Questions and Answers #7 Sequences

Questions and Answers Sheet 7 Sequences Question #1 Which of the sequences are bounded? 1) 4(−1)𝑛 2) 7n 3) 4) 1 13𝑛 𝑛 3𝑛+1 Answer: 1) 4(−1)𝑛 lim 𝑎𝑛 = lim 4 (−1)𝑛 = 𝑜𝑛𝑒 𝑛→∞ 𝑛→∞ Thus, it's not bounded 2) 7n lim 𝑎𝑛 = lim 7 𝑛 = ∞ 𝑛→∞ 𝑛→∞ It's not bounded 1 3) 𝑛 13 1 lim 𝑎𝑛 = lim 𝑛 = 0 𝑛→∞ 𝑛→∞ 13 It is bounded 𝑛 4) 3𝑛+1 𝑛 1 lim 𝑎𝑛 = lim = 𝑛→∞ 𝑛→∞ 3𝑛 + 1 3 Thus, it's bounded Question #2 Which of the following are geometric sequences? 1) 9,-92,94,-98,916,... 2) 6,-4,83,-169,3227,... 3) 4,6,8,10,12,... 4) 8,24,72,-216,648,... 5) 3,32,2,52,3,... Answer: 1) 9,-92,94,-98,916,... Here, 𝑎2 −92 = 𝑎1 9 𝑎3 94 −47 = = 𝑎2 −92 46 𝑎4 −98 −49 = = 𝑎3 94 47 As ratio is not the same, the sequence is not a geometric sequence. 2) 6,-4,83,-169,3227,... 𝑎 −4 −2 Here, 2 = = 𝑎1 6 3 𝑎3 83 −83 = = 𝑎2 −4 4 𝑎4 −169 −169 = = 𝑎3 83 83 The ratio is not the same, thus, the sequence is not a geometric sequence. 3) 4,6,8,10,12,... 𝑎2 6 3 = = 𝑎1 4 2 𝑎3 8 4 = = 𝑎2 6 3 𝑎4 10 5 = = 𝑎3 8 4 Ratio is not the same, thus, the sequence is not geometric 4) 8,24,72,-216,648,... 𝑎2 24 = =3 𝑎1 8 𝑎3 72 = =3 𝑎2 24 𝑎4 −216 = = −3 𝑎3 72 𝑎5 648 = = −3 𝑎4 −216 Ratio is not the same, so the sequence is not geometric 5) 3,32,2,52,3,... 𝑎2 32 32 = = 𝑎1 3 3 𝑎3 2 = 𝑎2 32 𝑎4 52 = 𝑎3 2 𝑎5 3 = 𝑎4 52 As ratio is not the same, the sequence is not geometric. Thus, none of the sequence is geometric. Question #3 Which answer is correct: "A sequence 𝑠𝑛 is said to be bounded above if there is some number N such that..." a) 𝑠𝑛+1 ≥ 𝑁 b) 𝑠𝑛 ≥ 𝑁 c) 𝑠𝑛 = 𝑁 d) 𝑠𝑛 ≤ 𝑁 Answer: A sequence 𝑎𝑛 is said to be bounded above if there exists a real number K such that 𝑎𝑛 ≤ 𝐾 for all 𝑛 ∈ 𝑁 Here, the sequence is 𝑠𝑛 and the real number is N . Thus, 𝑠𝑛 is said to be bounded above if 𝑠𝑛 ≤ 𝑁 for all 𝑛 ∈ 𝑁 . Therefore, the correct answer is (d). Question #4 Consider the following sequences. a) Find the first four terms of the sequence. b) Based on part (a), determine a plausible limit of the sequence. 𝑎𝑛 = 𝑛2 𝑛2 −1 ; 𝑛 = 2,3,4, … Answer: 1)Given: 𝑎𝑛 = 𝑛2 ; 𝑛 = 2,3,4, … 𝑛2 − 1 Put n=2 𝑎2 = 22 4 = 2 2 −1 3 Put n=3 𝑎3 = 32 9 = 32 − 1 8 Put n=4 𝑎4 = 42 16 = 2 4 − 1 15 Put n=5 𝑎5 = 52 25 = 52 − 1 24 Thus, first four terms: 4 9 16 25 , , , 3 8 15 24 2) As the value of n increases, the value of 𝑎𝑛 approaches to 1. Therefore, the limit of the sequence is 1. Question #5 Define whether sequence is arithmetic or geometric and write the n-th term formula 1) 11,17,23,... 