Sequences

A sequence is a function whose domain is $N$
$f : N \to R, f (n)$

Convergence of a Sequence

A sequence ( $a_{n}$ ) converges to $a \in R$ if for every $ϵ > 0$ there exists a $N \in N$ such that if $n > N$ then $| a_{n} - a | < ϵ$

$ϵ$ -Neighbourhood of a

V_{ϵ} = {x \in R : | x - a | < ϵ}

So an equivalent definition is:
$lim_{n \to \infty} a_{n = a}$ iff $\forall ε > 0, B_{ε} = {m : a_{n} \notin V_{ε} (a)}$ is finite.

Thm 2.2.7 Uniqueness of Limits

If $(a_{n}) \to a$ and $(a_{n}) \to b$ then $a = b$

Proof

There exists and $N_{a}$ and $N_{b}$ where
for all $\frac{ϵ}{2} > 0$ $| a_{n} - a | < \frac{ϵ}{2}$ and $| a_{n} - b | < \frac{ϵ}{2}$

$| a_{n} - a - a_{n} + b | \leq | a_{n} - a | + | a_{n} - b | < ϵ$
$| b - a | < ϵ$

Algebraic and Order Limit Theorems

Algebraic Theorems

$lim_{n \to \infty} a_{n} = a$ and $lim_{n \to \infty} b_{n} = b$ then

\begin{aligned} (i) & lim_{n \to \infty} (c \cdot a_{n}) = c \cdot a, for all c \in R \\ (ii) & lim_{n \to \infty} (a_{n} + b_{n}) = a + b \\ (iii) & lim_{n \to \infty} (a_{n} \cdot b_{n}) = a \cdot b \\ (iv) & lim_{n \to \infty} (\frac{a_{n}}{b_{n}}) = \frac{a}{b}, if b \neq 0 \end{aligned}

Proofs

Proof i
There is an $N$ so that $n > N$

| a_{n} - a | < \frac{ϵ}{| c |}

\begin{array}{r} | c a_{n} - c a | = | c | | a_{n} - a | < | c | \frac{ϵ}{| c |} = ϵ \end{array}

Proof ii

We know for some $N_{a}$ that $| a_{n} - a | < \frac{ε}{2}$ and some $N_{b}$ that $| b_{n} - b | \leq \frac{ϵ}{2}$ . Take $max {N_{a}, N_{b}}$ and the rest is trivial

Proof iii

\begin{array}{rcl} | a_{n} b_{n} - a b | & = & | a_{n} b_{n} - a b_{n} + a b_{n} - a b | \\ \leq & | a_{n} b_{n} - a b_{n} | + | a b_{n} - a b | \\ = & | b_{n} | | a_{n} - a | + | a | | b_{n} - b | . \end{array}

We know there is a $N_{b}$ so that

n \geq N_{b} ⟹ | b_{n} - b | < \frac{1}{| a |} \frac{ϵ}{2}

Also $b_{n}$ converges so it is bounded $| b_{n} | \leq M$

n \geq N_{a} ⟹ | a_{n} - a | < \frac{1}{M} \frac{ϵ}{2}

$n \geq max {N_{a}, N_{b}}$

\begin{aligned} | a_{n} b_{n} - a b | & \leq | b_{n} | | a_{n} - a | + | a | | b_{n} - b | \\ \leq M | a_{n} - a | + | b_{n} - b | \\ \leq M (\frac{ϵ}{2 M}) + | a | (\frac{ϵ}{| a | 2}) = ϵ \end{aligned}

Proof iv
$(b_{n}) \to b \overset{(i i i)}{⟹} (\frac{1}{b_{n}}) \to \frac{1}{b}$ if $b \neq 0$

Since $(b_{n}) \to b$ there is a $N_{1}$ so that $| b_{n} - b | < | b | / 2$ for any $n > N_{1}$ so $b_{n} < | b | / 2$

there is an $N_{2}$ where $| b_{n} - b | < \frac{ϵ | b |^{2}}{2}$ so take $max {N_{1}, N_{2}}$

| \frac{1}{b_{n}} - \frac{1}{b} | = | b - b_{n} | \frac{1}{| b | | b_{n} |} < \frac{ϵ | b |^{2}}{2} \frac{1}{| b |^{\frac{| b |}{2}}} = ϵ .

