Limits

Sequence Convergence

Let

(x_{n}) : ℝ \to ℝ

be a sequence and let

x \in ℝ

, then we say that

(x_{n})

converges to

x

and write

(x_{n}) \to x

when the following holds true

\forall ϵ \in ℝ^{+}, \exists n \in ℕ_{0} such that \forall n \in ℕ_{0}, n \geq N ⟹ | x_{n} - x | < ϵ

Note that you may come across various other notations for this definition in other resources, the main ones to note will be: $x_{n} \to x$ as $n \to \infty$ , or even more shortened as $x_{n} \to x$ , sometimes limit notation will be used: $\lim_{n \to \infty} x_{n} = x$ or just $\lim x_{n} = x$

Convergent Sequence Doesn't Require Strict Inequality

(a_{n}) \to L

is equivalent to

\forall ϵ \in ℝ^{+}, \exists N \in ℕ_{0}, st \forall n \in ℕ_{0}, n \geq N ⟹ | a_{n} - L | \leq ϵ

⟹

Suppose

(a_{n}) \to L

, now let

ϵ \in ℝ

, and consider

\frac{ϵ}{2}

, then since

(a_{n}) \to L

we get some

N \in ℕ_{0}

such that for every

n \geq N

we have

| a_{n} - L | < \frac{ϵ}{2} \leq ϵ

as needed.

⟸

Assume the other direction, and we want to prove

(a_{n}) \to L

so let

ϵ \in ℝ^{+}

therefore we get an

N \in ℕ_{0}

such that for all

n \geq N

we have

| a_{n} - L | \leq \frac{ϵ}{2} < ϵ

as needed.

Limit Notation

Suppose that

x_{n} \to x

then we write

\lim_{n \to \infty} x_{n} : = x

Note that if you have an arbitrary sequence $(y_{n})$ which you don't know if it converges to anything and you write $\lim_{n \to \infty} y_{n} = L$ for some $L \in ℝ$ , this is two assertions, first that $(y_{n})$ converges to something, and the value it converges to is $L$ .

Eventually Increasing Index Mapping Maintains Limit

Suppose that

\lim_{n \to \infty} x_{n} = x

and

f : ℕ_{1} \to ℕ_{1}

is an eventually increasing function, then

\lim_{n \to \infty} x_{f (n)} = x

Reciprocal Power Goes to Zero

Let

(x_{n}) : ℕ_{0} \to ℝ

be the sequence such that

x_{n} = \frac{1}{n^{α}}

for some

α \in ℝ^{> 0}

, then

(x_{n}) \to 0

Let

ϵ \in ℝ^{> 0}

, we need to pick an

N \in ℕ_{1}

such that for

n \geq N

we have that

| \frac{1}{n^{α}} - 0 | < ϵ

, it's clear that

\frac{1}{n^{α}} > 0

so therefore we can drop the absolute values so we just need an

n

that satisfies

\frac{1}{n^{α}} < ϵ ⟺ \frac{1}{ϵ^{\frac{1}{α}}} < n

By setting

N = ⌈ \frac{1}{ϵ^{\frac{1}{α}}} ⌉ + 1

then we have that for any

n \geq N

\frac{1}{ϵ^{\frac{1}{α}}} < ⌈ \frac{1}{ϵ^{\frac{1}{α}}} ⌉ + 1 = N \leq n

So that

\frac{1}{ϵ^{\frac{1}{α}}} < n

as needed.

Limit of Sine Doesn't Exist at Infinity

Show that

\lim_{n \to \infty} \sin (\frac{n π}{2})

doesn't exist using the definition of limit.

To show that the limit doesn't exist we will use the negation of the definition and show that is true, so we have to prove that

\forall L \in ℝ, \exists ϵ \in ℝ^{+} st \forall N \in ℕ_{0}, \exists n \in ℕ_{0} st (n \geq N \land | a_{n} - L | \geq ϵ)

So let

L \in ℝ

and take

ϵ = \frac{1}{4}

let

N \in ℕ_{0}

Before finding our $n$ value, consider sequence $a_{n} = \sin (\frac{n π}{2})$ , and that $| a_{n} - a_{n + 1} | = 1$ .

Now to find our $n$ value we look at the requirement that $| a_{n} - L | \geq ϵ$ which means that it's not true that $| a_{n} - L | < ϵ$ , so that in other words $a_{n} \notin (L - ϵ, L + ϵ)$ , due to the fact that $ℓ ((L - ϵ, L + ϵ)) = \frac{1}{2}$ then given that $| a_{N} - a_{N + 1} | = 1$ then exactly one of $a_{N}, a_{N + 1}$ is not in $(L - ϵ, L + ϵ)$ therefore we can set $n$ respectively to force $| a_{n} - L | \geq ϵ$ as needed.

