Math 101 – SOLUTIONS TO WORKSHEET 22 SEQUENCES 1. Skill 1: expression for sequences (1) For each of the following sequences, write a formula for the general term (a) {1, 2, 3, 4, 5, 6, · · · } Solution: an = n. 1 1 1 , 25 , 36 ,··· (b) 1, 14 , 19 , 16 Solution: an = n12 (c) {3, 7, 11, 15, 19, · · · } Solution: an = 4n − 1. 7 7 7 7 7 (d) 79 , 27 , 81 , 243 , 729 , 3187 ,··· 7 Solution: a = . n+1 n 3 5 3 7 1 9 5 4 5 6 7 8 9 10 (e) 21 , 12 , 38 , 14 , 32 , 32 , 128 , 32 , 512 , 512 · · · = 12 , 24 , 38 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 ,··· Solution: an = 2nn . (f) {1, −1, 1, −1, 1, −1, 1, −1, 1, −1, 1, −1, · · · } Solution: an = (−1)n−1 . 2 5 4 7 6 9 8 11 (g) 0, 38 , 27 , 64 , 125 , 216 , 343 , 512 , 729 , 1000 ,··· n Solution: an = n+(−1) n3 2. Skill 2: limits of sequences (2) Determine ∞ if the sequences is convergent of divergent. If convergent, evaluate the limit. (a) n1 n=1 Solution: limn→∞ n1 = limx→∞ x1 = 0 . n o∞ n (b) n+1 n=1 Solution: limn→∞ (c) n n+1 = limx→∞ ∞ {sin(n)}n=5 1 1 1+ x = 1 1+0 = 1. Solution: ∞ The function oscillates and the sequence is divergent. sin( n1 ) n=1 Solution: Since the sine function is continouous we have limn→∞ sin sin 0 = 0 . (3) Further problems n (a) Does limn→∞ √n+1000 exist? (d) √ √ √ n n n = √n+1000 = √ n1000 −−−−→ ∞. n+1000 1+ n n→∞ 1 √ n q q 1 = = . We can see that n 1 1000 n+1000 + 1000 n + n2 n2 n2 Solution: No. We have Solution: No. 1 n = sin limn→∞ 1 n = √ the denominator tends to 0 so the sequence diverges to ∞. (b) limn→∞ 2nn = Solution: We have limn→∞ 2nn = limx→∞ 2xx = limx→∞ (log12)2x = 0 by l’Hopital. ∞ (c) (Math 103 final, 2014) Consider the sequence {an }n=1 = 1, 0, 21 , 0, 0, 31 , 0, 0, 0, 14 , 0, 0, 0, 0, 51 , · · · . Decide whether limn→∞ an = 0. Solution: Yes, the limit is zero. Date: 2/3/2016, Worksheet by Lior Silberman. This instructional material is excluded from the terms of UBC Policy 81. 1 3. Tool: Squeeze Theorem (4) Determine if the sequences is convergent of divergent. If convergent, evaluate the limit. ∞ 1 n (a) (Final 2013) (−1) sin n n=1 . Solution: For n ≥ 1, sin n1 ≥ 0 so 1 1 1 n − sin ≤ (−1) sin ≤ sin . n n n We have seen in 1(d) that limn→∞ sin n1 = 0 and it follows that limn→∞ − sin − limn→∞ sin n1 = 0 as well. By the squeeze theorem we conclude that 1 n = 0. lim (−1) sin n→∞ n o∞ n sin(n) (why do we have n ≥ 2 here?) (b) (Final 2011) log(n) n=2 Solution: Since limn→∞ log(n) = limx→∞ log(x) = ∞, we have limn→∞ every n we have −1 ≤ sin n ≤ 1 so that 1 sin n 1 − ≤ ≤ . log n log n log n Since limn→∞ − log1 n = − limn→∞ 1 log n 1 log n 1 n = = 0. Also, for = 0 also, we have by the squeeze theorem that sin n = 0. n→∞ log n lim (c) (Math 105 Final 2012) an = 1 + n! sin(n3 ) (n+1)! . Solution: We have (n + 1)! = n!(n + 1) so an = 1 + −1 ≤ sin(n3 ) ≤ 1 so that sin(n3 ) n+1 , and for every n we have sin(n3 ) 1 1 ≤1+ ≤1+ . 1− n+1 n+1 n+1 1 = 1 ± limx→∞ x1 = 1 and it follows from the squeeze theorem that Now limn→∞ 1 ± n+1 lim 1 + n→∞ n! sin(n3 ) = 1. (n + 1)! 2