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Theorem 9.1

Convergent sequences are bounded.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Let $(s_n) $ be a convergent sequence, and let $s = \lim s_n$.

Arbitrarily choose an $\epsilon > 0 $. Here we choose $\epsilon = 1 $.

Since $(s_n) $ is convergent, exists $N \in \N$ so that

$$n > N \implies \abs{ s_n - s } < 1 $$

$$\begin{align*} \abs{(s_n - s) + s } &\leq \abs{ s_n - s } + \abs{ s } \br \abs{ s_n } &\leq \abs{ s_n - s } + \abs{ s } \br \abs{ s_n } - \abs{ s } &\leq \abs{ s_n - s } \br \end{align*}$$

$\because \abs{ s_n - s } < 1 $

$$\abs{ s_n } - \abs{ s } < 1 $$

i.e. For all $n > N $, $\abs{ s_n } < 1 + \abs{ s } $.

Define $M = \max{ \set{ \abs{ s } + 1, \abs{ s_1 }, \abs{ s_2 }, …, \abs{ s_N }}}$.

$M$ is an upper bound of $(s_n)$.

Example

Show that $\frac{ 2n + 1 }{ 3n -1 } \to \frac{ 2 }{ 3 }$.

Fix $\epsilon > 0 $, want to find $ N $ s.t. $n > N \implies \abs{ \frac{ 2n+1 }{ 3n-1 } - \frac{ 2 }{ 3 }} < \epsilon$.

$\begin{align*} \abs{ \frac{ 2n+1 }{ 3n-1 } - \frac{ 2 }{ 3 }} &< \epsilon \br \frac{ 5 }{ 3n-1 } &< \epsilon \end{align*}$

$N = \frac{5 + \epsilon}{3 \epsilon}$.

Theorem

$s_n \to s \implies \abs{ s_n } \to \abs{ s }$.

Theorem : The Subsequence Theorem

Let $\set{ t_n } $ be a subsequence of $\set{ s_n } $.

$s_n \to s $, $t_n \to s \implies t_n = s_{f(n)}$.

where $f(n)$ is an increasing function from $\N $ to $\N$ called the selection function.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Consider $\epsilon > 0 $, then $\exists N , \abs{s_n - s} < \epsilon$.

$n \geq N$, some $f(n)$ is increasing, then $f(n) \geq n$.

$f(1) \geq 1$, assume $f(n) \geq n $.

$f(n + 1) > f(n) \because f$ is increasing.

$\therefore f(n + 1) > n $

$\therefore f(n + 1) \geq n +1 $(because $n$ is an integer)

$\abs{ s_n - s } < \epsilon, n > N $

$t_n = s_{f(n)} $

since f(n) > n, then

$n \geq N \implies f(n) \geq N$

$\therefore \abs{ t_n - s } = \abs{ s_{f(n)} - s } < \epsilon $

$\therefore t_n \to s $ is true $\forall s_n n \geq N$

Example

$s_n \to s $, then $\set{ s_n } $ is bounded.

Can know this just like the $inf \set{ \abs{ s_n }}$ problem

assume false

Then $\set{ s_n } $ either has no upper bound (-sup \set{ s_n } = \infty) or has no lower bound.

Consider $\abs{ S_n } $ we know $\abs{ s_n } \to s $

so $\set{ \abs{ s_n }}$ has no upper bound.

Then for any $M $, there is some $n $ with $\abs{s_n} > M $

Take a subsequence

$t_1 $ where $t_1 = s_{n_1} $ where $\abs{ s_{n_1}} $ $t_2 $ where $t_2 = s_{n_2} $ where $\abs{ s_{n_2}} > 2$

$\abs{ t_k } > k $

Then $t_1, t_2, t_3 … $

is subseq

that does not $\to s $

Theorem : Binomial Theorem

$$\binom{ n }{ k } = \frac{ n! }{ k! (n - k)! } $$

where $n! = n(n-1) … 3 \times 2 \times 1 $

Note

It’s obvious that $\binom{ n }{ k } \in \Q$, how prove that it is also in $\Z $.