Non-Zero Limit Implies Eventually Non-Zero Sequence

Suppose that

\lim_{n \to \infty} a_{n} = L \neq 0

then there exists some

N \in ℕ_{0}

such that

a_{n} \neq 0

for all

n \geq N

Suppose for the sake of contradiction that that was not true, so that for every $N$ we had some $n \geq N$ such that $a_{n} = 0$ , let's derive a contradiction.

Since $(a_{n}) \to L$ then specifically for $ϵ = \frac{L}{2}$ we have some $N^{'} \in ℕ_{0}$ such that for every $n \geq N^{'}$ we have $| a_{n} - L | \leq \frac{L}{2}$ , therefore by our contradiction assumption we also know there is some $k \geq N^{'}$ such that $a_{k} = 0$ so therefore because the limit exists we have $| a_{k} - L | \leq \frac{L}{2}$ but $a_{k} = 0$ so that $| L | \leq \frac{L}{2}$ which is impossible so therefore we have a contradiction and so the original statement holds true.

Sequence Goes to Positive Infinity

We say that a sequence

(x_{n})

goes to positive infinity and notate it as

(x_{n}) \to \infty

when

\forall M \in ℝ^{+}, \exists N \in ℕ_{0} such that \forall n \in ℕ_{0}, n \geq N ⟹ x_{n} \geq M

You can probably guess the next definition.

Sequence Goes to Negative Infinity

We say that a sequence

(x_{n})

goes to negative infinity and notate it as

(x_{n}) \to - \infty

when

\forall M \in ℝ^{-}, \exists N \in ℕ_{0} such that \forall n \in ℕ_{0}, n \geq N ⟹ x_{n} \leq M

Note that these definitions formalize our idea that if something approaches infinity, namely that this object is greater (or smaller) than every possible other object.

Sequence Goes to Infinity

We say that a sequence goes to infinity if it goes to positive infinity or negative infinity, and write

(x_{n}) \to \pm \infty

Log Goes to Infinity

Show that

{(\ln (n))}_{n = 1}^{\infty} \to \infty

Let

M \in ℝ^{+}

and take

N = ⌈ e^{M} ⌉

and let

n \geq N

therefore

n = ⌈ e^{M} ⌉ + k

for some

k \in ℕ_{0}

since

\ln

is increasing then we know that

M = \ln (e^{M}) \leq \ln (⌈ e^{M} ⌉) \leq \ln (⌈ e^{M} ⌉ + k)

so that

M \leq \ln (n)

as needed therefore it goes to infinity

Two Limits that get Arbitrarily Close Have the Same Limit

Suppose htat

(a_{n}), (b_{n})

are sequence of real numbers such that

| a_{n} - b_{n} | < \frac{1}{n}

. If

L = \lim_{n \to \infty} a_{n}

exists, then

\lim_{n \to \infty} b_{n} = L

as well

Let $ϵ \in ℝ^{+}$ we need to pick an $N \in ℕ_{0}$ such that for every $n \geq N$ we have that $| b_{n} - L | < ϵ$ .

Since $\lim_{n \to \infty} a_{n} = L$ then we can set $ϵ^{'} = f (ϵ)$ for any function $f : ℝ^{+} \to ℝ^{+}$ of our choosing, to get some $N^{'}$ such that for any $n \geq N^{'}$ we have $| a_{n} - L | < f (ϵ)$

Morover note that for this given $ϵ$ we can always find a $K \in ℕ_{1}$ such that $\frac{1}{K} < g (ϵ)$ where $g : ℝ^{+} \to ℝ^{+}$ is of our choosing, By picking $N = \max (N^{'}, K)$ we know that

\begin{matrix} | b_{n} - L | & a m p; = | b_{n} - a_{n} + a_{n} - L | \\ a m p; \leq | b_{n} - a_{n} | + | a_{n} - L | \\ a m p; \leq \frac{1}{n} + f (ϵ) \\ a m p; \leq \frac{1}{K} + f (ϵ) \\ a m p; \leq g (ϵ) + f (ϵ) \end{matrix}

to obtain the desired inequality one would have to pick $f, g$ such that $f (ϵ) + g (ϵ) < ϵ$ , a simple choice of $f (x) = g (x) = \frac{x}{4}$ yields $| b_{n} - L | < \frac{ϵ}{2} < ϵ$