Order Theorems

Assume $lim_{n \to \infty} a_{n} = a$ and $lim_{n \to \infty} b_{n} = b$

\begin{array}{r} (i) & \forall n, a_{n} \geq 0 ⟹ a \geq 0 \\ (ii) & \forall n, a_{n} \leq b_{n} ⟹ a \leq b \\ (iii) & \forall n, c \leq b_{n} ⟹ c \leq b \\ a_{n} \leq c ⟹ a \leq c \end{array}

Proofs

Proof i
For contradiction assume that $a < 0$
$ϵ = | a |$ so for some $N$ , $n \geq N$ , $| a_{n} - a | < | a |$ so $a_{N} < 0$ which is a contradiction

Proof ii
by the algebraic limit theorems $(b_{n} - a_{n}) \to b - a$ . Since $b_{n} - a_{n} \geq 0$ using i we know $b - a \geq 0$

Proof iii
Take $a_{n} = c$ for all $n \in N$ and apply ii

Bounded Sequence

A sequence ( $x_{n}$ ) is bounded if there exists a number $M > 0$ so that $| x_{n} | \leq M$ for all $n \in N$

Thm 2.3.2 Convergent then bounded

Proof

Assume $(x_{n}) \to l$ . So $\forall ϵ > 0 \exists N$ so that when $n > N ⟹ | x_{n} - l | < ϵ$

So let $ϵ = 1$ , for some $N$ , all $n > N$ we have

| x_{n} - l | < 1 ⟹ | x_{n} | < | l | + 1

Now the set of $x_{i}$ not in $V_{ε = 1}$ is finite. Let

M = max {| x_{1} |, | x_{2} |, \dots, | x_{N - 1} |, | l | + 1}

therefore $| x_{n} | \leq M$ for all $n \in N$

Monotone Sequence

A sequence ( $a_{n}$ ) is increasing if $a_{n} \leq a_{n + 1}$ for all $n \in N$ and decreasing if $a_{n} \geq a_{n + 1}$ and monotonic if either.

Thm Squeeze

$y_{m} \leq x_{n} \leq z_{n}$ for all $n \in N$
Assume $(y_{n}) \to L$ and $(z_{n}) \to L$ . Then $(x_{n}) \to L$

Proof

$\exists ϵ > 0$ since $(y_{n}) \to L$ there exists an $N_{y} \in N$ so that for all $n > N_{y}$ , $L - ϵ < y_{n} < L_{ϵ}$

Since $(z_{n}) - L$ there exists an $N_{z} \in N$ so that for all $n > N_{y}$ we have $L - ϵ < z_{n} < L + ϵ$

let $N = max {N_{y}, N_{z}}$
$n \geq N$
$L - ϵ < y_{n} \leq x_{n} \leq z_{n} < L + ϵ$

Thm Ratio Test

Let $x_{n} > 0$ for all $n \in N$ and $lim \frac{x_{n + 1}}{x_{n}} = L$ exists. If $L < 1$ then $(x_{n})$ converges and $(x_{n}) \to 0$

Proof

Let $r \in R$ so that $L < r < 1$ . Set $ϵ = r - L$ , there exists $k \in N$ so that $\frac{x_{n + 1}}{x_{n}} < L + ϵ = r$ for all $n > k$

So, $0 < x_{k + 1} < r x_{k}$ so

\begin{aligned} x_{k + 1} & < r x_{k} \\ x_{k + 2} & < r x_{k + 1} < r^{2} x_{k} \\ x_{k + m} & < r^{m} x_{k} \\ (0<r<1) & r^{m} & \to 0 \end{aligned}

So $(x_{k + m}) \to 0$
$(x_{n}) \to 0$

Thm 2.4.2 Monotone Convergence

If $(a_{n})$ is monotonic bounded then it converges if and only if it is bounded

Proof

Assume $a_{n}$ is increasing and bounded and set $s = sup {a_{n} : n \in N}$ ,

s - ϵ < a_{N} \leq a_{n} \leq s < s + ϵ

by lemma 1.3.8

so $| a_{n} - s | < ϵ$

by thm 2.3.2 it is bounded.

