MATH 152 Chapter 10.1

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Sequences

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

Notation

$\left\{a_{1},a_{2},a_{3},\dots \right\}=\left\{a_{n}\right\}=\left\{a_{n}\right\}_{n=1}^{\infty }$

When sequence is denoted by formula:

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

Infinite Sequences

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

$\lim _{n\to \infty }{a_{n}}=L\quad \quad {\text{or}}\quad \quad a_{n}\to L\;{\text{as}}\;n\to \infty$

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

$a_{n}<a_{n+1}\,\!$

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

$f'\left(a\right)>0$

A sequence is decreasing if

$a_{n}>a_{n+1}\,\!$

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

$f'\left(a\right)<0$

Every bounded monotonic sequence converges to a certain number:

Example

$a_{n}={\tfrac {1000^{n}}{n!}}$. 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

${\begin{aligned}\lim _{n\to \infty }\left(a_{n}+b_{n}\right)=&\lim _{n\to \infty }a_{n}+\lim _{n\to \infty }b_{n}\\\lim _{n\to \infty }(ca_{n})=&c\lim _{n\to \infty }a_{n}\\\lim _{n\to \infty }a_{n}b_{n}=&\lim _{n\to \infty }a_{n}\times \lim _{n\to \infty }b_{n}\\\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=&{\dfrac {\lim _{n\to \infty }a_{n}}{\lim _{n\to \infty }b_{n}}}\quad \lim _{n\to \infty }b_{n}\neq 0\\{\mbox{if}}\ \lim _{n\to \infty }\left|a_{n}\right|=&0\quad {\mbox{then}}\ a_{n}=0\end{aligned}}$

Example 1

Find the limit of the sequence

$a_{n}=\left(1+{\tfrac {3}{n}}\right)^{4n}$

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

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

${\begin{aligned}e^{\lim _{x\to \infty }4x\ln {\left(1+{\frac {3}{x}}\right)}}=&e^{4\lim _{x\to \infty }{\frac {\ln {\left(1+{\frac {3}{x}}\right)}}{\frac {1}{x}}}}\\=&e^{4\lim _{x\to \infty }{\frac {{\frac {1}{1+3/x}}\,-{\frac {3}{x^{2}}}}{-{\frac {1}{x^{2}}}}}}\\=&e^{4\lim _{x\to \infty }{\frac {3}{1+3/x}}}\\=&e^{12}\end{aligned}}$

Example 2

Find the limit of the sequence with mathematical justification:

$a_{n}={\frac {\left(-1\right)^{n}}{n!}}$

Definition of factorial (!) operator

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

Let's look at an analogous function:

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

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

$0\leq \left|a_{n}\right|\leq {\frac {1}{n}}\quad 0\leq \lim _{n\to \infty }\left|a_{n}\right|\leq 0$

Therefore, if $\left|a_{n}\right|\to 0$ by squeeze theorem, $a_{n}\to 0$

Proof By Induction

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

• 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 $a_{n+1}={\tfrac {1}{3}}\left(a_{n}+4\right)$:

1. Show $a_{n}$ is increasing and $a_{n}<2$ for all $n$
2. Find $\lim _{n\to \infty }a_{n}$

Solve 1 by induction

Show that $a_{0}<2$ ($a_{0}=1$; true)

Induction hypothesis: assume $a_{k}<2$:

${\begin{array}{rcl}a_{k}&<&2\\a_{k}+4&<&6\\{\frac {1}{3}}\left(a_{k}+4\right)&<&2\\a_{k+1}&<&2\\\therefore \,a_{n}&<&2\,{\mbox{for all}}\,n\,{\mbox{by induction}}\end{array}}$

Show that $a_{0}<a_{1}$ ($1<{\tfrac {5}{3}}$; true)

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

${\begin{array}{rcl}a_{k}&<&a_{k+1}\\a_{k}+4&<&a_{k+1}+4\\{\frac {1}{3}}\left(a_{k}+4\right)&<&{\frac {1}{3}}\left(a_{k+1}+4\right)\\a_{k+1}&<&a_{k+2}\\a_{n}&<&a_{n+1}\,{\mbox{by induction, so}}\,a_{n}\,{\mbox{is increasing}}\end{array}}$

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 $n\to \infty$, $a_{n+1}={\tfrac {1}{3}}\left(a_{n}+4\right)$

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