Bounded Sequence

Suppose that

(x_{n}) : ℕ_{0} \to ℝ

, is a sequence, then we say that it is bounded when there exists some

M \in ℝ^{+}

such that for any

i \in ℕ_{0}

we have

| x_{i} | \leq M

If a Sequence Converges it is Bounded

Suppose that

(x_{n}) \to x

then

(x_{n})

is bounded

Let $a \in ℝ$ since $(x_{n})$ converges then we know that there is some $N$ such that for all $n \geq N$ we have $| x_{n} - x | \leq a$ , therefore $| x_{n} | \leq a + | x |$

Set $I : = {0, . . ., N - 1}$ and we define $L : = \max ({| x_{i} | : i \in I})$ , then it should be clear that for any $k \in I$ we have $| x_{k} | \leq L$

Finally, we can see that the value $M : = \max (L, a + | x |)$ has the property so that for any $j \in ℕ_{0}$ $| x_{j} | \leq M$ so that $(x_{n})$ is bounded.

Pointwise Function Sequence

Suppose that

(f_{1}, f_{2}, f_{3}, \dots)

are functions, of the form

f_{i} : X \to Y

then it's pointwise function sequence at $x \in X$ is defined as the sequence

(f_{1} (x), f_{2} (x), f_{3} (x), \dots)

denoted by

(f_{n} (x))

Pointwise Function Convergence

Suppoes that

(f_{n})

is a sequence of functions of the form

f_{i} : X \to ℝ

and we have another function

f : X \to ℝ

then we say that

(f_{n})

converges pointwise to

f (x)

whenever the pointwise function sequence

(f_{n} (x))

converges to

f (x)

Squeeze Theorem

Suppose that

(a_{n}), (b_{n}), (c_{n}) : ℕ_{1} \to ℝ

are all sequences such that for all

k \in ℕ_{1}

we have

a_{k} \leq b_{k} \leq c_{k}

Moreover suppose that

\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} c_{n} = L

, then

\lim_{n \to \infty} b_{n} = L

Let $ϵ \in ℝ^{+}$ , because the limits of $(a_{n}), (c_{n})$ exist, then we get $N_{a}, N_{c} \in ℕ_{0}$ such that for any $n \geq N_{a}$ we have $| a_{n} - L | < ϵ$ and for any $m \geq N_{c}$ we have $| c_{m} - L | < ϵ$ .

Recall that these two inequalities are equivalent to $L - ϵ < a_{n} < L + ϵ$ and $L - ϵ < c_{m} < L + ϵ$ . Therefore by setting $N = \max (N_{a}, N_{c})$ then for any $n \geq N$ both of the aformentioned inequalities hold true, showing that $L - ϵ < a_{n} \leq b_{n} \leq c_{n} < L + ϵ$ which is to say that $| b_{n} - L | < ϵ$

limsup

Suppose that

(a_{n})

is a sequence, then we define another sequence

s_{n} : = \sup ({a_{k} : k \geq n})

and then we define the limsup of

(a_{n})

lim sup (a_{n}) : = \lim_{n \to \infty} s_{n}

liminf

Suppose that

(a_{n})

is a sequence, then we define another sequence

i_{n} : = \inf ({a_{k} : k \geq n})

and then we define the liminf of

(a_{n})

lim inf (a_{n}) : = \lim_{n \to \infty} i_{n}

limsup Exists iff the Sequence is Bounded Above

As per title.

A Sequence Eventually gets Close to Limsups Limit

Let

(a_{n}) : ℕ_{1} \to ℝ

and suppose that

lim sup (a_{n}) \to L

, then for any

ϵ \in ℝ^{+}

there exists an

N \in ℕ_{1}

such that for any

n \geq N

a_{n} < L + ϵ

Let

ϵ \in ℝ^{+}

then there exists an

N \in ℕ_{1}

such that

| \sup ({a_{k} : k \geq n}) - L | < ϵ

which is true if and only if

L - ϵ < \sup ({a_{k} : k \geq n}) < L + ϵ

thus since

a_{n} \leq \sup ({a_{k} : k \geq n})

by transitivity we know that

a_{n} < L + ϵ

Convergence iff Lim Sup and Inf are Equal

(a_{n}) \to L ⟺ lim sup (a_{n}) = L = lim inf (a_{n})

Define $s_{n} : = \sup ({a_{k} : k \geq n})$ and $i_{n} : = \inf ({a_{k} : k \geq n})$ .

Now suppose that $lim sup (a_{n}) = lim inf (a_{n}) = L$ , since $a_{k} \in {a_{k} : k \geq n}$ , then as they are defined as upper and lower bounds, we have $i_{n} \leq a_{n} \leq s_{n}$ for all $n \in ℕ_{1}$ then by the squeeze theorem $(a_{n}) \to L$ .

