# 12.1 Content - Sequences & Series

Quick Instructions
Add Series and Sequences content to this page. See specific details on the home page. This unit has been split into parts A & B. Content between the two must be cohesive as it is one continuous, connected material. They are split simply to better organize the page.

Series and Sequences (Unit 12A) Learning Targets
• Write and use explicit and recursive sequences
• Use summation (sigma) notation.
• To find sums of finite series and sums of convergent infinite geometric series.
• To determine the interval of convergence of series in geometric form.
• Use mathematical induction to prove conjectures.

# Types of Rules

Explicit Rule = rule to evaluate the nth term of a sequence directly

Recursive Rule = rule to evaluate the nth term only by knowing the values of the previous (n-1)th and start terms

# Sequences

## Arithmetic

Arithmetic Sequence = a sequence of numbers in which the difference between consecutive terms is constant
Ex: 20, 24, 28, 32, 36 …

Equation to find the nth term in an arithmetic sequence:
$$a_{n}=a_{1}+(n-1)d$$

‍‍Example:
Write an explicit rule for the sequence: 20, 24, 28, 32, 36...
$$a_{n}=a_{1}+(n-1)d$$
$a_{1} = 20$
$d=4$
$$a_{n}=20+(n-1)4$$
$$a_{n}=4n+16$$

Geometric
Geometric Sequence = a sequence of numbers in which the common ratio between consecutive terms is constant
Ex: 3, 9, 27, 81, 243 …

Equation to find the nth term in a geometric sequence:
$$a_{n}=a_{1}r^{n-1}$$

Example:
Write an explicit rule for the sequence: 3, 6i, -12, -24i...
$$a_{n}=a_{1}r^{n-1}$$
$a_{1}=3$
$r=2i$
$$a_{n}=3(2i)^{n-1}$$

# Summation

Sigma (Summation) Notation = a compact way of expressing the sum of a sequence of numbers
i.e. Evaluate f(n) for all integers starting at n = i and stopping at n = k, then summing the results

$\sum_{n=i}^{k}f(n)$

f(n): function
n = i : start
n = k: stop

## Find Summation by Hand

$\sum_{n=3}^{7}(-1)^{^{n}}(n^{3})$

n = 3 --> $(-1)^{^{3}}(3^{3})$ = -27
n = 4 ... = 64
n = 5 ... = -125
n = 6 ... = 216
n = 7 ... = -343

-27 + 64 + (-125) +216 +(-343) = -215

$\sum_{n=3}^{7}(-1)^{^{n}}(n^{3})$ = -215

## Find Summation with a Calculator

TI 83/84 Command: sum(seq(rule, variable, start, stop))
~sum: 2nd->STAT->MATH->5
~seq: 2nd->STAT->OPS->5
TI 89 Command: F3, 4 then Σ(rule, variable, start, stop)

# Partial Sums and Infinite Sums

Partial Sums & Series = partial sums refer to the summation of the n amount of terms in a series

## Arithmetic Partial Sum

Partial sum of any arithmetic sequence:
$$S_{n}= \frac{n(a_{1}+a_{n})}{2}$$
*also known as the nth partial sum (or finite sum) of an arithmetic sequence

## Geometric Partial Sum

Partial sum of any geometric sequence:
$$S_{n}= \frac{a_{1}(1-r^{n})}{1-r}$ , r \neq 1$
*also known as the nth partial sum (or finite sum) of a geometric sequence

## Arithmetic Infinite Sum

Infinite sum of an arithmetic sequence:
‍‍Does not exist! Logically, unlike a geometric sequence, arithmetic sequences are linear and not asymptotic. Therefore, you never reach a single number; it gets higher and higher.

## Geometric Infinite Sum

‍‍Infinite sum of a geometric sequence:
$S_{\infty }=\frac{a}{1-r} , |r|<1$

## Key Terms

### ‍Converge

When the partial sums of a geometric series approach a finite number; when |r|< 1
*An infinite geometric series converges when |r|<1
If |r|<1, then the sum of an infinite geometric series can be found.

### Diverge

When the partial sums of a geometric series do NOT approach a finite number; when
$|r| \geq 1$
*An infinite geometric series diverges when
$|r| \geq 1$

Example - Geometric Series Infinite Sum:
Write an infinite geometric series, using sigma notation, whose result is
$f(x)=\frac{3}{x+4}$
(Find a rule whose limit as n approaches infinity results in that equation)
$\sum_{n=1}^{\infty }??? = \frac{3}{x+4}$
--> Recall:
An infinite geometric series summation has the equation:
$S_{\infty }=\frac{a}{1-r}$
So you set that equal to the f(x) equation to fit the f(x) equation to the format of the infinite series formula (i.e. get the denominator to become 1 - r)
$\frac{a}{1-r}=\frac{3}{x+4}=\frac{\frac{3}{4}}{\frac{x}{4}+\frac{4}{4}}=\frac{\frac{3}{4}}{\frac{4}{4}-(-\frac{x}{4})}=\frac{a}{1-r}$
and you get a = 3/4 and r = -x/4
With this, you plug in the a value and r value in the geometric sequence formula.
--> Recall:
Equation to find the nth term in a geometric sequence:
$$a_{n}=a_{1}r^{n-1}$$
$\sum_{n=1}^{\infty}\frac{3}{4}(\frac{-x}{4})^{n-1}$
But in PRECALC HONORS AND CALC BC, we don't like this. So we manipulate the algebra to isolate for the x value.
$\frac{3(-1)^{n-1}(x)^{n-1}}{(4)^{1}(4)^{n-1}}=\frac{3(-1)^{n-1}(x)^{n-1}}{(4)^{n}}$
So now the x value is alone and that's the answer.
If asked to identify the interval over which the series converges, |r| < 1. Since r = -x/4
$-1<\frac{-x}{4}<1$
-4 < x < 4
The series converges over the interval (-4, 4).

