3.6 Properly Divergent Sequences

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Summary

Definition

• Properly divergent sequence (tends to)

Theorem

• $X,Y$ sequence of real numbers

1. $X$ monotone $\to$ ($X$ properly divergent $\leftrightarrow$ $X$ unbounded)
2. $X$ unbounded increasing $\to$ $\lim (x_n)=+\infty$
3. $X$ unbounded decreasing $\to$ $\lim (x_n)=-\infty$
4. $X>Y$ $\to$
   • $\lim X=+\infty$ $\to$ $\lim Y=+\infty$
   • $\lim Y=-\infty$ $\to$ $\lim X=-\infty$
5. $X,Y$ positive, $\lim \frac{X}{Y}=L\in \mathbb{P}$ $\to$ $\lim X=+\infty \leftrightarrow \lim Y=+\infty$

3.6.1 Definition: Properly divergent sequence

Let $(x_n)$ be a sequence of real numbers

1. We say that $(x_n)$ tends to $\pm \infty$ and write $\lim (x_n)=+\infty$, if for every $\alpha \in \mathbb{R}$ there exists a natural number $K(\alpha )$ such that $x_n>\alpha$, $\forall n\ge K(\alpha )$.

2. We say that $(x_n)$ tends to $-\infty$ and write $\lim (x_n)=-\infty$, if for every $\beta \in \mathbb{R}$ there exists a natural number $K(\beta )$ such that $x_n<\beta$, $\forall n\ge K(\beta )$

We say that $(x_n)$ is properly divergent in case we have either $\lim (x_n)=+\infty$ or $\lim (x_n)=-\infty$



3.6.3 Theorem

A monotone sequence of real numbers is properly divergent if and only if it is unbounded.

Furthermore,

1. If $(x_n)$ unbounded increasing sequence, then $\lim (x_n)=+\infty$
2. If $(x_n)$ unbounded decreasing sequence, then $\lim (x_n)=-\infty$

3.6.4 Theorem

Let $(x_n)$ and $(y_n)$ be two sequences of real numbers and suppose that $x_n\ge y_n,\forall n\in \mathbb{N}$.

1. If $\lim (x_n)=+\infty$, then $\lim (y_n)=+\infty$
2. If $\lim (y_n)=-\infty$, then $\lim (x_n)=-\infty$

Concept

If $\lim X=+\infty$, and $X>Y$, $\lim Y$ must be higher and therefore $Y$ must converge to $+\infty$ too.

TODO reword, give example

Can also apply for $\forall n\ge m$.

Useful application of this theorem is proving a sequence properly diverges, knowing another sequence related to it properly diverges.

Lemma

If $\lim (y_n)=\infty$ and $c>0$ then $\lim (c{y}_n)=\infty$

3.6.5 Theorem: Ratio test

Let $(x_n)$ and $(y_x)$ be two sequences of positive real numbers and suppose that for some $L\in \mathbb{P}$ we have $\lim (x_n/y_n)=L$.

Then $\lim (x_n)=+\infty$ if and only if $\lim (y_n)=+\infty$

Examples

Prove lim(n) is infinity

Prove $\lim (n)=\infty$

We need to prove that for all $\alpha \in \mathbb{R}$ there exists $K(\alpha )>\alpha$ such that $x_n>\alpha$, $\forall n\ge K(\alpha )$

Let $a\in \mathbb{R}$, $(x_n)$ a sequence of real numbers.

According to Archimedean Property, there exists $K(\alpha )\in \mathbb{N}$ such that $K(\alpha )>\alpha$.

Consequently, for all $x_n=n\ge K(\alpha )>\alpha$, we have $x_n>\alpha$.

$\therefore$ By definition of properly divergent sequence, we have $\lim (n)=\infty$.

Prove lim(n^2) is infinity

Prove $\lim (n^2)=\infty$

Let $a\in \mathbb{R}$, $(x_n)$ a sequence of real numbers.

Consider all cases for $\alpha$:

• If $\alpha \le 0$, then $x_n=n^2>0>\alpha ,\forall n\in \mathbb{N}$.
• If $\alpha >0$, then according to Archimedean Property, there exists $K(\alpha )\in \mathbb{N}$ such that $K(\alpha )>\sqrt{\alpha }$. Consequently, for all $x_n=n^2\ge (K(\alpha ){)}^2>\alpha$, we have $x_n>\alpha$.

$\therefore$ By definition of properly divergent sequence, we have $\lim (n)=\infty$.