Now suppose that $lim sup (a_{n}) = lim sup (a_{n}) = L$ , and let's prove $(a_{n}) \to L$ , let $ϵ \in ℝ^{+}$ , therefore we get an $N \in ℕ_{1}$ such that for every $n \geq N$ we have $| a_{n} - L | < ϵ$ iff $L - ϵ \leq a_{n} \leq L + ϵ$ , thus $L + ϵ$ and $L - ϵ$ are upper and lower bounds of ${a_{k} : k \geq n}$ respectively, since the sup and the inf are the lest and greatest of their bounds respectively so we have that $\sup ({a_{k} : k \geq n}) \leq L + ϵ$ and $L - ϵ \leq \inf ({a_{k} : k \geq n})$ combining the two we obtain $L - ϵ \leq \inf ({a_{k} : k \geq n}) \leq \sup ({a_{k} : k \geq n}) \leq L + ϵ$ which is equivalent to both $| \inf ({a_{k} : k \geq n}) - L | \leq ϵ and | \sup ({a_{k} : k \geq n}) - L | \leq ϵ$ therefore $lim sup (a_{n}) = lim inf (a_{n}) = L$

Sum Law

Suppose that

\lim_{x \to a} f (x)

and

\lim_{x \to a} g (x)

exist, then

\lim_{x \to a} [f (x) + g (x)] = \lim_{x \to a} f (x) + \lim_{x \to a} g (x)

TODO

Cauchy with Converging Subsequence Implies Entire Sequence Converges

Let

(x_{n})

be a cauchy sequence and

(x_{σ (n)})

a subsequence such that

\lim_{k \to \infty} x_{σ (k)} = L

then

\lim_{n \to \infty} x_{n} = L

Let

ϵ \in ℝ^{+}

, since

(x_{n})

is cauchy we obtain some

N_{1}

such that for any

n, m \geq N, | a_{n} - a_{m} | \leq \frac{ϵ}{2}

, on the other hand we see that there is some

N_{2}

such that for all

n \geq N_{2}

we have

| x_{σ (n)} - L | \leq \frac{ϵ}{2}

, so let

N = \max (N_{1}, N_{2})

and let

n \geq N

, consider

σ (n)

, we will want this value to be greater or equal to

N_{1}

so we can use it in-place of

m

, if it turns out that

σ (n) < N_{1}

then since

σ

is increasing there exists some

k > n \geq N_{2}

such that

σ (k) \geq N_{1}

, then note:

\begin{matrix} | x_{n} - L | & a m p; = | x_{n} - x_{σ (k)} + x_{σ (k)} - L | \\ a m p; \leq | x_{n} - x_{σ (k)} | + | x_{σ (k)} - L | \\ a m p; < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ \end{matrix}

Strictly Increasing Natural Number Function Surpasses All Values

Suppose

f : ℕ_{1} \to ℕ_{1}

is increasing, then for any

K \in ℝ

there exists an

n \in ℕ_{1}

such that

f (n) > K

Before we start this proof, note that for any $n \in ℕ_{1}$ we have that $f (n) < f (n + 1)$ which means that $f (n) + 1 \leq f (n + 1)$ which is to say that $1 \leq f (n + 1) - f (n)$ , running with this we can generalize this to say that for any $j \leq k \in ℕ_{1}$ $f (n + j) - f (n + k) \geq j - k$

With that in mind, if $f (1) > K$ we are done, so then suppose that $f (1) \leq K$ , let $α = K + 1 - f (1) \in ℕ_{1}$ but then we know that $f (1 + α) - f (1) \geq α = K + 1 - f (1)$ adding $f (1)$ to both sides tells us that $f (1 + α) \geq K + 1 > K$ as needed.

For all, There Exists Past a Point Yields Ordered Satisfaction

Suppose that you know the following statement holds true:

\forall x \in ℕ_{1}, \exists N_{x} \in ℕ_{1}, f_{x} : ℕ_{1} \to ℕ_{1} st, n \geq N_{x} ⟹ P (x, f_{x} (n))

where

f_{x}

is strictly increasing. Then there exists an order sequence of values in

ℕ_{1}

where

(α_{1} \leq α_{2} \leq α_{3} \leq α_{4} \leq \dots)

such that

P (1, α_{1}) \land P (2, α_{2}) \land P (3, α_{3}) \land \dots

Plugging in $x = 1$ we obtain some $N_{1}, f_{1}$ such that for all $n \geq N_{1}$ we have $P (1, f_{1} (n))$ , let $α_{1} = f_{1} (N_{1})$ note that since $N_{1} \geq N_{1}$ we have $P (1, f_{1} (N_{1})) = P (1, α_{1})$

