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Section 9.1 – Sequences Sequence A sequence {𝑎𝑛 } is a list of numbers written in an explicit order. 𝑎𝑛 = {𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , … } First Term Second Term nth Term Generally, we will concentrate on infinite sequences, that is, sequences with domains that are infinite subsets of the positive integers. Recursive Formula A formula that requires the previous term(s) in order to find the value of the next term. Example: Find a Recursive Formula for the sequence below. 2, 4, 8, 16, … 𝑎1 = 2 𝑎𝑛 = 2 ∙ 𝑎𝑛−1 Explicit Formula A formula that requires the number of the term in order to find the value of the next term. Example: Find an Explicit Formula for the sequence below. 2, 4, 8, 16, … The Explicit Formula is also known as the General or nth Term equation. an 2 n Arithmetic Sequences A sequence which has a constant difference between terms. The rule is linear. Example: 1, 4, 7, 10, 13,… n a(n) 1 1 2 4 3 7 4 10 5 13 +3 +3 +3 +3 (generator is +3) Discrete Explicit Formula an 3n 2 Recursive Formula a1 1 an an 1 3 Sequences typically start with n=1 White Board Challenge Write an equation for the nth term of the sequence: Then find the n=0 term. n=0 n=1 n=2 n=3 n=4 40, 36, 32, 28, 24, … –4 First find the generator a(0) is not in the sequence! Do not include it in tables or graphs! an 4n 40 Geometric Sequences A sequence which has a constant ratio between terms. The rule is exponential. Example: 4, 8, 16, 32, 64, … n t(n) 1 4 2 8 3 16 4 32 5 64 x2 x2 x2 x2 (generator is x2) Discrete 0 Explicit Formula an 2 2 n 1 2 3 4 5 6 Recursive Formula a1 4 an 2 an 1 Sequences typically start with n=1 White Board Challenge Write an equation for the nth term of the sequence: Then find the n=0 term. n=0 3 , 5 n=1 n=2 n=3 n=4 3, 15, 75, 375, … a(0) is not in the sequence! Do not include it in tables or graphs! x5 First find the generator 3 n an 5 5 New Sequences The previous sequences were the only ones taught in Algebra 2. But, it is possible for a sequence to be neither arithmetic nor geometric. Example: Find a formula for the general term of the sequence below n=1 n=2 n=3 n=4 n=5 3 4 5 6 7 ,− , , − , ,… 5 25 125 625 3125 1 n1 n2 n 5 White Board Challenge Example: Find a formula for the general term of the sequence below n=1 n=2 n=3 n=4 n=5 1 1 1 1 1, , , , , … 3 5 7 9 1 an 2n 1 Monotonic Sequence A sequence is monotonic if it is either increasing (if 𝑎𝑛 < 𝑎𝑛+1 for all 𝑛 ≥ 1) or decreasing (if 𝑎𝑛 > 𝑎𝑛+1 for all 𝑛 ≥ 1). Example 1: Find the first 4 terms of 𝑎𝑛 = to see how the sequence is monotonic. 1 11 , 2 2 1 1 2 2 3 3 31 , 3 4 , , , 4 5 , 4 4 1 𝑛 𝑛+1 Example 2 Prove the sequence 𝑎𝑛 = 3 𝑛+5 is decreasing. If the sequence is decreasing, 𝒂𝒏 > 𝒂𝒏+𝟏 for all 𝒏. 3 IF: an n5 3 3 THEN: an 1 n 1 5 n 6 Since the denominator is smaller: 3 3 n5 n6 Therefore, 𝒂𝒏 is decreasing. OR an an 1 Bounded Sequence A sequence {𝑎𝑛 } is bounded above if there is a number 𝑀 such that 𝑎𝑛 ≤ 𝑀 for all 𝑛 ≥ 1 A sequence {𝑎𝑛 } is bounded below if there is a number 𝑚 such that 𝑚 ≤ 𝑎𝑛 for all 𝑛 ≥ 1 If it is bounded above and below, then {𝑎𝑛 } is a bounded sequence. Example Determine if the sequences below bounded below, bounded above, or bounded. 