Subsequences

Let ( $a_{n}$ ) be a sequence and let $n_{1} < n_{2} < \dots < n_{5}$ be a sequence of increasing natural numbers. The then sequence

(a_{n_{1}}, a_{n_{2}}, \dots)

is a subsequence of ( $a_{n}$ ) and is denoted by $a_{n_{k}}$ where $k \in N$ indexes the subsequence.

Thm Convergence and subsequences (divergence)

The subsequences of a convergent sequence converges to the same limit as the original sequence.

Proof

Assume $(a_{n}) \to a$ and let $(a_{n_{k}})$ be a subsequence.

For any $ε > 0$ there exist an $N \in N$ so that $| a_{n} - a | < ε$ for any $n > N$ . $n_{k} \geq k$ for all $k$ , then $| a_{n_{k}} - a | < ε$ for any $k \geq N$

Thm Bolzano Weierstrass

Every bounded Sequence contains a convergent sequence

Proofs

Using Nested Interval Property

Let ( $a_{n}$ ) be bounded so that there exists an $M > 0$ satisfying $| a_{n} | \leq M$ for all $n \in N$ .
Take $[- M, 0]$ and $[0, - M]$ . At least one interval must have an infinite number of terms outside the epsilon neighbourhood.

Let $I_{1}$ be the the interval with infinite elements. Let $a_{n_{1}}$ be a term in the sequence $(a_{n})$ such that $a_{n_{1}} \in I_{1}$ .

Bisect $I_{1}$ and let $I_{2}$ have the infinite number of terms.

Prick $a_{n_{2}}$ from the original sequence with $n_{2} > n_{1}$ and $a_{n_{2}} \in I_{2}$ In general, construct the closed interval $I_{k}$ taking a half of $I_{k - 1}$ containing an inﬁnite number of terms of $(a_{n})$ and then select $n_{k} > n_{k - 1} > \cdot \cdot \cdot > n_{2} > n_{1}$ so that $a_{n_{k}} \in I_{k}$

I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \dots

By the the nested interval property there exists at least one $x \in R$ for every $I_{k}$ . WTS that $(a_{n_{k}}) \to x$

lets $ε > 0$ the length of $I_{k}$ is $M {(\frac{1}{2})}^{k - 1}$ which converges to 0. Choose $N$ so that $k \geq N$ the length of $I_{k} < ε$ . Since $x, a_{n_{k}} \in I_{k}$ it follows that $| a_{n_{k}} - x | < ε$

Using Monotone Convergence Theorem

Lemma:
Every Sequence has a monotone subsequence
Let element $a_{n}$ of $(a_{n})$ be called a castle if $a_{n} \geq a_{k}$ for all $k \geq n$

Case 1: There are infinitely many castles
$a_{n 1}, a_{n_{2}}, \dots$ then $a_{n_{1}} \geq a_{n_{2}} \geq a_{n_{3}} \geq \dots$
and so $(a_{n_{i}})$ is a decreasing sequence

Case 2: Only finitely many castles in ( $a_{n}$ )
There $\exists$ $a_{n_{1}} > a_{k}$ for every castle $a_{k}$ where any element past $a_{n_{1}}$ is not a castle.
so $\exists$ $a_{n_{2}} > a_{n_{1}}$ and $a_{n_{2}}$ is not a castle. $a_{n_{3}} > a_{n_{2}}$ and so on, so strictly increasing.

Monotone + bounded = convergent

Limit Superior

Let $(x_{n})$ be bounded
is the infimum of the set $V$ of $v \in R$ such that $v < x_{n}$ for at most finitely many $n \in N$
denoted $lim sup a_{n}$ or $lim^{―} x_{n}$

Alternatively $lim sup x_{n} = inf sup x_{m}$ for $n = 1, 2, 3, \dots m = n, n + 1, \dots$

Limit Inferior

Let $(x_{n})$ be bounded
is the supremum of the $W$ of $w \in R$ such that $x > a_{n}$ for at most finitely many $n \in N$

Alternatively $lim inf x_{n} = sup inf x_{m}$ for $n = 1, 2, 3, \dots m = n, n + 1, \dots$

