Real Analysis – Sequences IV | From Suits to Spectra

After having stated and/or proved some important theorems about sequences in the previous post we will know introduce some auxiliary notions that will help us continuing our study of sequences.

— 5.3. Relationships Between Sequences —

Definition 19 Let us consider ${ u_n}$ and ${ v_n}$ . Furthermore let us suppose that there exists another sequence, ${ h_n}$ , such as ${ u_n = h_n v_n}$ .

If ${ \lim h_n=1}$ we’ll say that ${ u_n}$ is asymptotically equal to ${ v_n}$ and denote it by ${ u_n \sim v_n}$ . If ${ v_n \neq 0}$ we can write ${ h_n = \dfrac{u_n}{v_n}}$ .

As an example let us consider the sequence ${ u_n=3n^2-5n+1}$ . It is easy to see that in this case we have ${ u_n \sim 3n^2}$ .

We can write ${ 3n^2-5n+1=3n^2\left(1-\dfrac{5}{3n}+\dfrac{1}{3n^2}\right)}$ .

In this case it is ${ h_n=1-\dfrac{5}{3n}+\dfrac{1}{3n^2}}$ and we have ${ \lim h_n = \lim \left( 1-\dfrac{5}{3n}+\dfrac{1}{3n^2} \right) = 1}$ .

Theorem 23 Consider the sequences ${ a_n}$ , ${ b_n}$ , ${ c_n}$ , and ${ d_n}$ .

If ${ a_n \sim b_n}$ and ${ \lim a_n = a}$ then we also have ${ \lim b_n = a}$
If ${ a_n \sim c_n}$ and ${ b_n \sim d_n}$ then ${ u_n b_n \sim c_n d_n}$ and ${ \dfrac{a_n}{b_n} \sim \dfrac{c_n}{d_n}}$ .

Proof: By definition of ${ a_n \sim b_n}$ it is ${ a_n=h_n b_n}$ . Applying limits to both sides of the previous equation we have ${ \lim a_n =\lim (h_n b_n)= \lim h_n \lim b_n= 1\cdot \lim b_n}$ where ${ \lim h_n = 1}$ by hypothesis. So what we have is ${ \lim b_n =\lim a_n=a}$ .

Let us write ${ a_n= h_n c_n}$ and ${ b_n= t_n d_n}$ with ${ \lim h_n = \lim t_n = 1}$ . Then ${ a_n b_n = h_n t_n c_n d_n}$ and applying limits what we have is ${ \lim ( a_n b_n )= \lim (h_n t_n)\lim ( c_n d_n )}$ with ${ \lim (h_n t_n)= \lim h_n \lim t_n=1\times 1 =1}$ . So ${ \lim ( a_n b_n )= \lim ( c_n d_n )}$ as we intended to prove.

The division part of the enunciate is proven with the same kind of reasoning. $\Box$

Definition 20 Let ${ u_n}$ and ${ v_n}$ be two sequences and let us suppose that we can write ${ u_n= h_n v_n}$ with some sequence ${ h_n}$ :

If ${ \lim h_n = 0}$ we’ll say that ${ u_n}$ is negligible to ${ v_n}$ and denote it by ${ u_n = o(v_n)}$ . Or we can say in a more colloquial way that ${ u_n}$ is little-o of ${ v_n}$ .
If ${ h_n}$ is bounded we’ll say that ${ u_n}$ and ${ v_n}$ have the same order of magnitude (or say that ${ u_n}$ is big-o to ${ v_n}$ ) and denote it by ${ u_n = O(v_n)}$ .

Let us now try to give a more intuitive meaning to these three notions introduced so far:

First of the notion ${ u_n \sim v_n}$ expresses the fact the difference between ${ u_n}$ and ${ v_n}$ tends to ${ 0\,}$ as ${ n \rightarrow \infty}$ . That is to say that the two sequences get closer and closer together.

The notion of ${ u_n = O(v_n)}$ expresses the fact the both sequences differ only by a scale factor. That is to say that they have the same kind of behavior at ${ \infty}$ .

The meaning of the sentence the same kind of behavior will be made clearer as real analysis gets unfolded in this blog.

The notion of ${ u_n = o(v_n)}$ tell us at the ${ u_n}$ gets smaller and smaller when compared to ${ v_n}$ when we get to ${ \infty}$ . In a more formal way: if ${ v_n \neq 0 \quad \lim \dfrac{u_n}{v_n}=0}$

Let us now give some examples in order to make things a little bit easier to grasp:

$\displaystyle \dfrac{1}{n^3}=o \left(\dfrac{1}{n}\right)$

This is easy to see if we write ${ \dfrac{1}{n^3}=\dfrac{1}{n^2}\dfrac{1}{n}}$ . Taking ${ h_n = \dfrac{1}{n^2}}$ we see that it is effectively ${ \lim h_n=0}$

$\displaystyle \dfrac{\sin n}{n}=O\left(\dfrac{1}{n}\right)$

In this case we write ${ \dfrac{\sin n}{n}=\sin n \dfrac{1}{n}}$ and take ${ h_n=\sin n}$ . Since ${ \sin n}$ is a bounded function we get the intended result.

