Questions and Answers #9 Sequences

Questions and Answers Sheet 9 Sequences Question #1 𝑛 Prove that lim 𝑛⇒∞ 2𝑛−1 = 1 2 Answer: Let 𝑆𝑛 be a sequence of real numbers, which is said to be convergent to limit L as n approaches to ∞ if ∀∈> 0∃𝑁 ∈I such that |𝑆𝑛 − 𝐿| <∈, ∀𝑛 ≥ 𝑁 𝑛 Given lim 𝑛⇒∞ 2𝑛−1 Here, 𝑆𝑛 = 𝑛 2𝑛−1 = 1 2 ,𝐿 = 1 2 Thus n 1 − | 2n − 1 2 2n − 1(2n − 1) | | 2(2n − 1) 2n − 2n + 1 | | 2(2n − 1) 1 | | 4n − 2 1   4n − 2 1   4n 1 1  4n  n  =N  4 | S n − L || 1 Select 𝑁 = [ ] 4∈ 1 ∀∈> 0∃𝑁 = [ ] such that | 4∈ Hence, lim 𝑛 𝑛⇒∞ 2𝑛−1 = Question #2 Prove: 𝑛+1 lim ln ( )=0 𝑛 𝑛⇒∞ Answer: 1 2 n 1 − |, n  N 2n − 1 2 Here 𝑎𝑛 = ln ( 𝑛+1 𝑛 ) and a=0 Then, for any given ξ > 0 Let |𝑎𝑛 − 𝑎| < ξ n +1 ) − 0 | n n +1 n +1  ln(   n n n 1 1 1 e +  e 1 +  e  e − 1 n n n n | ln( Thus, by Archimedian Property ∃𝑁 ∈ 𝑁 such that 1 < 𝑒 ξ − 1, ∀𝑛 ≥ 𝑁 𝑛 Therefore, lim 𝑎𝑛 = 𝑎 𝑛⇒∞ 𝑛+1 lim ln ( )=0 𝑛⇒∞ 𝑛 Question #3 3𝑛√5 −2𝑛3 +𝑛 𝑊𝑛 = 6𝑛√7 +4𝑛3 −8 , 𝑌𝑛 = 6𝑛5 +2𝑛2 −1 3𝑛2 (2𝑛3 +1) , 𝑍𝑛 = 1 − sin ( Answer: Given 𝑊𝑛 = 𝑌𝑛 = 3𝑛√5 − 2𝑛3 + 𝑛 6𝑛√7 + 4𝑛3 − 8 6𝑛5 + 2𝑛2 − 1 3𝑛2 (2𝑛3 + 1) 2𝑛π 𝑍𝑛 = 1 − sin ( ) 5 Now, 1) lim 𝑊𝑛 = lim 𝑛⇒∞ 3𝑛√5 − 2𝑛3 + 𝑛 𝑛⇒∞ 6𝑛√7 + 4𝑛3 − 8 2𝑛π 5 ) lim Wn = lim n  n −2n3 + n + 3n 4n3 − 8 + 6n 5 7 1 3n 5 + 3 ) 2 n n lim Wn = lim n  n 8 6n 7 ) n3 (4 − 3 + 3 ) n n n3 (−2 + 1 3n 5 −2 + 2 + 3 n n lim Wn = lim n  n 8 6n 7 4− 3 + 3 n n −2 + 0 + 0 lim Wn = n  4−0+0 1 lim Wn = − n  2 2) lim 𝑌𝑛 = lim 6𝑛5 +2𝑛2 −1 𝑛⇒∞ 3𝑛2 (2𝑛3 +1) 𝑛⇒∞ 2 1 3 − 𝑛5 ) 𝑛 lim 𝑌𝑛 = lim 1 𝑛⇒∞ 𝑛⇒∞ 3𝑛2 𝑛3 (2 + 3 ) 𝑛 𝑛5 (6 + 2 1 − ) 𝑛3 𝑛5 1 3𝑛5 (2 + 3 ) 𝑛 𝑛5 (6 + lim 𝑌𝑛 = lim 𝑛⇒∞ 𝑛⇒∞ 2 1 − ) 𝑛3 𝑛5 1 3 (2 + 3 ) 𝑛 (6 + lim 𝑌𝑛 = lim 𝑛⇒∞ 𝑛⇒∞ lim 𝑌𝑛 = 6+0−0 3(2 + 0) 𝑛⇒∞ lim 𝑌𝑛 = 1 𝑛⇒∞ 3) lim 𝑍𝑛 = lim [1 − sin ( 𝑛⇒∞ 𝑛⇒∞ 2𝑛π 5 )] As we don't know the value of the second term sin ( 2𝑛π 5 ) when 𝑛 ⇒ ∞𝑍𝑛 diverges. Question #4 Show directly from the definition that if (𝑥𝑛 ) and (𝑦𝑛 ) are Cauchy sequences, then (𝑥𝑛 + 𝑦𝑛 ) and (𝑥𝑛 𝑦𝑛 ) are Cauchy sequences. If 𝑥𝑛 ≔ √𝑛, show that (𝑥𝑛 ) satisfies lim|𝑥𝑛+1 – 𝑥𝑛 | = 0, but that it is not a Cauchy sequence. Let p be a given natural number and give an example of a sequence (𝑥𝑛 ) that is not a Cauchy sequence, but that satisfies lim|𝑥𝑛+𝑝 – 𝑥𝑛 | = 0. Answer: 1) We have to show that 𝑥𝑛 + 𝑦𝑛 and 𝑥𝑛 𝑦𝑛 are Cauchy sequences if 𝑥𝑛 and 𝑦𝑛 are Cauchy sequence We know that these sequences are said to be Cauchy sequences if for given t>0 , there exists a positive integer "m", such that |𝑎𝑛 − 𝑎𝑚 | < 𝑡∀𝑛 ≥ 𝑚 First term let you prove that (𝑥𝑛 + 𝑦𝑛 ) is a Cauchy sequence. Given that 𝑥𝑛 and 𝑦𝑛 are CAuchy sequences. Let t>0 be any given number Positive integers: 𝑚1 and 𝑚2 𝑡 |𝑥𝑛 − 𝑥𝑚 | < ∀𝑛 ≥ 𝑚1 2 𝑡 and |𝑦𝑛 − 𝑦𝑚 | < ∀𝑛 ≥ 𝑚2 2 Let 𝑚 = 𝑚𝑎𝑥[𝑚1 , 𝑚2 ] |𝑥𝑛 − 𝑥𝑚 | < 𝑡 𝑡 𝑎𝑛𝑑 |𝑦𝑛 − 𝑦𝑚 | < ∀𝑛 ≥ 𝑚 2 2 ∴ |(𝑥𝑛 + 𝑦𝑛 ) − (𝑥𝑚 + 𝑦𝑚 )| = |(𝑥𝑛 − 𝑥𝑚 ) + (𝑦𝑛 − 𝑦𝑚 )| ≤ |𝑥𝑛 − 𝑥𝑚 | + |𝑦𝑛 − 𝑦𝑚 | < 𝑡 𝑡 + =𝑡 2 2 ∴ |(𝑥𝑛 + 𝑦𝑛 ) − (𝑥𝑚 + 𝑦𝑚 )| < 𝑡∀𝑛 ≥ 𝑚 ∴ (𝑥𝑛 + 𝑦𝑛 ) is a Cauchy sequence Now, prove that (𝑥𝑛 𝑦𝑛 ) is a Cauchy sequence |𝑥𝑛 𝑦𝑛 − 𝑥𝑚 𝑦𝑚 | = |𝑥𝑛 𝑦𝑛 − 𝑥𝑛 𝑦𝑚 + 𝑥𝑛 𝑦𝑚 − 𝑥𝑚 𝑦𝑚 | |𝑥𝑛 (𝑦𝑛 − 𝑦𝑚 ) + 𝑦𝑚 (𝑥𝑛 − 𝑥𝑚 )| ≤ |𝑥𝑛 ||𝑦𝑛 − 𝑦𝑚 | + |𝑥𝑛 − 𝑥𝑚 ||𝑦𝑚 |(1) Since 𝑥𝑛 is a Cauchy sequence and every Cauchy sequence is bounded ∴ 𝑥𝑛 is bounded Therefore, a number k>0 such that |𝑥𝑛 | < 𝑘∀𝑛 ∈ 𝑁 Also 𝑦𝑛 is a Cauchy sequence Therefore, positive integer m, such that |𝑦𝑛 − 𝑦𝑚 | < 𝑡 2𝑘 ∀𝑛 ≥ 𝑚1 (2) Now, same for 𝑥𝑛 Cauchy sequence Positive integer 𝑚2 such that |𝑥𝑛 − 𝑥𝑚 | < Now, let 𝑚 = 𝑚𝑎𝑥[𝑚1 , 𝑚2 ] 𝑡 2|𝑦𝑚 | ∀𝑛 ≥ 𝑚2 |𝑥𝑛 − 𝑥| < 𝑡 2|𝑦𝑛 | and |𝑦𝑛 − 𝑦𝑚 | < 𝑡 2𝑘 ∀𝑛 ≥ 𝑚(3) Using equation (3) and |𝑥𝑛 | < 𝑘 in equation (1) |𝑥𝑛 𝑦𝑛 − 𝑥𝑚 𝑦𝑚 | < 𝑘 × 𝑡 𝑡 + × |𝑦𝑚 |∀𝑛 ≥ 𝑚 2𝑘 2|𝑦𝑚 | ⇒ |𝑥𝑛 𝑦𝑛 − 𝑥𝑚 𝑦𝑚 | < 𝑡∀𝑛 ≥ 𝑚 ∴ (𝑥𝑛 𝑦𝑛 ) is a Cauchy sequence Question #5 There are three sequences. Which of them are arithmetic 2,4,6,8,10,.. 