Now for any $k \in ℕ_{2}$ we similarly get an $N_{k}, f_{k}$ such that for all $n \geq N_{k}$ we have $P (k, f_{k} (n))$ , now consider the value $f_{k} (N_{k})$ , we'd like to set $α_{k}$ to it, but we have no idea if $f_{k} (N_{k}) \geq α_{k - 1}$ , but since we do know that $α_{k - 1} \in ℕ_{1}$ then since $f_{k}$ is increasing we know that there is a value $N_{k} + m$ for some $m \in ℕ_{1}$ such that $f_{k} (N_{k} + m) \geq α_{k - 1}$ , since $N_{k} + m \geq N_{k}$ then clearly $P (k, f_{k} (N_{k} + m))$ holds true, and thus we define $α_{k} = f (N_{k} + m)$ , so that $P (k, α_{k})$ also holds. Notice that we've also just proven that $α_{k} \geq α_{k - 1}$ making them increasing.

Limit to a Limit

Suppose

(x_{n}) : ℕ_{1} \to ℝ

and that

L_{k}

are real numbers such that

\lim_{k \to \infty} L_{k} = L

, if for each

k \in ℕ_{1}

there is a subsequence of

(x_{n})

converging to

L_{k}

show that some subsequence of

(x_{n})

converges to

L

We will construct a subsequence $(x_{σ (n)})$ such that $| x_{σ} (k) - L | < \frac{1}{k}$ . By doing so, we will have solved the problem, because for any $ϵ \in ℝ^{+}$ there is always a $k \in ℕ_{1}$ such that $\frac{1}{k} < ϵ$ which shows that $| x_{σ (k)} - L | < ϵ$ by transitivity.

Let $k \in ℕ_{1}$ since $(L_{n}) \to L$ then by considering $\frac{1}{2 k}$ we know that there is an $N_{k}^{'} \in ℕ_{1}$ such that for all $n \geq N_{k}^{'}$ we have $| L_{n} - L | < \frac{1}{2 k}$ , now consider any $m_{k} \geq N_{k}^{'}$ , thus we know that $| L_{m_{k}} - L | < \frac{1}{2 k}$ .

Now since $m_{k} \in ℕ_{1}$ then there exists a subsequence $x_{σ_{m_{k}} (n)} \to L_{m_{k}}$ , therefore using $ϵ = \frac{1}{2 k}$ we get a $N_{k}$ such that for all $n \geq N_{k}, | x_{σ_{m_{k}} (n) - L_{m_{k}}} | < \frac{1}{2 k}$ , therefore we have

\begin{matrix} | x_{σ_{m_{k}} (n)} - L | & a m p; = | x_{σ_{m_{k}} (n)} - L_{m} + L_{m} - L | \\ a m p; \leq | x_{σ_{m_{k}} (n)} - L_{m_{k}} | + | L_{m_{k}} - L | \\ a m p; < \frac{1}{2 k} + \frac{1}{2 k} = \frac{1}{k} \end{matrix}

Initially it might seem as simple as taking the subsequence $x_{σ_{m_{1}} (N_{1})}, x_{σ_{m_{2}} (N_{2})}, x_{σ_{m_{3}} (N_{3})}, \dots$ but there's no guarentee that this subsequence has increasing indices (which is a requirement of a subsequence), therefore we need more work.

Condensing what we've proven above, we've shown that $\forall k \in ℕ_{1}, \exists N_{k} \in ℕ_{1}, σ_{m_{k}} : ℕ_{1} \to ℕ_{1} st \forall n \geq N_{2}, P (k, σ_{m} (n))$

Where $P (k, v) : = | x_{v} - L | < \frac{1}{k}$ , and since $σ_{m_{k}}$ defines a subsequence then it is strictly increasing, therefore there exists an ordered sequence $(α_{1} \leq α_{2} \leq α_{3} \leq \dots)$ such that $P (k, α_{k})$ which is to say that $| x_{α_{k}} - L | < \frac{1}{k}$ so that the following subsequence works $(x_{α_{1}}, x_{α_{2}}, \dots)$ or in another notation with $σ (n) = α_{n}$ we are using the subsequence $(x_{σ (n)})$

Telescoping Triangle Inequality

Let

j \leq k \in ℕ_{0}

and

(a_{n})

be a sequence, then we have the following

| a_{j} - a_{k} | \leq | a_{j} - a_{j + 1} | + | a_{j + 1} - a_{j + 2} | + \dots + | a_{k - 1} - a_{k} |

which is obtained by using the fact that we can insert

0 = - a_{i} + a_{i}

into the absolute value and then use the triangle inequality to break it, the general result is proven using induction using this fact in the induction step.

Series of Differences Converges Implies Cauchy

Let

(a_{n})

be a sequence such that

\lim_{N \to \infty} \sum_{n = 1}^{N} | a_{n} - a_{n + 1} | < \infty

, show that

(a_{n})

is cauchy.

