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.

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.

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$