33 - Sequences

A sequence is a function $f : N \to R$ . $a_{n} = f (n)$ .

Example:
3, 9, 27, ...
$(a_{n})_{n \in N}$

Arithmetic sequence:

$a_{n} = c + (n - 1) d$
Example: $c = 4$ , $d = 5$
$a_{1} = 4, a_{2} = 9, a_{3} = 14, . . .$

Geometric sequence:

$a_{n} = c \cdot r^{n - 1}$
Example:
$c = 2, r = \frac{1}{3}$
$a_{1} = 2, a_{2} = \frac{2}{3}, a_{3} = \frac{2}{9}, . . .$

Limits:
$lim_{n \to \infty} \frac{1}{n} = 0$

Choose an epsilon $ε$ for which to be within the limit.

Call the $x$ value where this happens $N$ . All $x$ after $N$ must be within $ε$ .

$\forall ε > 0, \exists N \in R, \forall n \in N, n > N ⟹ | a_{n} - L | < ε$ .

!Screenshot 2023-12-04 at 10.18.42 AM.png
!Screenshot 2023-12-04 at 1.00.14 PM.png

Proof of limit:
The sequence $(a_{n})_{n \in N}$ defined by $a_{n} = 3 - \frac{4}{n}$ converges to $L = 3$ .

$\forall ε > 0, \exists N \in R, \forall n \in N n > N ⟹ | (3 - \frac{4}{n}) - 3 | < ε$ .

Scratch work.
Let $ε > 0$ .
$| (3 - \frac{4}{n}) - 3 | < ε$
$\frac{4}{n} < ε$
$4 < n ε$
$n > \frac{4}{ε}$
Set $N = \frac{4}{ε}$

Proof.
Let $ε > 0$ . Fix $N = \frac{4}{ε}$ . Let $n \in N, n > N$
Then

\begin{aligned} | 3 - \frac{4}{n} - 3 | & = | - \frac{4}{n} | \\ = \frac{4}{n} \\ < \frac{4}{N} \\ = \frac{4}{(\frac{4}{ε})} \\ = ε \end{aligned}

$◻$

Prove that $lim_{n \to \infty} (\sqrt{n^{2} + 1} - n) = 0$

Scratch work

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

Proof. Let $ε > 0$ . Fix $N = \frac{1 - ε^{2}}{2 ε}$ . Let $n > N, n \in N$ . Then

\begin{aligned} n & > \frac{1 - ε^{2}}{2 ε} \\ 2 ε n & > 1 - ε^{2} \\ 2 ε n + ε^{2} & > 1 \\ n^{2} + 2 ε n + ε^{2} & > n^{2} + 1 \\ (n + ε)^{2} & > n^{2} + 1 \\ n + ε & > \sqrt{n^{2} + 1} \\ ε & > \sqrt{n^{2} + 1} - n \end{aligned}

Thus $ε > | \sqrt{n^{2} + 1} - n - 0 |$ . $◻$

E 33.8
$\neg (\forall ε > 0, \exists n \in R, n > N ⟹ | a_{n} - L | < ε)$ =
$\exists ε > 0, \forall N \in R, \exists n \in N, n > N \land | a_{n} - L | \geq ε$

HW 33

Ben Finch

E 33.1
a)

$5, 2, - 1, - 4, - 7, - 10$
$a_{n} = 8 - 3 n$

4, 8, 16, 32, 64, 128
$a_{n} = 4 \cdot 2^{n - 1}$

$\frac{1}{2}, \frac{5}{4}, 2, \frac{11}{4}, \frac{14}{4}, \frac{17}{4}$
$a_{n} = \frac{1}{2} + \frac{3}{4} (n - 1)$

$\frac{1}{2}, \frac{1}{3}, \frac{2}{9}, \frac{4}{27}, \frac{8}{81}, \frac{16}{243}$
$a_{n} = \frac{1}{2} \cdot {(\frac{2}{3})}^{n - 1}$