2,4,8,16,32,... a, a+2, a+4, a+6, a+8, ... Answer: 1) 2,4,6,8,10,.. 𝑎1 = 2, 𝑑 = 𝑇2 − 𝑇1 = 2 This sequence is arithmetic Let's check 𝑇5 = 𝑎1 (𝑛 − 1)𝑑 = 2 + (4)2 = 10 2) 2,4,8,16,32,... 2 × 2 = 4 , 4 × 2 = 8, thus, this sequence is geometric 3) a, a+2, a+4, a+6, a+8, ... d=a+2-a=2 This is arithmetic sequence Thus, 1 and 3 sequences are arithmetic Question #6 Use the approach in Gauss's Problem and find the following sums of arithmetic sequences. 293+290+287+284+..+2 Answer: Gauss's approach: Consider the given sequence 𝑎1 + 𝑎2 + 𝑎3 + ⋯ 𝑎𝑛−2 + 𝑎𝑛−1 + 𝑎𝑛 Pair up the terms as 𝑎1 + 𝑎𝑛 , 𝑎2 + 𝑎𝑛−1 , 𝑎3 + 𝑎𝑛−2 , … Each pairwise sum in the sequence will be the same aand there are 𝑛 2 such pairs Thus, we get 𝑛 × (𝑎1 + 𝑎𝑛 ) 2 𝑛 Hence, 𝑎1 + 𝑎2 + 𝑎3 + ⋯ 𝑎𝑛−2 + 𝑎𝑛−1 + 𝑎𝑛 = × (𝑎1 + 𝑎𝑛 ) 2 The common difference of the terms of the sequence is d=290-293=-3 To find n use the formula 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 2=293+(n-1)(-3) 𝑛−1= −291 −3 n=98 Therefore, the sum of the sequence is 𝑛 98 × (𝑎1 + 𝑎𝑛 ) = × (293 + 2) = 14455 2 2 Question #7 Use the approach in Gauss’s Problem and find the sums of following arithmetic sequences (a) 1+2+3+4+...+999 (b) 1+3+5+7+...+999 (c) 8+17+26+35+...+881 (d) 692+685+678+671+...+6 Answer: (a) 1+2+3+4+...+999 Here, first term 𝑎0 = 1, 𝑎1 = 2, 𝑎2 = 3 𝑑 = 𝑎1 − 𝑎0 = 2 − 1 = 1 Thus, it's an arithmetic sequence 𝑎𝑛 = 𝑎0 + (𝑛 − 1)𝑑 999 = 1 + (𝑛 − 1)(1) 998 = 𝑛 − 1 ⇒ 𝑛 = 999 𝑛 Sum of the series: 𝑆𝑛 = [ ]2𝑎 + (𝑛 − 1)𝑑 2 𝑆𝑛 = 999 2 [2 + 998] 𝑆𝑛 = 499500 (b) 1+3+5+7+...+999 𝑎0 = 1, 𝑎1 = 3, 𝑎2 = 5, 𝑎𝑛 = 999 𝑑 = 𝑎1 = 𝑎0 = 3 − 1 = 2 Thus, the sequence is arithmetic 999 = 1 + (𝑛 − 1)2 = 998 =𝑛−1 2 𝑛 = 500 The sum of sequence: 𝑆𝑛 = 500 2 [2 + 499 × 2] 𝑆𝑛 = 250000 (c) 8+17+26+35+...+881 𝑎0 = 8, 𝑎1 = 17, 𝑎𝑛 = 881 𝑑 = 17 − 8 = 9 881 = 8 + (𝑛 − 1)(9) 𝑛−1= 881 − 8 9 𝑛 = 97 The sum of sequence: 𝑆𝑛 = 97 2 [2(8) + 96(9)] 𝑆𝑛 = 42680 Question #8 Use the approach in Gauss’s Problem to find the following sums of arithmetic sequences: 1) 1+2+3+4+....