Example:

$(a_{n}) = 2 - \frac{1}{2}, 6 + \frac{1}{3}, 2 - \frac{1}{4}, 6 + \frac{1}{5}, \dots$

inf $a_{n} = 2 - \frac{1}{2}$

throw away 2 terms
inf $a_{n} = 2 - \frac{1}{4}$

throw away 2 $n$ terms
inf $a_{n} = 2 - \frac{1}{2 n}$
take the collection and lim inf = 2

Now:

\begin{array}{r} {a_{k} : k \geq 2} \\ sup : 6 + \frac{1}{3} \\ {a_{k} : k \geq 4} \\ sup : 6 + \frac{1}{5} \\ {a_{k} : k \geq 2 n} \\ sup : 6 + \frac{1}{2 n + 1} \\ inf_{n \in N} sup_{m = n, n + 1, \dots} = 6 \end{array}

Thm Limit with limsup and liminf

Let $a_{n}$ be bounded then $lim (a_{n}) = a$ iff $lim sup (a_{n}) = lim inf (a_{n}) = a$

Cauchy:

if the terms of $(a_{n})$ become "closer" as $n \to \infty$ then $a_{n}$ should converge

A sequence is a Cauchy Sequence if $\forall ε > 0$ there exists $k \in N$ so that if n, $m \geq k$ then $| a_{n} - a_{m} | < ε$

note m and n are independent can assume $m > n$

Thm Convergent iff Cauchy

Proof

$⟹$

Converges then Cauchy

If $(a_{n}) \to l$ then $a_{n}$ is Cauchy
Let $ε > 0$ and $K$ be such that $| a_{n - L} | < \frac{ε}{2}$ for $n \geq K$

If $m, n \geq k$

\begin{aligned} | a_{n} - a_{m} | = | a_{n} - L + L - a_{m} | & \leq | a_{n} - L | + | a_{n} - L | \\ < \frac{ε}{2} + \frac{ε}{2} = ε \end{aligned}

Lemma Cauchy then Bounded

If $(a_{n})$ is Cauchy then it is Bounded

From the definition of a Cauchy Sequence then there exists $k \in N$ so that for $n \geq k$ that $| a_{n} - a_{k} | < ε \overset{s e t}{=} 1$

By triangle

$| a_{n} | \leq | a_{n} - a_{k} | + | a_{k} |$
$| a_{n} | \leq | a_{k} | + 1$ for all $n \geq k$
Set $M = max {| a_{1} |, | a_{2} |, \dots, | a_{k} |, | a_{k} + 1 |}$

Then $| a_{n} | \leq M$ for all $n \in N$

Since $(a_{n})$ is Cauchy it is bounded so by BW there exists and convergent subsequence $(a_{n_{k}}) \to L$

Let $ε > 0$ since $a_{n}$ is Cauchy, there exists $M \in N$ so that if $m, n \geq M$ , then $| a_{n} - a_{m} | < ε / 2$

Also since $(a_{n_{k}}) \to L$ there exists $n_{l} \geq M$ so that $| a_{n_{l}} - L | \leq \frac{ε}{2}$

So if

\begin{aligned} n \geq M \\ | a_{n} - L | & = | a_{n} - a_{n_{l}} + a_{n_{l}} - L | \\ \leq | a_{n} - a_{n_{l}} | + | a_{n_{l}} - L | \\ \leq \frac{ε}{2} + \frac{ε}{2} = ε \end{aligned}

Example

$(1 + (- 1)^{n})$

There exists some value $ε_{0} > 0$ so that for every $k \in N$ there exists at least one value $n \geq k$ and $m \geq k$ so that

| a_{n} - a_{m} | > ε_{0}

Have $a_{2 n} = 2$ and $a_{2 n + 1} = 0$ . Take $ε_{0} = 1$ then for any $k \in N$ choose $n$ even and $m = n + 1$ then

| a_{n} - a_{n + 1} | = 2 > 1

Thm Archimedean and Cauchy to Completeness

Suppose every Cauchy Sequence in $R$ converges to an element in $R$ and that the Archimedean property holds. Then $R$ satisfies the completeness property.

completeness -> MCT -> BW -> Cauchy