Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 9 July 2014

Lecture 10

Sequences

Let $Y$ be a set. A sequence $(y_1,y_2,y_3,\dots )$ in $Y$ is a function $$\begin{array}{ccc}{\mathbb{Z}}_{>0}& ⟶& Y\\ n& ⟼& y_n\text{.}\end{array}$$

Let $Y\subseteq ℝ\text{.}$ Let $(y_1,y_2,\dots )$ be a sequence in $Y\text{.}$ The sequence $(y_1,y_2,\dots )$ is increasing if $(y_1,y_2,\dots )$ satisfies if $i\in {\mathbb{Z}}_{>0}$ then $y_i\le y_{i+1}\text{.}$ The sequence $(y_1,y_2,\dots )$ is decreasing if $(y_1,y_2,\dots )$ satisfies if $i\in {\mathbb{Z}}_{>0}$ then $y_i\ge y_{i+1}\text{.}$ A sequence $\left(a_n\right)$ is monotone if $\left(a_n\right)$ is increasing or decreasing.

Let $Y$ be a metric space. Let $(y_1,y_2,\dots )$ be a sequence in $Y\text{.}$ The sequence is bounded if the set $$\{y_1,y_2,\dots \}$$ is bounded.

The sequence $(y_1,y_2,\dots )$ is contractive if $(y_1,y_2,\dots )$ satisfies: There exists $\alpha \in (0,1)$ such that if $i\in {\mathbb{Z}}_{>0}$ then $d(y_i,y_{i+1})\le \alpha d(y_{i-1},y_i)\text{.}$

The sequence $(y_1,y_2,\dots )$ is Cauchy if $(y_1,y_2,\dots )$ satisfies if $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(y_m,y_n)<\epsilon \text{.}$

Let $\ell \in Y\text{.}$ The sequence $(y_1,y_2,\dots )$ converges to $\ell$ if $(y_1,y_2,\dots )$ satisfies if $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $d(y_n,\ell )\le \epsilon \text{.}$

Let $(y_1,y_2,\dots )$ be a sequence in $ℝ$ (or more generally, any totally ordered set with the order topology). The upper limit of $(y_1,y_2,\dots )$ is $$\text{lim}\hspace{0.17em}\text{sup}\hspace{0.17em}{y}_n=\underset{n\to \infty }{\text{lim}}\hspace{0.17em}\text{sup}\{y_n,y_{n+1},\dots \}\text{.}$$ The lower limit of $(y_1,y_2,\dots )$ is $$\text{lim}\hspace{0.17em}\text{inf}\hspace{0.17em}{y}_n=\underset{n\to \infty }{\text{lim}}\hspace{0.17em}\text{inf}\{y_n,y_{n+1},\dots \}\text{.}$$

If $y_n={(-1)}^n(1-\frac{1}{n})$ then $$\text{lim}\hspace{0.17em}\text{sup}\hspace{0.17em}{y}_n=1\text{and}\text{lim}\hspace{0.17em}\text{inf}\hspace{0.17em}{y}_n=-1\text{.}$$

Let $(y_1,y_2,\dots )$ be a sequence in $ℝ\text{.}$ Then

(a)	$\text{lim}\hspace{0.17em}\text{sup}\hspace{0.17em}{y}_n=\text{sup}\left\{\text{cluster points of}\hspace{0.17em}(y_1,y_2,\dots )\right\},$ and
(b)	$\text{lim}\hspace{0.17em}\text{inf}\hspace{0.17em}{y}_n=\text{inf}\left\{\text{cluster points of}\hspace{0.17em}(y_1,y_2,\dots )\right\}\text{.}$

Boundedness, sup, inf, limsup and liminf

Let $a_n=\frac{\text{log}\hspace{0.17em}n}{\sqrt{n}}\text{.}$

(a)	Show that $a_n$ is bounded.
(b)	Find $N\in {\mathbb{Z}}_{>0}$ such that $a_n$ is decreasing for $n\in {\mathbb{Z}}_{>0}$ such that $n>N\text{.}$
(c)	Show that $\underset{n\to \infty }{\text{lim}}\frac{\text{log}\hspace{0.17em}n}{\sqrt{n}}=0\text{.}$

Since $$\begin{array}{ccc}{\displaystyle \frac{d}{dx}\left(\frac{\text{log}\hspace{0.17em}x}{\sqrt{x}}\right)}& =& {\displaystyle \frac{d}{dx}\left(x^{-\frac{1}{2}}\text{log}\hspace{0.17em}x\right)}\\ & =& {\displaystyle x^{-\frac{1}{2}}\frac{1}{x}+\frac{-1}{2}{x}^{-\frac{3}{2}}\text{log}\hspace{0.17em}x}\\ & =& {\displaystyle x^{-\frac{3}{2}}(1-\frac{1}{2}\text{log}\hspace{0.17em}x)}\\ & =& {\displaystyle \frac{1}{2}{x}^{-\frac{3}{2}}(2-\text{log}\hspace{0.17em}x)}\end{array}$$ is less than $0$ for $x>e^2,$ is equal to $0$ for $x=e^2,$ and is greater than $0$ for $x<e^2,$ the function $f\left(x\right)=\frac{\text{log}\left(x\right)}{\left|x^{\frac{1}{2}}\right|}$ is decreasing for $x>e^2$ and has a maximum at $e^2\text{.}$

