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$
    
Notice:
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}$$
Let $\{x_n\}:=\{\frac{1}{n^2}\}$ and let $\{y_n\}=\{\frac{1}{n^2+n}\}$.
Then $\forall~n$,
$$0 \leq \frac{1}{n^2+n} \leq \frac{1}{n^2} \implies 0 \leq x_n \leq y_n$$
    
By the convergence of the p-series, we know that $\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$.
$\therefore$ By the Comparison Test, $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$ also converges.

.

**e) $\sum_{n=1}^{\infty} ne^{-n^2}$**

$$\sum_{n=1}^{\infty}ne^{-n^2}$$
    
Let $\{x_n\}:=\{ne^{-n^2}\}$. Note that $\forall~n$, $x_n>0$.
Then, 
$$L:=\lim \sup \lvert x_n \rvert^{\frac{1}{n}}=\lim \sup (x_n)^{\frac{1}{n}}=\lim \sup (ne^{-n^2})^\frac{1}{n}=\lim \sup (n^{\frac{1}{n}}e^{-n})>0$$
    
By the Ratio Test for monotonicity we see,
$$\frac{n^{\frac{1}{n+1}}e^{-n-1}}{n^{\frac{1}{n}}e^{-n}}=n^{\frac{-1}{n(n+1)}}e^{-1}<1$$

Thus, $\{(x_n)^{\frac{1}{n}}\} \searrow$.
    
Then, since the first term $(x_1)^{\frac{1}{1}}=e^{-1}<1$,
$$0<\lim \sup \lvert x_n \rvert^{\frac{1}{n}}=L<1$$

$\therefore$ By the Root Test, $\sum_{n=1}^{\infty}ne^{-n^2}$ converges.

.

**Exercise 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.**

*Proof*
$\sum_{n=1}^{\infty} x_n$ converges.
This implies that $\sum_{n=2}^{\infty} x_n$ also converges since finite terms do not change the behavior of a series.
Then,
$$\sum_{n=2}^{\infty} x_n = x_2+x_3+x_4+ x_5 +... = \sum_{n=1}^{\infty} x_{2n}+x_{2n+1}$$.

Since the series is convergent, rearranging the terms does not change the convergence.

Thus $\sum_{n=1}^{\infty} (x_{2n}+x_{2n+1})$ is also convergent. $\blacksquare$

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

Consider the sequence $a_n = (-1)^n$
Then,
$$a_{2n} = (-1)^{2n} = 1$$
$$a_{2n+1} = (-1)^{2n+1} = -1$$
$$\implies a_{2n}+a_{2n+1} = 1+-1 = 1-1 = 0$$
$$\implies \sum_{n=1}^{\infty} (a_{2n}+a_{2n+1}) = \sum_{n=1}^{\infty} 0 = 0$$

Thus this series is convergent, but $\sum_{n=1}^{\infty} a_n$ diverges.

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**Week 12**

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Rewrite lemma 3.1.7 and rework the proof for the case when S=(a,b), and c is a point in the interior of S.

**Lemma 3.1.7**
Let $S⊂{\mathbb R}$ and $c$ be a cluster point of $S$. Let $f:S → {\mathbb R}$ be a function.
Then $f(x) → L$ as $x → c$, iff for every sequence ${x_n}$ of numbers such that $x_n ∈ S\ \{c\} \forall ~ n$ s.t. $lim x_n = c$, we have that the sequence $\{f(x_n)\}$ converges to $L$.

*Proof*
Suppose $f(x) → L$ as $x → c$ and $\{x_n\}$ is a sequence s.t. $x_n \in (a,b)~\backslash \{c\}$ and $\lim x_n=c$.

Let $\epsilon>0$. 
Note before the next step that since all points in $[a,b]$ are cluster points of $(a,b)$, $c$ is a cluster point of $(a,b)$.
Thus, we can choose a $\delta>0$ s.t. if $x \in (a,b)~\backslash \{c\}$ and $|x-c|<\delta$, then $|f(x)-L|<\epsilon$.