E 33.2
a) $\forall ε > 0, \exists N \in R, n > N ⟹ | (3 - \frac{4}{n}) - 3 | < ε$
b) $\exists ε > 0, \forall N \in R, \exists n \in N, n > N \land | 6 - 3 | \geq ε$

E 33.3
a)
Proof. We have two cases:
Suppose $max (a, b) = a$ . Thus $m a x (a, b) \geq a$ . Then $a \geq b$ so $m a x (a, b) \geq b$ .
Suppose $max (a, b) = b$ . Thus $m a x (a, b) \geq b$ . Then $b > a$ , so $m a x (a, b) > a$ . $◻$

b)
Proof. We have two cases:
Suppose $m i n (a, b) = a$ . Thus $m i n (a, b) \leq a$ . Then $a \leq b$ so $m i n (a, b) \leq b$ .
Suppose $m i n (a, b) = b$ . Thus $m i n (a, b) \leq b$ . Then $b < a$ so $m i n (a, b) < a$ .

c)
Proof. Assume $x > m a x (a, b)$ . We have two cases:
Suppose $m a x (a, b) = a$ . Then $x > a \geq b$ .
Suppose $m a x (a, b) = b$ . Then $x > b > a$ . $◻$

E 33.4

Scratch work:

\begin{aligned} | \frac{2}{n^{2}} | & < ε \\ \frac{2}{n^{2}} & < ε \\ 2 & < ε n^{2} \\ \frac{2}{ε} & < n^{2} \\ N & = \sqrt{\frac{2}{ε}} \end{aligned}

Proof.
Let $ε > 0$ . Fix $N = \sqrt{\frac{2}{ε}}$ . Let $n > N, n \in N$ . Then

\begin{aligned} n & > \sqrt{\frac{2}{ε}} \\ n^{2} & > \frac{2}{ε} \\ ε n^{2} & > 2 \\ ε & > \frac{2}{n^{2}} \end{aligned}

Thus $ε > | \frac{2}{n^{2}} - 0 |$ and so the $lim_{n \to \infty} \frac{2}{n^{2}} = 0$ . $◻$

E 33.5
Scratch work:

\begin{aligned} | \frac{3 n - 5}{2 n + 4} - \frac{3}{2} | & < ε \\ \frac{3 n - 5}{2 n + 4} - \frac{3}{2} & < ε \\ \frac{2 (3 n - 5) - 3 (2 n + 4)}{2 (2 n + 4)} & < ε \\ \frac{6 n - 10 - 6 n - 12}{4 n + 8} & < ε \\ - \frac{22}{4 n + 8} & < ε \\ - 22 & < ε (4 n + 8) \\ - 22 & < 4 ε n + 8 ε \\ - 4 ε n & < 22 + 8 ε \\ 4 ε n & > - 22 - 8 ε \\ n & > \frac{- 22 - 8}{4 ε} \\ N & = \frac{- 22 - 8}{4 ε} \end{aligned}

Proof.
Let $ε > 0$ . Fix $N = \frac{- 22 - 8}{4 ε}$ . Let $n > N, n \in N$ . Then

\begin{aligned} n & > \frac{- 22 - 8}{4 ε} \\ 4 ε n & > - 22 - 8 ε \\ - 4 ε n & < 22 + 8 ε \\ - 22 & < 4 ε n + 8 ε \\ - 22 & < ε (4 n + 8) \\ - \frac{22}{4 n + 8} & < ε \\ \frac{6 n - 10 - 6 n - 12}{4 n + 8} & < ε \\ \frac{2 (3 n - 5) - 3 (2 n + 4)}{2 (2 n + 4)} & < ε \\ \frac{3 n - 5}{2 n + 4} - \frac{3}{2} & < ε \end{aligned}

Thus $ε > | \frac{3 n - 5}{2 n + 4} - \frac{3}{2} |$ . $◻$

E 33.6
The sequence does converge with a limit at $1$ .