$n!$

step 1: choose $k $ from $n$ : $\binom{ n }{ k } $

step 2: order them : $k! $

step 3: order the rest : $(n - k)! $

step 4: Put $k $ first than $n-k $ second

$n! = \binom{ n }{ k } k! \pare{n-k}! $

Hence, $\frac{ n! }{ k! (n - k)! }$ is an integer, because it is the ways to choose $k $ from $n $.

Theorem

$$\binom{ n }{ k } = \binom{ n-1 }{ k } + \binom{ n-1 }{ k -1 } $$

Definition: Binomial Theorem

$$(1 + x)^n = \sum_{ k=0 }^{ n } \binom{ n }{ k }x^k$$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$$P(n): (1 + x)^n = \binom{ n }{ 0 } + \binom{ n }{ 1 }x + \binom{ n }{ 2 }x^2 + … + \binom{ n }{ n }x^n$$

The basis:

When $n = 0, (1 + x)^0 = \binom{ 0 }{ 0 } = 1$

When $n = 1, (1 + x)^1 = \binom{ 1 }{ 0 } + \binom{ 1 }{ 1 }x = 1 + x$

The Inductive step:

$\begin{align*} (1 + x)^{n + 1} &= (1+ x)^n (1 + x) \br &= (1 + x)\pare{\binom{ n }{ 0 } + \binom{ n }{ 1 }x + \binom{ n }{ 2 }x^2 + … + \binom{ n }{ n }x^n} \br = \binom{ n }{ 0 } + &\binom{ n }{ 1 }x + \binom{ n }{ 2 }x^2 + … + \binom{ n }{ n }x^n + \br &\binom{ n }{ 0 } x + \binom{ n }{ 1 }x^2 + … + \binom{ n }{ n-1 } x^n + \binom{ n }{ n } x^{n+1} \br &= \binom{ n+1 }{ 0} + \binom{ n+1 }{ 1 }x + … + \binom{ n+1 }{ n }x^n + \binom{ n+1 }{ n+1 } x^{n+1} \br &= \sum_{ k=0 }^{ n+1 } \binom{ n + 1 }{ k } x^k \end{align*}$

Definition: Infinity Limits

For a sequence $(s_n) $, we write lim $s_n = + \infty$ provided for each $M > 0 $ there is a number $N $ s.t. $n > N $ implies $s_n > M $.

Henceforth we will say $(s_n)$ HAS a limit or the limit exists provided $(s_n)$ converges or diverges to $+\infty $ or diverges to $-\infty $.

Theorem

$s, t$ are not $ +\infty$ or $-\infty$.

1. $s_n \to s \implies \set{ s_n } $ is bounded.

2. $s_n \to s \implies k s_n \to ks $.

3. $s_n \to s, t_n \to t \implies s_n + t_n \to s + t$.

4. $s_n \to s, t_n \to t \implies s_nt_n \to st$.

5. $s_n \to s, s_n \neq 0, s \neq 0 \implies \inv{\pare{s_n}} \to \inv{ s }$.

6. $s_n \to s, t_n \to t, s_n, s \neq 0 \implies \frac{ t_n }{ s_n } \to \frac{ t }{ s } $.

7a. $If p > 0, \frac{ 1 }{ n^p} \to 0 $.

7b. $If \abs{ a } < 1, a^n \to 0$.

7c. $\sqrt[ n ]{ n } \to 1$.

7d. $If a > 0, \sqrt[ n ]{ a } \to 1 $.

Theorem 9.9

Let $(s_n)$ and $(t_n) $ be sequences such that $\lim s_n = +\infty $ and $\lim t_n > 0 $ ($\lim t_n$ can be finite or $+\infty$). Then $\lim s_n t_n = +\infty $.

Theorem 9.10

Let $(s_n)$ be a sequence where $ s_n > 0, s_n \in \R $.

$$\lim s_n = +\infty \iff \lim \frac{ 1 }{ s_n } = 0 $$