+999 2) 1+3+5+7+....+103 3) 4+9+14+19+....+504 4) 8881+872+863+854+....+8 Answer: 1) Using Gauss take the sum of terms from 1,2,...,998 then add the sum to 999 1+2+3+...+499 998+997+996+...+500 999+999+999+...+999 Thus, sum is 999(499)+999=499500 2) 1+3+5+7+...+103 1+3+5+7+...+51 103+101+99+...+53 104+104+104+104+...+104 Thus, sum is 26(104)=2704 3) 4+9+14+19+...+504 Take the sum of terms from 100 terms then add the 101 term to sum 4+9+14+...+249 499+494+489+...+254 503+503+503+...+503 Thus, sum is 50(503)+504=25654 Question #9 Use the following generating function to create a sequence using n=0, 1, 2, 3, 4 𝑎𝑛 = 2𝑛+1 Answer: Given 𝑎𝑛 = 2𝑛+1 and a=0,1,2,3,4 Put n=0 𝑎0 = 2 𝑛 = 1, 𝑎1 = 4 𝑛 = 2, 𝑎2 = 8 𝑛 = 3, 𝑎3 = 16 𝑛 = 4, 𝑎4 = 32 Thus, the sequence is <2,4,8,16,32> Question #10 We have the sequence {0, 1, 2, 3, 3, 4, 6, 9, 8, ...} for which we have to find the summation of terms to the nth term, 𝐶𝑛 However, the number of terms in the sub sequence, 𝑁 ≠ 𝑛 , the number of terms in the larger sequence. What is a way to represent N in terms of n in order to arrive at a single formula for 𝐶𝑛 ? I've determined that it can be 𝑁 = 𝑛/3 but this is only true for cases where n is a multiple of 3. How do I find those two cases where n isn't a multiple of 3? Answer: 𝑁= (𝑛+𝑖) 3 , where 𝑖 = 0,1,2 Question #11 We have the sequence 31, 27, 23, 19 . How you can get the next three numbers Answer: Given: 31, 27, 23, 19 𝑎1 = 31, 𝑎2 = 27, 𝑎3 = 23, 𝑎4 = 19 Here, 𝑎2 − 𝑎1 = 27 − 31 = −4 𝑎3 − 𝑎2 = 23 − 27 = −4 𝑎4 − 𝑎3 = 19 − 23 = −4 Therefore, the sequence is in aruthmetic progression, where first term 𝑎1 = 31, d-common\ difference=-4 𝑎5 = 𝑎 + (5 − 1)𝑑 𝑎5 = 31 + 4 × (−4) = 31 − 16 = 15 𝑎6 = 𝑎 + (6 − 1)𝑑 𝑎6 = 𝑎 + 5 × (−4) = 