Scratch Work:

\begin{aligned} | \frac{n + 1}{n} - 1 | & < ε \\ \frac{n + 1}{n} - 1 & < ε \\ \frac{n + 1 - n}{n} & < ε \\ \frac{1}{n} & < ε \\ n > \frac{1}{ε} \\ N & = \frac{1}{ε} \end{aligned}

Proof.
Let $ε > 0$ . Fix $N = \frac{1}{ε}$ . Let $n > N, n \in N$ . Then

\begin{aligned} n > \frac{1}{ε} \\ \frac{1}{n} & < ε \\ \frac{n + 1 - n}{n} & < ε \\ \frac{n + 1}{n} - 1 & < ε \end{aligned}

Thus $ε > | \frac{n + 1}{n} - 1 |$ . So the sequence converges with a limit at 1. $◻$

E 33.7

Let $ε > 0$ . We want to find $N \in N$ such that for all $n > N$ , $| a_{n} - c | < ε$ holds.

Since $a_{n} = c$ for all $n$ , we can see that:
$| a_{n} - c | = | c - c | = 0 < ε$

The inequality $0 < ε$ holds for any positive $ε$ , as $0$ is less than any positive number.

Therefore, we can choose any $N$ , for example $N = 1$ , and the condition $| a_{n} - c | < ε$ will be satisfied for all $n > N$ , because the left-hand side of the inequality is always $0$ , which is less than any positive $ε$ .

This proves that the constant sequence $c, c, c, \dots$ converges to $c$ . $◻$

E 33.8
Proposition. $\exists ε > 0, \forall N \in R, \exists n \in N, n > N \land | n - 3 | \geq ε$

Proof.
Fix $ε = 1$ . We want to show that for any real number $N$ , there exists a natural number $n > N$ such that $| n - 3 | \geq 1$ .

For any given $N$ , choose $n$ to be the smallest integer greater than $N$ . Then $n > N$ .

If $n \geq 4$ , then $| n - 3 | = n - 3 \geq 1$ .
If $n \leq 2$ , then $| n - 3 | = 3 - n \geq 1$ .

Therefore, no matter how large we choose $N$ , we can always find an $n > N$ such that $| a_{n} - L | \geq ε$ , which means the sequence $a_{n} = n$ does not converge to $L = 3$ . $◻$

E 33.9
Scratch work:

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

Proof. Let $ε > 0$ . Fix $N = \frac{1 - ε^{2}}{2 ε}$ . Let $n > N, n \in N$ . Then

\begin{aligned} n & > \frac{1 - ε^{2}}{2 ε} \\ 2 ε n & > 1 - ε^{2} \\ 2 ε n + ε^{2} & > 1 \\ n^{2} + 2 ε n + ε^{2} & > n^{2} + 1 \\ (n + ε)^{2} & > n^{2} + 1 \\ n + ε & > \sqrt{n^{2} + 1} \\ ε & > \sqrt{n^{2} + 1} - n \end{aligned}

Thus $ε > | \sqrt{n^{2} + 1} - n - 0 |$ . $◻$

E 33.10
a)
Proof.
Let $ε > 0$ be given. Since $| r | < 1$ , the sequence $r^{n}$ converges to 0. Hence, there exists an $N$ such that for all $n > N$ , $| r^{n} | < \frac{ε}{| c |}$ . Then for all $n > N$ :

\begin{aligned} | a_{n} - 0 | & = | c \cdot r^{n - 1} | \\ = | c | \cdot | r^{n - 1} | \\ < | c | \cdot \frac{ε}{| c |} \\ = ε \end{aligned}

Thus, $a_{n}$ converges to 0.

b)
Proof.
Suppose $| r | \geq 1$ . Then $a_{n} = c \cdot r^{n - 1}$ cannot converge to 0, because for $| r | = 1$ , $a_{n}$ would stay at $| c |$ , and for $| r | > 1$ , $a_{n}$ would increase unbounded. Since $a_{n}$ is given to converge to 0, we must have $| r | < 1$ . $◻$

Proof.
Let $c > 0$ and $r > 1$ . Since $r > 1$ , $r^{n} \to \infty$ as $n \to \infty$ . Thus, $a_{n} = c \cdot r^{n - 1}$ also tends to infinity as $n$ increases, so $a_{n}$ diverges. $◻$