Let $ϵ \in ℝ^{+}$ and set $S_{m} : = \sum_{i = 1}^{k - 1} | a_{i - 1} - a_{i} |$ since we we assumed that $\lim_{N \to \infty} S_{N}$ exists, then the sequence is cauchy so we get some $N^{'}$ such that for any $a, b \geq N, | S_{a} - S_{b} | \leq ϵ$ and take $N = N^{'}$ and let $n, m \geq N$ , and without loss of generality assume that $n \leq m$ $\begin{matrix} | a_{n} - a_{m} | & a m p; \leq | a_{n} - a_{n + 1} | + | a_{n + 1} - a_{n + 2} | + \dots + | a_{m - 1} - a_{m} | \\ a m p; = S_{m} - S_{n} \\ a m p; = | S_{m} - S_{n} | \leq ϵ \end{matrix}$ where we've used the generalized triangle inequality and the fact that $S_{k}$ is increasing, as each higher index adds a non-negative element to the sum.

Cauchy Sequence Implies Subsequence with Convergent Differences

{(x_{n})}_{n = 1}^{\infty}

is cauchy, show that it has a subsequence

(x_{σ (n)})

such that

\sum_{k = 1}^{\infty} | x_{σ (k)} - x_{σ (k + 1)} | < \infty

Recall that a geometric series of halves is bounded, and thus we're inspired to try to make each $| x_{σ (k)} - x_{σ (k + 1)} | \leq \frac{1}{2^{k}}$ , call this property $α$ .

Let's construct this subsequence, for any $k \in ℕ_{1}$ consider $ϵ = \frac{1}{2^{k}}$ , then we get an $N_{k}$ such that for all $n, m \geq N_{k}, | a_{n} - a_{m} | \leq \frac{1}{2^{k}}$ , and consider the sequence $x_{N_{1}}, x_{N_{2}}, x_{N_{3}}, \dots$ , we'll show it satisfies $α$ .

Let $j \in ℕ_{1}$ , and consider $N_{j}, N_{j + 1}$ , in the case that $N_{j} \geq N_{j + 1}$ then note that $| x_{N_{j}} - x_{N_{j + 1}} | \leq \frac{1}{2^{j + 1}} \leq \frac{1}{2^{j}}$ (where $n = N_{j}, m = N_{j + 1}$ ), in the case that $N_{j} < N_{j + 1}$ we also get that $| x_{N_{j}} - x_{N_{j + 1}} | \leq \frac{1}{2^{j}}$ so $α$ is satisfied for this subsequence. Therefore $\sum_{k = 1}^{\infty} | x_{N_{k}} - x_{N_{k + 1}} | \leq \sum_{k = 1}^{\infty} \frac{1}{2^{k}} = 1$ as needed.

Zig-Zag Sequence has Nested Bound

Suppose that

(a_{n})

is a sequence such that

a_{2 n} \leq a_{2 n + 2} \leq a_{2 n + 3} \leq a_{2 n + 1}

for all

n \in ℕ_{0}

then for any

k, n \in ℕ_{0}

n \geq 2 k + 1 ⟹ a_{n} \in [a_{2 k}, a_{2 k + 1}]

Proven by induction.

Zig-Zag Sequence is Cauchy iff Differences go to Zero

Suppose that

(a_{n})

is a sequence such that

a_{2 n} \leq a_{2 n + 2} \leq a_{2 n + 3} \leq a_{2 n + 1}

for all

n \in ℕ_{0}

, then this sequence is cauchy iff

\lim_{n \to \infty} | a_{n} - a_{n + 1} | = 0

$⟹$ this direction is easy, let $ϵ \in ℝ^{+}$ , since $(a_{n})$ is cauchy we get some $N^{'}$ such that for all $n, m \geq N^{'}, | a_{n} - a_{m} | < ϵ$ , and so take $N = N^{'}$ then specifically for $n, n + 1 \geq N$ we have that $ϵ > | a_{n} - a_{n + 1} | = | | a_{n} - a_{n + 1} | - 0 |$ as needed.

$⟸$ Assume that $\lim_{n \to \infty} | a_{n} - a_{n + 1} | = 0$ and now we want to prove that $(a_{n})$ is cauchy so let $ϵ \in ℝ^{+}$ and so we obtain some $N$ such that for any $n \geq N$ we have that $| a_{n} - a_{n + 1} | \leq ϵ$ take $N^{'} = 2 N + 1$ and let $n, m \geq N^{'}$ by our lemma, we see that $a_{n}, a_{m} \in [a_{2 N}, a_{2 N + 1}]$ , but then again since $2 N, 2 N + 1 \geq N$ we see that $| a_{2 N + 1} - a_{2 N} | \leq ϵ$ and therefore $| a_{n} - a_{m} | \leq ϵ$ as needed.