2) 5,15,45,... Answer: (1) 11,17,23,... 17-11=6 , 23-17=6 , thus, r=6 and the sequence is arithmetic n-th term: a+(n-1)d =11+(n-1)6 =11+6n-6 =6n+5 (n-th term formula) (2) 5,15,45,... 15 5 = 3 and 45 15 = 3, thus, r=3 and the sequence is geometric n-th term: 𝑎𝑟 𝑛−1 = 5 × 3𝑛−1 (n-th term formula) Question #6 Define whether sequence is arithmetic or geometric and write the n-th term formula 1) 5,8,11,... 2) 5,20,80,... Answer: (1) Given 5,8,11,... First tern a=5 Now, 11-8=8-5 Common difference d=3 Thus, the sequence is arithmetic And the n-th term formula: 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = 5 + (𝑛 − 1)3 𝑎𝑛 = 3𝑛 + 2 Question #7 Determine whether the sequences is convergent or divergent 𝑎𝑛 = ln ( 𝑎𝑛 = 2𝑛2 − 𝑛 + 1 ) 2𝑛2 + 10𝑛 − 7 2𝑛 𝑛! Answer: 1) 𝑎𝑛 = ln ( 2𝑛2 −𝑛+1 2𝑛2 +10𝑛−7 lim 𝑎𝑛 = lim ln ( 𝑛→∞ 𝑛→∞ ) 2𝑛2 − 𝑛 + 1 ) 2𝑛2 + 10𝑛 − 7 1 1 + ) 𝑛 𝑛2 = lim ln ( ) 10 7 𝑛→∞ 𝑛2 (2 + − 2) 𝑛 𝑛 𝑛2 (2 − 1 1 + 𝑛 𝑛2 = lim ln ( ) 10 7 𝑛→∞ 2+ − 2 𝑛 𝑛 2− Substitute 𝑛 = ∞ = ln ( 1 1 + ∞ ∞) 10 7 2+ − ∞ ∞ 2− 2 = ln ( ) = ln 1 = 0 2 Thus, the sequence converges Question #8 Determine the pattern and find the missing terms in each of the following sequences. 8, 5, 4, 9, 1, 7, ... 2, 5, 10, 17, ...,37,... Answer: 1) Given sequence: 8,5,4,9,1,7, ... If we write the numbers in words (Eight, Five, Four, Nine, One, Seven,....) we can determine that the numbers are arranged in the alphabetical order. The numbers that are left: 2,3,6,0 . Now, write them in words (Two, Three, Six, Zero) and put them in the alphabetical order: Six, Three, Two, Zero. Therefore, the complete sequence is 8,5,4,9,1,7,6,3,2,0 Question #9 Determine the Z-transform for the sequence 𝑥[𝑛] = (−1)𝑛 𝑢[𝑛] Answer: 𝑥[𝑛] = (−1)𝑛 𝑢[𝑛] Apply Z-transform to the sequence ∞ 𝑛 𝑧[(−1) 𝑢[𝑛]] = ∑ (−1)𝑛 𝑢[𝑛]𝑧 𝑛 𝑛=−∞ n=0 𝑛 −1 Hence, 𝑧[(−1)𝑛 𝑢[𝑛]] = ∑∞ 𝑛=−∞(−1) 𝑧 ∞ = ∑ ((−1)𝑛 𝑧 −1 )𝑛 𝑛=−∞ ∞ = ∑ (−𝑧 −1 )𝑛 𝑛=−∞ = 1 − 𝑧 −1 + 𝑧 −2 − 𝑧 −3 + ⋯ 𝑧[(−1)𝑛 𝑢[𝑛]] = 1 1 + 𝑧 −1 Thus, z-transform of (−1)𝑛 𝑢[𝑛] 𝑧[(−1)𝑛 𝑢[𝑛]] = 1 (1 + 𝑧 −1 ) |𝑧| > 1 The z-transform of (−1)𝑛 𝑢[𝑛] has zero at z=0 and pole at z=-1 Question #10 Determine the Z-transform for the sequence 1 𝑛−2 𝑥[𝑛] = ( ) 𝑢[𝑛 − 2] 3 Answer: 1 𝑛−2 𝑥[𝑛] = ( ) 𝑢[𝑛 − 2] 3 1, n  2 0, n  2 𝑢[𝑛 − 2] =  Calculate the z-transform of the sequence ∞ 1 𝑛−2 𝑋(𝑧) = ∑ (( ) 𝑢(𝑛 − 2)) 𝑧 −𝑛 3 𝑛=−∞ 1 𝑛−2 = ∑∞ 𝑛=−∞ ( ) 3 𝑧 −𝑛 (Putting the value of