Example 1 - Interval of Convergence:
Evaluate the interval over which the given value r of a geometric series converges.
$r= \frac{1}{x+3}$
|r| <1
Put
$\frac{1}{x+3}$
into r:
$|\frac{1}{x+3}| <1$
Change the equation to:
$\frac{|1|}{|x+3|} <1$
Multiply |x+3| both sides:
1<|x+3|
Separate into two equation:
x+3>1 or x+3<-1
Solve x
x> -2 and x<-4
Write the interval notation:
$Therefore, (-\infty, -4) \cup (-2, +\infty)$

Example 2 - Interval of Convergence:
For what values of x will the following series converge? (Interval notation)
$\frac{1}{2}+\frac{5}{4(x-1)}+\{25}{8(x-1)^{2}}+\frac{125}{16(x-1)^{3}}+...$
By looking at the second term, divide it by the first term to get the ratio
$r=\frac{5}{2(x-1)}$
is the ratio (i.e. multiply that ratio by the first term and you get the second term, etc. etc.)
When you want to find where the series converges, |r|<1
so -1 < r < 1
$5<\left | 2(x-1) \right |$
You know the value of 2(x-1) has to be greater than 5 because the magnitude of the ratio has to be less than one.
2(x-1) > 5
2x > 7
x > 7/2
-5 > 2(x-1)
-3 > 2x
x < -3/2
So the interval at which the series converges is
$(-\infty ,\frac{-3}{2})\bigcup (\frac{7}{2},\infty )$

# Telescoping Sums

Telescoping Sum = a sum in which subsequent terms in a series cancel out, leaving only initial and final terms; used with decomposing partial fractions

Example - Telescoping Sums:

Given
$\sum_{k=1}^{n}\frac{1}{k(k+1)}$
Find
$S_{100}$
First, decompose the function into partial fractions using the "Cover Up" method
$\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}$
You get A = 1 and B = -1
So then,
$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n}(\frac{1}{k}-\frac{1}{k+1})=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+ ... +(\frac{1}{n}-\frac{1}{n+1})$
‍‍the middle terms cancel out (1/2 would cancel out -1/2, etc.) so only‍‍
$\frac{1}{1}-\frac{1}{n+1}$
is left. Now you can substitute 100 for n to solve for the sum of the first 100 terms.
$S_{100}=1-\frac{1}{100+1}=\frac{100}{101}$

# Proof by Induction

This is not a direct proof.

Goal: To apply "proof by mathematical induction" as a method to prove that a given series can be represented by a specific rule.

Step-by-Step Process:
1. Let’s show the rule is true for the base case; for n=1. (Prove that the first statement is true; usually check for k=1.)
2. If the result is true, then assume it will be true for n=k. Then, I will show that the rule also applies for n=k+1. (Now since the given statement is true for all n=k, we need to show that it is true for n=k+1.)
3. Write a concluding statement.

This is a really cool video that explains the basics of induction!

Example 1 - Proof by Induction:
→ Question: Prove by induction that the
$$S_{n}\, \textup{\textrm{of\, }}1+2+4+8+...= 2^{n}-1$$
$$\rightarrow \textup{\textrm{1+2+4+8+...+(\, )}}= 2^{n}-1$$
→ Put the rule of this series inside the parentheses.
→ Has a common ratio of 2 so this is a geometry series.
$$\rightarrow a_{n}= ar^{k-1}\: \leftarrow \textup{Recall this is the formula to find the nth term of a geometric sequence}$$
$$\therefore a_{n}= \left ( 1 \right )\left ( 2 \right )^{k-1}$$
$$\rightarrow \: \textup{1+2+4+8+...+(1)}(2)^{k-1}= 2^{n}-1$$
1. Check base case. n=1.
$$\rightarrow 2^{(1-1)}= 2^{(1)}-1$$
$$\: \rightarrow 2^{(0)}= 2-1$ $\[\rightarrow 1= 1\; \mathbf{TRUE}$$
2. Since base case true, we assume the rule is true for n=k and must show that it is also true for n=k+1.
$$\rightarrow \textup{1+2+4+8+...+}((1)(2)^{k-1})+((2)^{k-1+1})= 2^{k+1}-1$$
$$\rightarrow 2^{k}-1+2^{k}= 2^{k+1}-1$$
$$\Rightarrow 2^{(k+1)}-1= 2^{(k+1)}-1\; \mathbf{TRUE}$$
3. By induction, we have shown the rule
$$1+2+4+8+...+((1)(2)^{(k-1)})= 2^{n}-1$$
is true for all values of n.

Example 2 - Proof by Induction:
→ Question: Prove by induction that
is true.
Rewrite in form we like:

Follow the same steps 1-3 above as before.