Real Analysis – Sequences IV

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


4 Responses to “Real Analysis – Sequences IV”

  1. […] After introducing sequences and gaining some knowledge of some of their properties (I,II ,III , and IV) we are ready to embark on the study of real analysis while using concepts that are more in the […]

  2. PiterJankovich Says:

    My name is Piter Jankovich. oOnly want to tell, that your blog is really cool
    And want to ask you: is this blog your hobby?
    P.S. Sorry for my bad english

  3. Thanks for the comments. It is only a hobby.

  4. […] is a sequence of points in the compact interval (see definition 22 in post Real Analysis – Sequences IV) , by Corollary 27 (also in post Real Analysis – Sequences IV) there […]

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: