MATH 152 Chapter 10.1

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Sequences

sequence defined as list of numbers (terms) written in a definite order.

Notation

$\{a_1,a_2,a_3,\dots \}=\left\{a_n\right\}={\left\{a_n\right\}}_{n=1}^{\mathrm{\infty }}$

When sequence is denoted by formula:

${\left\{\frac{n}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}{a}_n=\frac{n}{n+1}\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\dots ,\frac{n}{n+1},\dots \}$

Infinite Sequences

Sequence $\left\{a_n\right\}$ has the limit $L$:

$\underset{n\to \mathrm{\infty }}{\lim }{a}_n=L\text{or}{a}_n\to L\text{as}n\to \mathrm{\infty }$

Definitions

If the limit exists, then the sequence can be bounded:

bounded above

a_n ≤ M for all n ≥ 1

all terms are less than the limit

bounded below

a_n ≥ M for all n ≥ 1

all terms are greater than the limit

Sequences can be increasing or decreasing. If sequence is increasing or decreasing, it is called monotonic.

A sequence is increasing if

$a_n<a_{n+1}$

$\frac{a_n}{a_{n+1}}<1$

$f^{\prime }\left(a\right)>0$

A sequence is decreasing if

$a_n>a_{n+1}$

$\frac{a_n}{a_{n+1}}>1$

$f^{\prime }\left(a\right)<0$

Every bounded monotonic sequence converges to a certain number:

Example

$a_n=\frac{1000^n}{n!}$. Show that $a_n$ is decreasing (for $n$ > some $N$) and bounded below. What is the limit of this sequence and why?

Review: Limit Laws

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }(a_n+b_n)=& \underset{n\to \mathrm{\infty }}{\lim }{a}_n+\underset{n\to \mathrm{\infty }}{\lim }{b}_n\\ \underset{n\to \mathrm{\infty }}{\lim }(c{a}_n)=& c\underset{n\to \mathrm{\infty }}{\lim }{a}_n\\ \underset{n\to \mathrm{\infty }}{\lim }{a}_n{b}_n=& \underset{n\to \mathrm{\infty }}{\lim }{a}_n\times \underset{n\to \mathrm{\infty }}{\lim }{b}_n\\ \underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=& \frac{\underset{n\to \mathrm{\infty }}{\lim }{a}_n}{\underset{n\to \mathrm{\infty }}{\lim }{b}_n}\underset{n\to \mathrm{\infty }}{\lim }{b}_n\ne 0\\ \text{if}\underset{n\to \mathrm{\infty }}{\lim }\left|a_n\right|=& 0\text{then}{a}_n=0\end{array}$

Example 1

Find the limit of the sequence

$a_n={(1+\frac{3}{n})}^{4n}$

Look at $f(x)={(1+\frac{3}{x})}^{4x}$:

$\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)={\left(1\right)}^{\mathrm{\infty }}$... That "1" is approaching 1, not equal to 1. We have to take a limit using L'Hospital's rule:

$\begin{array}{rl}{e}^{\underset{x\to \mathrm{\infty }}{\lim }4x\ln (1+\frac{3}{x})}=& e^{4\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+\frac{3}{x})}{\frac{1}{x}}}\\ =& e^{4\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{1}{1+3/x}-\frac{3}{x^2}}{-\frac{1}{x^2}}}\\ =& e^{4\underset{x\to \mathrm{\infty }}{\lim }\frac{3}{1+3/x}}\\ =& e^{12}\end{array}$

Example 2

Find the limit of the sequence with mathematical justification:

$a_n=\frac{{(-1)}^n}{n!}$

Definition of factorial (!) operator

$n!=n(n-1)(n-2)(n-3)\dots \left(3\right)\left(2\right)\left(1\right)$

Let's look at an analogous function:

$\left|a_n\right|=\frac{1}{n!}$

We can squeeze this function between 0 and $\frac{1}{n}$:

$0\le \left|a_n\right|\le \frac{1}{n}0\le \underset{n\to \mathrm{\infty }}{\lim }\left|a_n\right|\le 0$

Therefore, if $\left|a_n\right|\to 0$ by squeeze theorem, $a_n\to 0$

Proof By Induction

Given a statement about positive integers ($n\in {\mathbb{\mathbb{Z}}}^{+}$)

• If statement is true for $n=1$
• If statement is true for $n=k+1$ (given it is true for $n=k$)
• Then statement is true for all positive integers (like a domino effect; the first domino pushes down the second, and so on...

Recursively Defined Functions

Given $a_0=1$ and $a_{n+1}=\frac{1}{3}(a_n+4)$:

1. Show $a_n$ is increasing and $a_n<2$ for all $n$
2. Find $\underset{n\to \mathrm{\infty }}{\lim }{a}_n$

Solve 1 by induction

Show that $a_0<2$ ($a_0=1$; true)

Induction hypothesis: assume $a_k<2$:

$\begin{array}{rcl}{a}_k& <& 2\\ a_k+4& <& 6\\ \frac{1}{3}(a_k+4)& <& 2\\ a_{k+1}& <& 2\\ \therefore a_n& <& 2\text{for all}n\text{by induction}\end{array}$

Show that $a_0<a_1$ ($1<\frac{5}{3}$; true)

Induction hypothesis: assume $a_k<a_{k+1}$

$\begin{array}{rcl}{a}_k& <& a_{k+1}\\ a_k+4& <& a_{k+1}+4\\ \frac{1}{3}(a_k+4)& <& \frac{1}{3}(a_{k+1}+4)\\ a_{k+1}& <& a_{k+2}\\ a_n& <& a_{n+1}\text{by induction, so}{a}_n\text{is increasing}\end{array}$

Solve 2 by induction

So $a_n$ is convergent by the Monotonic Sequence Theorem. We'll call the limit $L$: $a_n\to L$, so $a_{n+1}\to L$ (same sequence; converges to same $L$)

So as $n\to \mathrm{\infty }$, $a_{n+1}=\frac{1}{3}(a_n+4)$

$\begin{array}{rcl}L& =& \frac{1}{3}(L+4)\\ 3L& =& L+4\\ 2L& =& 4\\ L& =& 2\end{array}$