31 − 20 = 11 𝑎7 = 𝑎 + (7 − 1)𝑑 𝑎7 = 31 + (6) × (−4) = 31 − 24 = 7 Thus, the next three terms are 15,11,7 Question #12 We know that convergent sequences approximate certain numbers. For example, the sequence 1 1 1 1 4 2 4 8 1 1 , + 2 2 , + + ,... etc approximates the number 1. Are there sequences that approximate sequences? Answer: Yes it is. That's part of the field of Functional Analysis. Within this field of Mathematics, the sequence spaces are studied. Question #13 What is the common ratio of the geometric sequence 135,45,15,5... ? Answer: A geometric sequence should be 𝑎, 𝑎𝑟, 𝑎𝑟 2 , 𝑎𝑟 3 , … … . The geometric sequence's common ratio is 𝑟 = 𝑢(𝑛+1) /𝑢𝑛 Then, we have 𝑟 = 45/135 = 1/3 𝑟 = 15/45 = 1/3 𝑟 = 5/15 = 1/3 Question #14 How to find and write the first four terms of the sequences below with the given general term? (1) 𝑎𝑛 = (−1)(𝑛+1) (𝑛 + 4) (2) 𝑎𝑛 = (−1)(𝑛−1) 𝑛2 Answer: 𝑎𝑛 = (−1)(𝑛+1) (𝑛 + 4) 5,6,7, and 8 are the first four terms in the sequence with the general term (n+4). (−1)(𝑛+1) simply means that terms will be positive when they are odd and negative when they are even. Therefore, 𝑎(𝑛=1−&𝑔𝑡;4) = (−1)(𝑛+1) (𝑛 + 4) = +5, −6, +7, −8 𝑎𝑛 = (−1)(𝑛−1) 𝑛2 The first 4 terms of the sequence with general term 𝑛2 are: 1,4,9,16 (−1)(𝑛−1) also means that odd terms will be positive and even terms will be negative. [Since, (−1)0 = 1] Thus, 𝑎(𝑛=1−&𝑔𝑡;4) = (−1)(𝑛−1) 𝑛2 = +1, −4, +9, −16 Question #15 4 (−1)𝑛 ∗ ((𝑛 + 1)!)/(2𝑛 − 2) ? Evaluate the series 𝑠𝑢𝑚(𝑛=2) Answer: Remember that 𝑘 𝑠𝑢𝑚(𝑛=1) 𝑎𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑘 . In other words, a series 𝑠𝑢𝑚𝑎𝑛 is the partial sum of the sequence 𝑎𝑛 . In our case, the series begins with n=2 and ends with k=4. 𝑎𝑛 = (−1)𝑛 ∗ ((𝑛 + 1)!)/(2𝑛 − 2) Therefore, 4 (−1)𝑛 ∗ ((𝑛 + 1)!)/(2𝑛 − 2) = (−1)2 ∗ (3!)/2 + (−1)3 ∗ (4!)/4 + (−1)4 ∗ (5!)/6 = 3 − 6 + 20 = 17 𝑠𝑢𝑚(𝑛=2)