12.1+Content+-+Sequences+&+Series

toc 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.
 * Quick Instructions**


 * 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 //n//th term of a sequence //directly//


 * Recursive Rule** = rule to evaluate the //n//th 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: math \[a_{n}=a_{1}+(n-1)d\] math

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

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

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

Write an explicit rule for the sequence: 3, 6i, -12, -24i... math \[a_{n}=a_{1}r^{n-1}\] math math a_{1}=3 math math r=2i math math \[a_{n}=3(2i)^{n-1}\] math
 * __Example:__**

=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

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

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

Find Summation by Hand
math \sum_{n=3}^{7}(-1)^{^{n}}(n^{3}) math

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

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

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

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

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

Geometric Partial Sum
__**Partial sum of any geometric sequence:**__ math \[S_{n}= \frac{a_{1}(1-r^{n})}{1-r}\], r \neq 1 math
 * 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:**__ math S_{\infty }=\frac{a}{1-r}, |r|<1 math

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

Diverge
When the partial sums of a geometric series do NOT approach a finite number; when math math math math
 * 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 math f(x)=\frac{3}{x+4} math (Find a rule whose limit as n approaches infinity results in that equation) math \sum_{n=1}^{\infty }??? = \frac{3}{x+4} math --> Recall: An infinite geometric series summation has the equation: math S_{\infty }=\frac{a}{1-r} math 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) math \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} math 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: math \[a_{n}=a_{1}r^{n-1}\] math math \sum_{n=1}^{\infty}\frac{3}{4}(\frac{-x}{4})^{n-1} math But in PRECALC HONORS AND CALC BC, we don't like this. So we manipulate the algebra to isolate for the x value. math \frac{3(-1)^{n-1}(x)^{n-1}}{(4)^{1}(4)^{n-1}}=\frac{3(-1)^{n-1}(x)^{n-1}}{(4)^{n}} math 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 math -1<\frac{-x}{4}<1 math -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. math r= \frac{1}{x+3} math Put math \frac{1}{x+3} math into r: math math Change the equation to: math \frac{|1|}{|x+3|} <1 math 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: math Therefore, (-\infty, -4) \cup (-2, +\infty) math
 * r| <1
 * \frac{1}{x+3}| <1

__**Example 2 - Interval of Convergence:**__ For what values of //x// will the following series converge? (Interval notation) math \frac{1}{2}+\frac{5}{4(x-1)}+\{25}{8(x-1)^{2}}+\frac{125}{16(x-1)^{3}}+... math By looking at the second term, divide it by the first term to get the ratio math r=\frac{5}{2(x-1)} math 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 math 5<\left | 2(x-1) \right | math 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 math (-\infty ,\frac{-3}{2})\bigcup (\frac{7}{2},\infty ) math

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

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

__**Example 2 - Proof by Induction: **__ <span style="font-family: Arial,Helvetica,sans-serif;">→ Question: Prove by induction that

<span style="font-family: Arial,Helvetica,sans-serif;">[[image:pch-wiki/CodeCogsEqn.gif]]
<span style="font-family: Arial,Helvetica,sans-serif;">is true. <span style="font-family: Arial,Helvetica,sans-serif;">Rewrite in form we like: <span style="font-family: Arial,Helvetica,sans-serif; line-height: 0px; overflow: hidden;">

Follow the same steps 1-3 above as before.


 * Primary authors of this page (as of 06/02/12):**


 * Jenny Li**
 * Tina Moazezi**

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