Monotone Increasing Sequence

A sequence

{(a_{n})}_{n = 1}^{\infty}

is said to be monotone increasing whenever we have:

a_{n} \leq a_{n + 1}

for any

n \in ℕ_{1}

Monotone Decreasing Sequence

A sequence

{(a_{n})}_{n = 1}^{\infty}

is said to be monotone decreasing whenever we have:

a_{n} \geq a_{n + 1}

for any

n \in ℕ_{1}

Bounded Above and Monotone Increasing Converges to Supremum

A monotone increasing sequence that is bounded above converges to

L = \sup (im ((a_{n})))

Since ${(a_{n})}_{n = 1}^{\infty}$ is bounded above, then so is $im ((a_{n})) \subseteq ℝ$ therefore since $ℝ$ has the least upper bound property, we know that $\sup (im ((a_{n})))$ exists, call it $L$ and we'll prove that $(a_{n})$ converges there.

Let $ϵ \in ℝ^{+}$ by definition of the supremum we know that $L - ϵ$ is not an upper bound for $A$ , in other words there exists some element $a_{N} \in im ((a_{n}))$ such that $L - ϵ < a_{N}$

Since $(a_{n})$ is increasing, we know that for any $n \in ℕ, n \geq N, a_{N} \leq a_{n}$ , moreover $L$ is an upperbound so that we get $L - ϵ < a_{N} \leq a_{n} \leq L < L + ϵ ⟹ L - ϵ < a_{n} < L + ϵ$ therefore $| a_{n} - L | < ϵ$ , therefore $\lim_{n \to \infty} a_{n} = L$ as needed.

Bounded Below and Monotone Decreasing Converges to Infimum

A monotone decreasing sequence that is bounded below converges to

L = \inf (im ((a_{n})))

(a_{n})

is decreasing and bounded below by

B

, then the sequence

(- a_{n})

is increasing and bounded above by

- B

. Thus the sequence

{(- a_{n})}_{n = 1}^{\infty}

has limit

L = \sup (im ((- a_{n})))

recall that

\sup (im ((- a_{n}))) = - \inf ({- x : x \in im ((- a_{n}))}) = - \inf (im (a_{n}))

So that

- L = \inf (im ((a_{n})))

. Then note that

\lim_{n \to \infty} - a_{n} = L

implies that

- L = \lim_{n \to \infty} a_{n}

so we conclude that

\lim_{n \to \infty} a_{n} = \inf (im ((a_{n})))

as needed.

Nested Square Root Twos

Consider the sequence defined as follows for

α \in ℝ^{+}

$a_{1} = \sqrt{α}$
$a_{n + 1} = \sqrt{α + a_{n}}$ for every $n \geq 1$

Show that it converges

One way to show that it converges would to be the use the MCT for monotone increasing sequences. We start by showing that it is monotone increasing by showing it is strictly increasing.

For the base case we need to verify that $a_{1} < a_{2}$ which is equivalent to showing that $\sqrt{α} < \sqrt{α + \sqrt{α}}$ this is true because $\sqrt{\cdot}$ is order preserving and we know that $α < α + \sqrt{α}$ because $\sqrt{α} > 0$ .

For the induction step assume that $a_{k} < a_{k + 1}$ and we'll prove that $a_{k + 1} < a_{k + 2}$ . Note that $a_{k + 2} = \sqrt{α + a_{k + 1}}$ by the induction hypothesis we can see that $\sqrt{α + a_{k + 1}} > \sqrt{α + a_{k}} = a_{k + 1}$ , thus we have that $a_{k + 2} > a_{k + 1}$ as needed.

We'll move on to showing that it is bounded now, to do so we claim that $\sqrt{α} + 1$ is an upper bound, we'll show it is by induction, so for $n = 1$ we see that $\sqrt{α} \leq \sqrt{α} + 1$ , now assuming it's true for $k \in ℕ_{1}$ we'll show that $a_{k + 1} \leq \sqrt{α} + 1$ . We note that $a_{k + 1} = \sqrt{α + a_{k}}$ and thus by the induction hypothesis we have $\sqrt{α + a_{k}} \leq \sqrt{α + \sqrt{α} + 1}$ , but note that $\sqrt{α + \sqrt{α} + 1} < \sqrt{α + 2 \sqrt{α} + 1} = \sqrt{{(\sqrt{α} + 1)}^{2}} = \sqrt{α} + 1$ thus chaining inequalities we obtain $a_{k + 1} \leq \sqrt{α} + 1$ as needed, the choice of $\sqrt{α} + 1$ was inspired by thinking ahead to induction, what we would have to prove and figuring out what it can be upper bounded.