u(n-2) ) Let 𝑛1 = 𝑛 − 2 in the second term ∞ 1 𝑛1 𝑋(𝑧) = ∑ ( ) × 𝑧 −𝑛1−2 3 𝑛=−∞ = 𝑧 −2 𝑧 −2 1 𝑛1 ∑∞ 𝑛=−∞(3) × 𝑧 −𝑛1 1 −1 𝑛1 ∑∞ 𝑛=−∞(3𝑧 ) Apply the sum of an infinite sequence = 𝑧 −2 ( 1 ) 1 1 − 𝑧 −1 3 1 1 3 3 Since the sequence converges if |𝑧| > , the radius of convergence is |𝑧| > Question #11 Determine whether each of the following sequences with the given n-th term is convergent or divergent. Find the limit of those sequences that converge. 1) 𝑎𝑛 = sin−1 ( 2) 𝑎𝑛 = 1 √2 cos 1 𝑛 ) 2𝑛+2 +5 3𝑛−1 1 3) 𝑎𝑛 = (1 + 𝑛)2𝑛 Answer: 1) 𝑎𝑛 = sin−1 ( 1 √2 cos 1 𝑛 ) ∞ 1 1 ∑ sin−1 ( cos ) 𝑛 √2 𝑛=1 Apply series divergence test lim 𝑎𝑛 ≠ 0, ∑∞ 𝑛=1 𝑎𝑛 diverges 𝑛→∞ 1 1 = lim sin−1 ( cos ) 𝑛→∞ 𝑛 √2 = 1 lim sin−1 (cos √2 𝑛→∞ 1 ) 𝑛 Apply limit = sin−1 ( = sin−1 ( = sin−1 ( 1 √2 1 √2 1 √2 cos 1 ) ∞ cos(0)) [cos(0) = 1] ) = π ≠0 4 Thus, 𝑎𝑛 diverges 2) 𝑎𝑛 = 2𝑛+2 +5 3𝑛−1 Now, ∑∞ 𝑛=0 ∞ 2𝑛+2 +5 3𝑛−1 = ∑∞ 𝑛=0 ( 2𝑛+2 3𝑛−1 + 5 3𝑛−1 ) ∞ = ∑( 𝑛=0 2𝑛+2 5 + ∑ 𝑛−1 ) 𝑛−1 3 3 𝑛=0 Apply ratio test 𝑎𝑛 = 2𝑛+2 2𝑛+3 , 𝑎𝑛+1 = 𝑛 𝑛−1 3 3 𝑎𝑛 = 5 5 ,𝑎 = 3𝑛−1 𝑛+1 3𝑛  lim | n → an+1 2n+3 3n  3−1 |= lim | n  n 2 | n → an 3 2 2 1 2 3 3 = 2 × = < 1, converges Similarly, 5 3n −1 lim | n  | n → 3 5 1 = < 1, converges 3 Thus, the sequence converges Question #12 Determine whether the sequence converges or diverges 𝑘 𝑛 𝑎𝑛 = (1 + ) 𝑛 Answer: Given 𝑘 𝑛 𝑎𝑛 = (1 + ) 𝑛 The sequence is infinite Calculate the limit 𝑘 𝑛 lim 𝑎𝑛 = lim (1 + ) 𝑛→∞ 𝑛→∞ 𝑛 𝑘 lim (1+ −1)𝑛 𝑛 = 𝑒 𝑛→∞ = 𝑒𝑘 Therefore, the sequence converges. Question #13 Determine whether the sequence converges or diverges 𝑎𝑛 = ln(𝑛3 ) 2𝑛 Answer: Given 𝑎𝑛 = ln(𝑛3 ) 2𝑛 We have to find lim 𝑎𝑛 𝑛→∞ lim 𝑎𝑛 = lim 𝑛→∞ 𝑛→∞ 3 log 𝑛 ∞ = 𝑛 ∞ Use l'Hospital Rule lim 𝑎𝑛 = lim 𝑛→∞ 𝑛→∞ 3 𝑑(log 𝑛) 𝑑𝑛 𝑑(𝑛) 𝑑𝑛 3 3 𝑛 lim 𝑎𝑛 = lim = 𝑛→∞ 𝑛→∞ 1 ∞ lim 𝑎𝑛 = 0 𝑛→∞ The sequence is convergent Question #14 Determine whether the sequence is arithmetic or geometric, and write the formula for the n-th term 5,12,19,... 4,12,36,... Answer: 1) 5,12,19,... 𝑑 = 𝑎2 − 𝑎1 = 12 − 5 = 7 𝑎 = 𝑎3 − 𝑎2 = 19 − 12 = 7 Thus, the sequence is arithmetic 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 5 + (𝑛 − 1)7 2) 4,12,36,... 𝑑= 𝑎2 12 = =3 𝑎1 4 𝑑= 𝑎3 36 = = 3𝐾 𝑎2 12 Thus, the sequence is geometric 𝑎𝑛 = 𝑎𝑟 𝑛−1 = 4 × 3𝑛−1 Question #15 Determine whether the sequences are arithmetic or geometric and write the n-th term formula 6,24,96,... 