Since $\text{log}\left(x\right)\ge 0$ for $x\ge 1$ and $\left|x^{\frac{1}{2}}\right|>0$ for $x\ge 1,$ $a_n=\frac{\text{log}\hspace{0.17em}n}{\left|n^{\frac{1}{2}}\right|}\ge 0$ for $n\in {\mathbb{Z}}_{>0}$ $$\begin{array}{c} 1 e 2 x 2 e y \end{array}$$ So $a_n$ is bounded by $\frac{2}{e}$ and $0$ $(0\le a_n\le \frac{2}{e})$ and $a_n$ is decreasing if $n>e^2$ (in particular, if $n>9\text{).}$

(c) $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}\frac{\text{log}\hspace{0.17em}n}{n^{\frac{1}{2}}}}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{\text{log}\hspace{0.17em}n}{{\left(e^{\text{log}\hspace{0.17em}n}\right)}^{\frac{1}{2}}}=\underset{n\to \infty }{\text{lim}}\frac{\text{log}\hspace{0.17em}n}{e^{\frac{1}{2}\text{log}\hspace{0.17em}n}}}\\ & =& {\displaystyle \underset{y\to \infty }{\text{lim}}\frac{y}{e^{\frac{1}{2}y}}=\underset{y\to \infty }{\text{lim}}\frac{y}{1+\frac{1}{2}y+\frac{1}{2!}{\left(\frac{1}{2}y\right)}^2+\cdots }}\\ & =& {\displaystyle \underset{y\to \infty }{\text{lim}}\frac{1}{\frac{1}{y}+\frac{1}{2}+\frac{1}{2^1}\frac{1}{2^2}y+\frac{1}{3!}\frac{1}{2^3}{y}^2+\cdots }}\\ & \le & {\displaystyle \underset{y\to \infty }{\text{lim}}\frac{1}{\frac{1}{2!}\frac{1}{2^2}y}=\underset{y\to \infty }{\text{lim}}\frac{8}{y}=0\text{.}}\end{array}$$

Analyse the sequence $a_n={(-1)}^{n-1}(1+\frac{1}{n})\text{.}$ $$\begin{array}{c} 1 2 3 4 5 6 7 n - 2 - 1 1 2 \end{array}$$ $a_n$ is bounded above by $2\text{,}$
$a_n$ is bounded below by $-2\text{,}$
$\text{lim sup}{a}_n=1$ and $\text{lim inf}\hspace{0.17em}{a}_n=-1,$
$\text{sup}\hspace{0.17em}{a}_n=2$ and $\text{inf}\hspace{0.17em}{a}_n=-\frac{3}{2}\text{.}$

$\underset{n\to \infty }{\text{lim}}{a}_n$ does not exist because, as $n$ gets larger and larger, $a_n$ oscillates between close to $\text{lim sup}\hspace{0.17em}{a}_n$ $\text{(}1,$ in this case) and $\text{lim inf}\hspace{0.17em}{a}_n$ $\text{(}-1,$ in this case).

If $a_n$ is a sequence in $ℝ$ (or a totally ordered set $X\text{)}$ such that

(a)	$a_n$ is increasing,
(b)	$a_n$ is bounded, and
(c)	$\text{sup}\hspace{0.17em}{a}_n$ exists,

then $\left(a_n\right)$ converges to $\text{sup}\hspace{0.17em}{a}_n\text{.}$

Note: In $ℝ,$ $\text{sup}\hspace{0.17em}{a}_n$ always exists, in $\mathbb{Q},$ $\text{sup}\hspace{0.17em}{a}_n$ does not always exist.

		Proof.
	Let $\ell =\text{sup}\hspace{0.17em}{a}_n\text{.}$ To show: $\underset{n\to \infty }{\text{lim}}{a}_n=\ell \text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $d(a_n,\ell )<\epsilon \text{.}$ Proof by contradiction. Assume that there exists $\epsilon \in {ℝ}_{>0}$ such that there does not exist $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $d(a_n,\ell )<\epsilon \text{.}$ So, if $N\in {\mathbb{Z}}_{>0}$ then there exists $n\in {\mathbb{Z}}_{>0}$ with $n>N$ such that $d(a_n,\ell )>\epsilon \text{.}$ so, if $N\in {\mathbb{Z}}_{>0}$ then $\ell -\epsilon >a_n\ge a_N\text{.}$ So $\ell -\epsilon$ is an upper bound of $\left(a_n\right)\text{.}$ Contradiction to $\ell =\text{sup}\hspace{0.17em}{a}_n\text{.}$ So $\underset{n\to \infty }{\text{lim}}{a}_n=\ell \text{.}$ $\square$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100322suggLect10.pdf and was given on 22 March 2010.

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