Geometry
# Topology

On Day 1, Zeno is 1,000,000 miles from his goal. Each day, he travels $\frac{9}{10}$ of the remaining distance to his goal.

Let $d_n =$ the distance remaining at the end of Day $n.$ Is there a day after which Zeno will always be less than $\frac{1}{4}$ mile from his goal?

**Hint.**

$d_1 = 1,000,000$

$d_2 = 100,000$

$d_3 = 10,000$

$d_4 = 1,000$

etc.

On Day 1, Zilo is at home, 1,000,000 miles from his goal. He progresses towards his goal on even days, but needs to return home on odd days to check in.

Specifically on even days, he travels to the closest point to his goal he's reached yet, and then proceeds an additional $\frac{9}{10}$ of the remaining distance to his goal. On odd days, he goes back to being 1,000,000 miles from his goal.

Let $d_n =$ the distance remaining at the end of Day $n.$ Is there a day after which Zilo will always be closer than $\frac{1}{4}$ mile from his goal?

**Hint.**

$d_1 = 1,000,000$

$d_2 = 100,000$

$d_3 = 1,000,000$

$d_4 = 10,000$

$d_5 = 1,000,000$

$d_6 = 1,000$

etc.

Consider the sequence $\{(-1)^n\} = \{-1, +1, -1, +1, -1, +1, \dots\}.$

What are the subsequential limit(s) of this sequence?

**Note.** A subsequence of a sequence $\{x_n\}= \{x_1, x_2, \ldots\}$ is a sequence
$\{x_{n_k}\}= \{x_{n_1}, x_{n_2}, \ldots\}$ where $n_1 < n_2 < \cdots$ are natural numbers.

Consider the sequence

$\{a_n\} = \{1,\frac{1}{2^2}, \frac{1}{3}, \frac{1}{4^2}, \frac{1}{5}, \frac{1}{6^2}, \frac{1}{7}, \frac{1}{8^2}, \frac{1}{9}, \frac{1}{10^2}, \ldots \}$

defined by $a_n = \begin{cases} \frac{1}{n} & \textrm{if } n \textrm{ is odd} \\ \frac{1}{n^2} & \textrm{if } n \textrm{ is even} \\ \end{cases}$

for each natural number $n.$ What is the smallest value of $N$ such that

$a_n < \frac{1}{64} \text{ whenever } n >N?$

Does the sequence

$\{a_n\} = \{5 + \frac{(-1)^n}{n}\}$

converge?