15,17,19,... Answer: a) First term: 6 = 6 × 4 Second term: 24 = 6 × 4 Third term: 96 = 6 × 4 Thus, it is a geometric sequence. Now, the n-th term formula: 𝑎𝑛 = 6 × 4𝑛−1 , where n=1,2,3,... b) First term: 15=13+2 Second term: 17 = 13 + 2 × 2 Third term: 19 = 13 + 2 × 3 Hence, this is an arithmetic sequence The n-th term formula: 𝑎𝑛 = 13 + 2𝑛 n=1,2,3... Question #16 Determine whether the sequences are arithmetic or geometric and write the n-th term formula 6,24,96,... 7,13,19,... Answer: 1) Given 6,24,96,... 24 96 = =4 6 24 Thus, it's a geometric sequence And r=4 a=6 𝑎𝑛 = 𝑎𝑟 𝑛−1 = 6(4𝑛−1 ) 2) Given 7,13,19,... 13-7=19-13=6 Thus, it's an arithmetic sequence And a=7 , d=6 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 = 7 + (𝑛 − 1)6 = 1 + 6𝑛 Question #17 Determine Z-transform for the sequence 𝑥[𝑛] = δ[𝑛 + 5] Answer: Given 𝑥[𝑛] = δ[𝑛 + 5] ∞ 𝑥[𝑧] = ∑ 𝑥[𝑛]𝑧 −𝑛 𝑛=−∞ 𝑥[𝑛] = δ[𝑛 + 5] ∞ 𝑥[𝑧] = ∑ δ[𝑛 + 5]𝑧 −𝑛 𝑛=−∞ ∞ = ∑ δ[𝑛 + 5]𝑧 5 𝑛=−∞ 𝑥[𝑧] = 𝑧 5 Z-transform ROC - entire Z-plane exept \pm\infty Since ROC includes z=1 , forier transform exist Question #18 Determine Z-transform for the sequence 1 𝑛 𝑥[𝑛] = ( ) 𝑢[3 − 𝑛] 4 Answer: 1 𝑛 𝑥[𝑛] = ( ) 𝑢[3 − 𝑛] 4 Apply Z transform of x[n] ∞ 𝑍(𝑧) = ∑ 𝑥[𝑛]𝑧 −𝑛 𝑛=−∞ ∞ 1 𝑛 = ∑ ( ) 𝑢[3 − 𝑛]𝑥 −𝑛 4 𝑛=−∞ 3 1 𝑛 = ∑ ( ) 𝑧 −𝑛 4 𝑛=−∞ 3 𝑛 1 = ∑ ( 𝑧 −1 ) 4 𝑛=−∞ 3 2 1 −1 1 1 1 1 = ( 𝑧 −1 ) + ( 𝑧 −1 ) + ( 𝑧 −1 ) + 1 + ( 𝑧 −1 ) + ⋯ 4 4 4 4 3 1 ( 𝑧 −1 ) 4 = −1 1 1 − ( 𝑧 −1 ) 4 1 −3 𝑧 64 = 1 − 4𝑧 Thus, 𝑋(𝑧) = 1 𝑧 −3 64 1−4𝑧 Region of convergence |4𝑧| < 1 ⇒ |𝑧| < ⇒ 1 − 4𝑧 = 0 ⇒ 𝑧 = 1 4 1 4 is pole Question #19 Find formulas that describe the sequences 1) 4,7,10,13,16,19,... 2) 5,7,11,19,35,67,... Answer: 1) 4,7,10,13,16,19,... This is an arithmetic sequence First term a=4 Common difference d=7-4=3 formula for the n-th term 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = 4 + (𝑛 − 1)3 𝑎𝑛 = 4 + 3𝑛 − 3 𝑎𝑛 = 3𝑛 + 1 2) 5,7,11,19,35,67,... It's an arithmetic sequence First term a=5 Common difference d=7-5=2 Formula for the n-th term 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = 5 + (𝑛 − 1)2 𝑎𝑛 = 5 + 2𝑛 − 2 𝑎𝑛 = 2𝑛 + 3 Question #20 Find 𝑎5 and 𝑎𝑛 for the geometric sequences: 𝑎1 = −5, 𝑟 = −7 Answer: We have: 𝑎1 = −5, 𝑟 = −7 First term (𝑎) = 𝑎1 = −5 Common ratio (r)=-7 We know that the n-th term of geometric sequence (𝑎𝑛 ) = 𝑎𝑟 𝑛−1 Hence the fifth term (𝑎5 ) = 𝑎𝑟 5−1 𝑎5 = 𝑎𝑟 4 = (−5)(−7)4 = (−5)(2401) = 12005 The n-th term (𝑎𝑛 ) = 𝑎𝑟 𝑛−1 𝑎𝑛 = (−5)(−7)𝑛−1 = (−1)1+𝑛−1 (5)(7)𝑛−1 = 5(−1)𝑛 (7)𝑛−1 𝑎𝑛 = 5(−1)𝑛 (7)𝑛−1