22 Sequences Practice

2.2.1, 2.2.2, 2.2.3, 2.2.4, 2.2.7, 2.2.8

2.2.1

Question

What happens if we reverse the order of the quantifiers in Convergence of a Sequence? Namely, define vercongence as:

\exists ε > 0 \forall N \in N (n \geq N \to | x_{n} - x | < ε)

Give an examples of a vercongent sequence. Is there an example of a vercongent sequence that is divergenct? Can a sequence verconge to two different values? What exactly is being described in this strange definition?

Let's do some scratch:

The idea is that we can make $ε$ be as big as we want! As long as the sequence doesn't get bigger and bigger away from $x$ , then the sequence should verconge. Notice that the $\forall N \in N$ essentially restricts that $| x_{n} - x | < ε$ happens for any $n \in N$ instead. Namely, the definition encourages an equivalent statement:

\exists ε > 0 \forall n \in N (| x_{n} - x | < ε)

Proof
a. An example of a vergongent sequence is $a_{n} = \frac{1}{n}$ , which verconges to $0$ . Another example is $a_{n} = (- 1)^{n}$ , which verconges to any $0, 1, 2, \dots$ since I can make $ϵ = 10$ , and then since all $| a_{n} - x | < 10$ since $| a_{n} - x |$ is either $| 1 - x |$ or $| 1 + x |$ , then it technically verconges.

b. $a_{n} = 2^{n}$ is divergenct. For contradiction, assume some vercongent limit $x$ , so then there's an $ε > 0$ where for $n \in N$ :

| x_{n} - x | = | 2^{n} - x | < ε

But plug in $n = 2 \log_{2} (ε)$ . Then:

| x_{n} - x | = | 2^{\log_{2} (ε + x)} - x | = | ε + x - x | = ε ≮ ε

c. We've given the example of $a_{n} = (- 1)^{n}$ that verconges to many possible values, so it is not unique.

c. The definition is essentially describing whether or not a sequence is bounded, except it's a bit more specific in that its' bounding it to the range of some $V_{ε} (x)$ .
☐

2.2.2

Question

Verify, using the Convergence of a Sequence definition, that the following sequences converge to the proposed limit:
a. $lim \frac{2 n + 1}{5 n + 4} = \frac{2}{5}$
b. $lim \frac{2 n^{2}}{n^{3} + 3} = 0$
c. $lim \frac{\sin (n^{2})}{\sqrt[3]{n}} = 0$

Let's do some scratch work.

a. We want to show, for any $ε > 0$ :

\begin{aligned} | \frac{2 n + 1}{5 n + 4} - \frac{2}{5} | & < ε \\ | \frac{5 (2 n + 1) - 2 (5 n + 4)}{5 (5 n + 4)} | & < ε \\ | \frac{- 3}{25 n + 20} | & < ε \\ \frac{3}{25 n + 20} & < ε & (25 n + 20 > 0) \\ \frac{3}{ε} - 20 & < 25 n \\ \frac{3}{25 ε} - \frac{4}{5} & < n \end{aligned}

Thus this suggests choosing $N > \frac{3}{25 ε} - \frac{4}{5}$ such that then $n \geq N$ works with this.

b. We want to show, for any $ε > 0$ :

\begin{aligned} | \frac{2 n^{2}}{n^{3} + 3} - 0 | & < ε \\ \frac{2 n^{2}}{n^{3} + 3} & < ε & n \geq 0 \end{aligned}

We seem to be stuck since we can't isolate $n$ , but we can try to find something bigger than our $\frac{2 n^{2}}{n^{3} + 3}$ . Notice:

\frac{1}{n} < \frac{2 n^{2}}{n^{3} + 3} \equiv n^{3} + 3 < 2 n^{3} \equiv \sqrt[3]{3} < n

Thus, since then we'd have $\frac{1}{n} < ε \leftrightarrow \frac{1}{ε} < n$ then we need $n > max (\frac{1}{ε}, \sqrt[3]{3})$ . We don't even have to use a cube root, but rather $2$ would be fine. Thus choose $N > max (\frac{1}{ε}, 2)$ .