Since $\{x_n\}$ converges to $c$, we can find an $m$ s.t. $\forall~n \geq m$, $|x_n-c|<\delta \implies |f(x_n)-L|<\epsilon$.
$\therefore \{f(x_n)\} \rightarrow L$. $\blacksquare$

.

For the other direction, we use proof by contrapositive. 

Suppose $f(x) \nrightarrow L$ as $x \rightarrow c$.
Then, $\forall~\delta>0$, $\exists~x∈(a,b)~\backslash \{c\}$, 
where $|x−c|<\delta$ and $|f(x)−L$ \geq \epsilon$.

As seen in the proof of the other direction, note that $c$ is a cluster point of $(a,b)$. 
Let $\delta= \frac{1}{n}$ to construct a sequence $\{x_n\}$.
We have that $\exists~\epsilon > 0$ s.t. $\forall ~ n$, $\exists~x_n \in (a,b)\backslash \{c\}$, 
where $|x_n−c|< \frac{1}{n}$ and $|f(x_n)−L| \geq \epsilon$. 
Then $\{x_n\}$ converges to $c$, but $\{f(x_n)\}$ does not converge to $L$. $\blacksquare$


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**Week 13**

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**Ex. 6.1.1**
Let $f$ and $g$ be bounded functions on $[a,b]$. Prove
$$||f+g||_u ≤ ||f||_u + ||g||_u$$

*Proof*
Given that $f, g$ are bounded functions on $[a,b]$ $\exists ~ A, B >0$ s.t.
$$|f(x)|<A ~ \forall ~ x \in [a,b]$$
$$|g(x)|<B ~ \forall ~ x \in [a,b]$$
So, $\sup |f(x)|$ exists and $x \in [a,b]$ and $\sup |g(x)|$ exists ($\in {\mathbb R}) ~ x \in [a,b]$
Then, $$||f||_u := \sup_{x\in[a,b]}|f(x)|, ||g||_u := \sup_{x\in[a,b]}|g(x)|$$
Now $\forall ~ \in [a,b]$
$$|(f+g)(x)|= |f(x)+g(x)| \leq |f(x)|+|g(x)| \leq ||f||_u + ||g||_u$$
Thus, $|(f+g)(x)| \leq ||f||_u + ||g||_u ~\forall ~ x\in[a,b]$
So, $$\sup_{x\in[a,b]}|(f+g)(x)| \leq ||f||_u + ||g||_u$$
i.e. $||f+g||_u \leq ||f||_u + ||g||_u$
$[||f+g||_u = \sup{|(f+g)(x)|:x\in[a,b]}]$ $\blacksquare$

.

**Ex. 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 $\{f_ng_n\}$ (the multiple) does not converge uniformly to $fg$ on $A$. 
Hint: Let $A := R$, let $f(x) := g(x) := x$. You can even pick $f_n = g_n$.

*Proof*
Let $A={\mathbb R}$, $f(x)=g(x)=x$ and $f_n(x)=g_n(x)=x+\frac{1}{n}$
Then for $\epsilon > 0$, choose $m > \frac{1}{\epsilon}$ so that $\forall ~n\geq{m}$
$$|f_n(x)-f(x)|=|x+\frac{1}{n}-x|=\frac{1}{m} < \epsilon$$
Thus, $\{f_n(x)\}$ converges uniformly to $f$
Similarly, we can prove $\{g_n(x)\}$ converges uniformly to $g$
Now $\epsilon > 0$ and $x \in A$
$$|f_n(x)g_n(x) - f(x)g(x)| = |(x+\frac{1}{n})^2 - x^2| = |x^2 + \frac{2x}{n} + \frac{1}{n^2} - x^2| = |\frac{2x}{n} + \frac{1}{n^2}| = \frac{2}{n}|x| + \frac{1}{n^2} > \epsilon$$
Hence, $\{f_n(x)g_n(x)\}$ does not converge uniformly to $f(x)g(x)$. $\blacksquare$