1. 𝑎𝑛 = 𝑛 Since 𝒏 = 𝟏, 𝟐, 𝟑, … : 𝒂𝒏 ≥1 Therefore, 𝒂𝒏 is bounded below. Since lim 𝒏 = ∞ : The sequence is not bounded above. 𝒏→∞ 2. 𝑎𝑛 = 𝑛 𝑛+1 Since 𝒏 = 𝟏, 𝟐, 𝟑, … : 𝒂𝒏 >0 Since 𝒏 lim 𝒏→∞ 𝒏+𝟏 = 𝟏 : 𝒂𝒏 <1. Therefore, 𝒂𝒏 is bounded. Limit of a Sequence A sequence {𝑎𝑛 } has the limit 𝐿 and we write: lim 𝑎𝑛 = 𝐿 𝑛→∞ or 𝑎𝑛 → 𝐿 as 𝑛 → ∞ if we can make the terms 𝑎𝑛 as close to 𝐿 as we like by taking 𝑛 sufficiently large. If lim 𝑎𝑛 exists, we say the sequence 𝑛→∞ converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent). Reminder: Properties of Limits Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the following limits: l i m f() x L x c l i m g () x K x c Constant Function lim bb Limit of x lim xc Limit of a Power of x Scalar Multiple x c x c lim x c n n x c l i m b f ( x ) b L x c Reminder: Properties of Limits Let b and c be real numbers, let n be a positive integer, and let f and g be functions with the following limits: l i m f() x L x c Sum/Difference Product Quotient Power l i m g () x K x c l i m f ( x )( g x ) L K x c l i m f ( x ) g ( xL ) K x c fx () L l i m , K 0 xc g ( x ) K l i m () L fx n x c n Example Determine if the sequences below converge or diverge. If the sequence converges, find its limit. 1. 𝑎𝑛 = 2. 𝑎𝑛 = 𝑛 𝑛+1 lim n 𝑛 10+𝑛 n lim n n 1 n 10 n L. H . lim 1 n 1 1 L. H . lim 1 101n 1/2 n 2 lim 2 10 n 3. 𝑎𝑛 = ln 𝑛 𝑛 lim n n L. H . ln n n Converges to 1 1 n lim 1 0 n Diverges Converges to 0 White Board Challenge Determine whether the sequence converges or diverges. If it converges, find its limit. 22 −2 32 −2 42 −2 52 −2 , , , , … 2 2 2 2 2 3 4 5 2 1) −2 (𝑛 + 𝑎𝑛 = (𝑛 + 1)2 Converges to 1 Absolute Value Theorem It is not always possible to easily find the limit of a sequence. Consider: 1 1 1 −1 𝑛 −1, 2 , −3 , 4 , … , 𝑛 , … The Absolute Value Theorem states: If lim 𝑎𝑛 = 0, then lim 𝑎𝑛 = 0. 𝑛→∞ 𝑛→∞ Example Determine if the sequences below converge or diverge. 1. 𝑎𝑛 = lim n 2. 𝑎𝑛 = −1 𝑛 𝑛 1n Because of the Absolute Value Theorem, Converges to 0 lim 1n 0 n n −1 𝑛 Since the limit does not equal 0, we can not apply the Absolute Value Theorem. It does not mean it diverges. Another test is needed. lim 1 lim1 1 n n n The sequence diverges since it does not have a limit: -1,1,-1,1,-1,… Theorem: Bounded, Monotonic Sequences Every bounded, monotonic sequence is convergent. Example: Investigate the sequence below. 𝑎1 = 2 𝑎𝑛+1 = 12(𝑎𝑛 + 6) a1 2 a2 4 a3 5 a4 5.5 a5 5.75 a5 5.85 The sequence appears to be monotonic: It is increasing. The sequence appears to be bounded: 2 ≤ 𝑎𝑛 ≤ 6 The limit of the sequence appears to be 6. Since the sequence appears to be monotonic and bounded, it appears to converge to 6.