Summary

You are allowed to do $old sequence \leq new bigger sequence$ , but you have to check that the inequality holds for all $n$ . This helps choose $N$ .

c. We want to show, for any $ε > 0$ :

\begin{aligned} | \frac{\sin (n^{2})}{\sqrt[3]{n}} - 0 | & < ε \\ \frac{| \sin (n^{2}) |}{\sqrt[3]{n}} & < ε & (\sqrt[3]{n} \geq 1) \\ \frac{| \sin (n^{2}) |}{\sqrt[3]{n}} & < ε \end{aligned}

But notice that:

\frac{| \sin (n^{2}) |}{\sqrt[3]{n}} \leq \frac{1}{\sqrt[3]{n}} < ε

This suggests that $n > \frac{1}{ε^{3}}$ so choose $N > \frac{1}{ε^{3}}$ .

Proof
a. Let $ε > 0$ . Choose $N \in N$ such that $N > \frac{3}{25 ε} - \frac{4}{5}$ . Then notice that:

\begin{aligned} | \frac{2 n + 1}{5 n + 4} - \frac{2}{5} | & = | \frac{5 (2 n + 1) - 2 (5 n + 4)}{5 (5 n + 4)} | \\ = | \frac{- 3}{25 n + 20} | \\ = \frac{3}{25 n + 20} & (25 n + 20 > 0) \\ \leq \frac{3}{25 (\frac{3}{25 ε} - \frac{4}{5}) + 20} \\ = \frac{3}{\frac{3}{ε} - 20 + 20} \\ = ε \end{aligned}

Thus $| a_{n} - \frac{2}{5} | < ε$ as desired.

b. Let $ε > 0$ . Choose $N \in N$ such that $N > max (2, \frac{1}{ε})$ . Then for any arbitrary $n > N$ :

\begin{aligned} | \frac{2 n^{2}}{n^{3} + 3} - 0 | & = \frac{2 n^{2}}{n^{3} + 3} \\ < \frac{1}{n} \\ \leq ε & (n \geq \frac{1}{ε}) \end{aligned}

We just have to show that $\frac{2 n^{2}}{n^{3} + 3} < \frac{1}{n}$ and then we are done:

\begin{aligned} \frac{1}{n} < \frac{2 n^{2}}{n^{3} + 3} & \equiv n^{3} + 3 < 2 n^{3} \\ \equiv 3 < n^{3} \\ \equiv \sqrt[3]{3} < n \\ \Leftarrow 2 < n \end{aligned}

c. Let $ε > 0$ . Choose $N \in N$ such that $N > \frac{1}{ε^{3}}$ . Then for $n \geq N$ :

\begin{aligned} | \frac{\sin (n^{2})}{\sqrt[3]{n}} - 0 | & = \frac{| \sin (n^{2}) |}{\sqrt[3]{n}} \\ \leq \frac{1}{\sqrt[3]{n}} \\ < ε & (\frac{1}{\sqrt[3]{n}} < ε \equiv \frac{1}{ε^{3}} < n) \end{aligned}

☐

2.2.3

Question

Describe what we would have to demonstrate in order to disprove each of the following statements:
a. At every college in the United States, there is a student who is at least seven feet tall.
b. For all colleges in the US, there exists a professor who gives every student a grade of either A or B.
c. There exists a college in the US where every student is at least six feet tall.

Proof
a. There would have to exist some college in the US such that every student is strictly less than seven feet tall.

b. There would exist a college in the US such that for any professor there they never gives A's nor B's.

c. We'd show that for any college in the US, there is a student who is strictly less than six feet tall.
☐

2.2.4

Question

Given an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case.
a. A sequence with an infinite number of ones that does not converge to one.
b. A sequence with an infinite number of ones that converges to a limit not equal to one.
c. A divergent sequence such that for every $n \in N$ it is possible to find $n$ consecutive ones somewhere in the sequence.

Proof
a. $(a_{n}) = (- 1)^{n}$ has an infinite number of $1$ 's, but doesn't converge to $1$ .
b. This is impossible. Unlike (a), if some subsequence $b_{n} ↛ 1$ then the other subsequence of terms not with $b_{n}$ , denoted $a_{n}^{'}$ , would converge to $1$ . These disagree with the requirement that $n \geq N$ since then the value $| a_{n} |$ would have to be in two different $ε$ -neighborhoods.
c. Use:

(a_{k}) = {\begin{cases} 0 & k \equiv 0 \mod n + 1 \\ 1 & k ≢ 0 \mod n + 1 \end{cases}

This diverges similar to (a) since two different subsequences approach different $V_{ε}$ 's.
☐