Since we've shown monotone increasing and bounded above, then the sequence converges to some $L \in ℝ$ , if that's true then $L = \sqrt{α + L}$ because $\lim_{n \to \infty} a_{n + 1} = \lim_{n \to \infty} \sqrt{α + a_{n}}$ therefore $L^{2} = α + L$ so that $L^{2} - L - α = 0$ so by the quadratic formula we have $L = \frac{1 \pm \sqrt{1 + 4 α}}{2}$ but since $a_{n} \geq 0$ for all $n \in ℕ_{1}$ then we know that $L \geq 0$ and therefore specifically it converges to $L = \frac{1 + \sqrt{1 + 4 α}}{2}$

Inequality From Sequence to Limit

Suppose

(a_{n}), (b_{n})

are converging sequences such that

a_{n} \leq b_{n}

for all

n \in ℕ_{1}

then

\lim_{n \to \infty} a_{n} \leq \lim_{n \to \infty} b_{n}

Nested Intervals

Suppose that

I_{n} = [a_{n}, b_{n}]

are nonempty closed intervals such that

I_{n + 1} \subseteq I_{n}

for every

n \in ℕ_{1}

, then

⋂_{n \in ℕ_{1}} I_{n} \neq \emptyset

Since $[a_{n + 1}, b_{n + 1}] \subseteq [a_{n}, b_{n}]$ then we know that $a_{n} \leq a_{n + 1} \leq b_{n + 1} \leq b_{n}$ for any $n \in ℕ_{1}$ and therefore we conclude that $(a_{n})$ is monotone increasing and that $(b_{n})$ is monotone decreasing. Also note that $a_{n} \leq b_{n} \leq b_{1}$ and $a_{1} \leq a_{n} \leq b_{n}$ so that by transitivity we obtain $a_{n} \leq b_{1}$ and $a_{1} \leq b_{n}$ which is to say that $a_{n}$ is bounded above and that $b_{n}$ is bounded below.

Therefore from both versions of the montone convergence theorem we see that $a_{n}$ and $b_{n}$ converge to $a : = \sup (im ((a_{n}))), b : = \inf (im (b_{n}))$ respectively and from the above proposition we see that $a \leq b$

By the definition of the supremum and infimum, we also have that $a_{k} \leq a \leq b \leq b_{k}$ for all $k \in ℕ_{1}$ meaning that $a, b$ belongs to $I_{k}$ , so that ${a, b} \subseteq ⋂_{n \in ℕ_{1}} I_{n}$ and so it must not be empty.

Bolzano-Weierstrass

Every bounded sequence of real numbers has a convergent subsequence.

Let $(a_{n})$ be bounded by $B$ , therefore $im ((a_{n})) \subseteq [- B, B]$ . If $I$ is an interval containing infinitely man points of ths equence $(a_{n})$ and $I = J_{1} \cup J_{2}$ then at least one of $J_{1}, J_{2}$ has infinitely many elements or we would have a contradiction.

With that in mind we construct the following sequence of intervals

$I_{1} = [a_{1}, b_{1}]$ where $a_{1} = - B$ and $b_{1} = B$
Consider L=[am,am+bm2] and R=[am+bm2,bm]
- If $L \cap im (a_{n})$ is infinite then $I_{m + 1} = L$
- If $R \cap im (a_{n})$ is infinite then $I_{m + 1} = R$
- If both are infinite just use $I_{m + 1} = L$

We now construct a subsequence as follows: Since every $I_{k} \cap im ((a_{n}))$ is infinite then we know there is some $a_{j} \in I_{k} \cap im ((a_{n}))$ , this defines a function $σ : ℕ_{1} \to ℕ_{1}$ such that $σ (k) = j$ and we use this to make our subsequence $(a_{σ (n)})$

We will now show that this subsequence converges. Note that by construction we see that $I_{k + 1} \subseteq I_{k}$ for all $k \in ℕ_{1}$ , and that each is non-empty, so that by the nested intervals lemma there exists an $L \in ⋂ n \in ℕ_{1} I_{n}$ , let's show that the subsequence converges there. Moreover note that $| I_{k} | = 2^{2 - k} B$

Let $ϵ \in ℝ^{+}$ then there exists some $k \in ℕ_{1}$ such that $2^{k - 2} B \leq ϵ$ and thus by definition of $σ$ we have $a_{σ (k)}, L \in I_{k}$ and thus $| a_{σ (k)} - L | \leq | I_{k} | = 2^{2 - k} B \leq ϵ$ which shows that $(a_{σ (n)}) \to L$