— 5.4. Final Comments on Sequences —

Definition 21 We’ll say that ${ u_{\alpha_n}}$ is a subsequence of ${ u_n}$ whenever ${ \alpha_n}$ is a sequence that tends to ${ \infty}$ .

Roughly speaking a subsequence, ${ u_{\alpha_n}}$ , of a given sequence, ${ u_n}$ , is sequence that doesn’t consider some of the indexes of the initial sequence.

A few examples of subsequences would be ${ u_{2n}}$ (where we don’t take into account the odd numbered indexes of the initial sequence), ${ u_{n^2}}$ (only taking into account the the perfect square indexes of the initial sequence).

Theorem 24 If a sequence has a limit, then all of its subsequences have the same limit.

Proof: By hypothesis ${ u_n \rightarrow a \in \overline{\mathbb{R}}}$ and let ${ u_{\alpha_n}}$ be a subsequence of ${ u_n}$ .

If ${ u_n}$ converges we know that ${ \forall \delta > 0 \exists l \in \mathbb{N}: \; n \geq l \Rightarrow u_n \in V(a,\delta)}$ .

Since ${ \alpha_n \rightarrow \infty \quad \exists k \in \mathbb{N}: \; n \geq k \Rightarrow u_{\alpha_n} > l}$ .

Thus ${ n \geq k \Rightarrow u_{\alpha_n} \in V(a,\delta)}$ .

By definition this is ${ \lim u_{\alpha_n}=a}$ $\Box$

We already saw that ${ u_n = \left (1+\dfrac{1}{n} \right )^n}$ was a converging sequence, then even though ${ v_n = \left (1+\dfrac{1}{n^2} \right )^{n^2}}$ appears to be a harder sequence we can say, without any effort, that ${ \lim \left (1+\dfrac{1}{n} \right )^n = \lim \left (1+\dfrac{1}{n^2} \right )^{n^2}}$ if we note that it is actually ${ v_n=u_{n^2}}$ and so ${ v_n}$ is a subsequence of a converging sequence.

Corollary 25 If a sequence has two subsequences with distinct limits then the sequence is divergent.

Proof: Follows directly from ${ p\Rightarrow q \Leftrightarrow \left( \sim q \Rightarrow \sim p \right)}$ . $\Box$

As an application from the previous corollary we have ${ u_n = (-1)^n}$ .

${ u_{2n}= (-1)^{2n}=1}$ and it is ${ \lim u_{2n}=1}$ .

${ u_{2n+1}=(-1)^{2n+1}=-1}$ and it is ${ \lim u_{2n+1}=-1}$ .

In conclusion ${ u_n=(-1)^n}$ is a divergent sequence.

Theorem 26 (Bolzano-Weierstrass) Each bounded sequence has a converging subsequence in ${ \mathbb{R}}$ .

Proof: This is only the sketch of a proof.

One way to do this is first to prove that all sequences have a monotone subsequence. Applying this result to a bounded sequence we’d have that the bounded sequence have a subsequence that is monotone and bounded (since the sequence is bounded). But by the Corollary 21 we know that a bounded and monotone sequence is convergent. $\Box$

Definition 22 Let ${ X \subset \mathbb{R}}$ . We’ll say that ${ X}$ is a compact interval if it is bounded and closed.

Corollary 27 Let ${ X}$ be a compact interval and ${ u_n : \mathbb{N} \rightarrow X}$ . Then ${ \exists \, u_{\alpha_n}: \quad \lim u_{\alpha_n}=x \in X}$ where ${ u_{\alpha_n}}$ is a subsequence of ${ u_n}$ .

Proof: Let ${ X= \lbrack a, b \rbrack}$ be the interval and ${ u_n}$ be a sequence of points in ${ X}$ . Since ${ a \leq u_n \leq b}$ , ${ u_n}$ is bounded. From the theorem 26 ${ u_n}$ has a converging subsequence ${ u_{\alpha_n}}$ .

For ${ u_{\alpha_n}}$ it also is ${ a \leq u_{\alpha_n} \leq b}$ . This implies ${ \lim a \leq \lim u_{\alpha_n} \leq \lim b \Rightarrow a \leq \lim u_{\alpha_n} \leq b\Rightarrow \lim u_{\alpha_n} \in X}$ $\Box$

Real Analysis – Sequences IV

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