2.2.7

Question

Here are two useful definitions:
i. A sequence $(a_{n})$ is eventually in a set $A \subseteq R$ if there exists an $N \in N$ such that $a_{n} \in A$ for all $n \geq N$ .
ii. A sequence $(a_{n})$ is frequently in a set $A \subseteq R$ if, for every $N \in N$ there exists an $n \geq N$ such that $a_{n} \in A$ .
Answer the following questions:
a. Is the sequence $(- 1)^{n}$ eventually or frequently in the set ${1}$ .
b. Which definition is stronger? Does frequently imply eventually or does eventually imply frequently?
c. Give an alternate rephrasing of Convergence of a Sequence - Topological Version, using either frequently or eventually. Which is the term we want?
d. Suppose an infinite number of terms of a sequence $(x_{n})$ are equal to 2. Is $(x_{n})$ necessarily eventually in the interval $(1.9, 2.1)$ ? Is it frequently in $(1.9, 2.1)$ ?

Proof
a. It's definitely not eventually since if there was such an $N$ then either $a_{n}, a_{n + 1}$ is $- 1$ which is not in $A = {1}$ . It is frequently since if we let $N$ be arbitrary, then choosing $n = N$ or $n = N + 1 \geq N$ gives either $a_{n} = 1 \in {1}$ .

b. I'll show that eventually is stronger than (implies) frequently. Given that $\exists N \in N \forall n \geq N (a_{n} \in A)$ we'll show $\forall N^{'} \in N \exists n \geq N^{'} (a_{n} \in A)$ . Let $N^{'} \in N$ be arbitrary. By the Well-Ordering Principle of $N$ , then we have two possibilities:

$N < N^{'}$ . Then choose any $n \geq N^{'}$ , since then $n \geq N$ . Then by our given then $a_{n} \in A$ as required.
$N \geq N^{'}$ . Then choose any $n \geq N$ , since then by our given $a_{n} \in A$ as required.

Convergence - Topological Version (eventually/frequently definition)

A sequence $(a_{n})$ converges to $a$ if $(a_{n})$ is eventually in the set $V_{ε} (a)$ .

d. Notice that then $(x_{n})$ is not eventually by the counterexample $(x_{n}) = 2 (- 1)^{n}$ . However, it must be frequently since eventually we can always find a $2$ in the sequence (by definition) which is in $(1.9, 2.1)$ .
☐

2.2.8

Question

For some additional practice with nested quantifiers, consider the following invented definition:

Let's call a sequence $(x_{n})$ zero-heavy if $\exists M \in N$ such that for all $N \in N$ then $\exists n$ (where $N \leq n \leq N + M$ ) where $x_{n} = 0$ .

(this is saying that for a consecutive subsequence from $n = N$ to $n = N + M$ there's at least one $0$ in that subsequence)

a. Is the sequence $(0, 1, 0, 1, 0, \dots)$ zero-heavy?
b. If a sequence is zero-heavy, does it necessarily contain an infinite number of zeros? If not, provide a counter example.
c. If a sequence contains an infinite number of zeros, is it necessarily zero-heavy? If not, provide a counter example.
d. Form the logical negation of the above definition. That is, find the definition of a sequence being not zero-heavy.

Proof
a. It is zero-heavy. Specifically choose $M = 1$ (you can always choose something bigger, but this is easiest to prove). Then let $N \in N$ . Then:

N \leq n \leq N + 1

so if $a_{N} = 0$ then choose $n = N$ . Instead if $a_{N} = 1$ then by the construction of $(x_{n})$ then choose $n = N + 1$ so then $a_{N + 1} = 0$ .

b. It must have infinitely many zeroes. Suppose a zero-heavy sequence $(x_{n})$ and assume it only has finitely many zeroes. Then consider the sequence $(x_{n})_{n \geq N}$ where $N$ is the point after the last zero in the sequence. Then this new sequence has no zeroes, so it definitely isn't zero-heavy. As a result, then for our $N$ then no $M$ can exist such that any $N \leq n \leq N + M$ ever makes $x_{n} = 0$ . As a result, then the old subsequence must not be zero-heavy, which is a contradiction. As such, then it must have infinitely many zeroes.

c. This isn't true. Consider:

(x_{n}) = {\begin{cases} 1 & n^{2} \notin N \\ 0 & n^{2} \in N \end{cases}

not zero-heavy

A sequence $(x_{n})$ is not zero-heavy if $\forall M \in N$ then $\exists N \in N$ where $\forall n (N \leq n \leq N + M)$ we have it that $x_{n} \neq 0$ .

☐

#todo/school : Finish these problems 📅 2024-10-11 ✅ 2024-10-11