## Monotonicity

Monotone Sequence:

• ($$a_n$$) is increasing if $$a_n \leq a_{n+1} \forall n$$
• ($$a_n$$) is decreasing if $$a_n \geq a_{n+1} \forall n$$

Theorem: Any bounded monotone sequence converges.

Proof: $$(a_n)$$ bounded increasing. Let $$A=\{a_n | n \in \mathbb{N}\}$$. We note that $$A$$ is bounded, so there exists $$Z := \sup A$$.

Claim: $$z = \lim_{n \to \infty} a_n$$.

Idea: We must find an $$\epsilon$$ such that $$(z - \epsilon, z + \epsilon)$$ contains $$z$$.

1. There is some $$a_n$$ such that $$a_n \in (z - \epsilon, z + \epsilon)$$
• This is true because if no such element exists, then $$z - \epsilon$$ must be $$\sup A$$. This contradicts the claim that $$z = \sup A$$.
• A consequence is that we know there exists at least one $$a_n$$ within $$(z - \epsilon, z)$$.
2. Then we will show $$\forall \epsilon > 0, \exists N > 0 \textrm{ s.t. } n > N \implies \vert a_n - a \vert < \epsilon$$.

Proof: $$\forall \epsilon > 0$$, we claim that $$\exists n > 0 \textrm{ s.t. } a_N \in (z - \epsilon, z + \epsilon)$$. Then, $$\forall n > N$$, we have $$z - \epsilon < a_N \leq a_n \leq z$$. This implies that $$\vert a_n - z \vert < \epsilon\ \forall n > N$$.

We note that sequences do not have to be monotone to converge.

$(1, \frac{-1}{2}, \frac{1}{3}, \frac{-1}{4}, ...)$

converges to $$0$$.

## Some Limit Theorems

If $$\lim_{n \to \infty} a_n = a, \lim{n \to \infty} b_n = b$$ thm.

1. $$r \neq 0, \lim_{n \to \infty} ra_n = ra$$
2. $$\lim_{n \to \infty}a_n + b_n = a+b$$
3. $$\lim_{n \to \infty} a_nb_n = ab$$
4. If $$b_n \neq 0, \forall n, b\neq 0, then \lim_{n \to \infty} \frac{a_n}{b_n} = \frac{a}{b}$$.

### Example

Let us take $$a_n = \frac{4n^2 - 7n}{n^2 + 1}$$. We can divide top and bottom by $$n^2$$. Then, the top and bottom are both sequences. We take of their limits.

$\lim_{n \to \infty} a_n = \frac{\lim (4 - \frac{7}{n})}{\lim (1 - \frac{1}{n^2})} = 4$

### Example 2

Let us take:

$a_1 = 1$ $a_2 = 1 + \frac{1}{1 + \frac{1}{1}}$ $a_{n+1} = 1 + \frac{1}{1 + \frac{1}{a_n}}$ $a_1 = 1$ $a_2 = \frac{3}{2}$ $a_3 = \frac{8}{5}$ $a_4 = \frac{21}{13}$

Proof that $$(a_n)$$ is bounded and increasing:

• Bounded: $$1 < a_n < 2$$.
• Increasing: $$a_n \leq 1 + \frac{1}{1 + \frac{1}{a_n}} = \frac{2a_n+1}{a_n+1}$$