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Lecture 23 : Sequences Formula for An Sequences A Sequence is a list of numbers written in order. fa1; a2; a3;::: g The sequence may be infinite. The n th term of the sequence is the n th number on the list. On the list above a1 = 1st term, a2 = 2 nd term, a3 = 3 rd term, etc.... I Example In the sequence f1; 2; 3; 4; 5; 6;::: g, we have th a1 = 1; a2 = 2;::: . The n term is given by an = n. Some sequences have patterns, some do not. I Example If I roll a 20 sided die repeatedly, I generate a sequence of numbers, which have no pattern. I Example The sequences f1; 2; 3; 4; 5; 6;::: g and f1; −1; 1; −1; 1;::: g have patterns. Annette Pilkington Lecture 23 : Sequences Formula for an Sometimes we can give a formula for the n th term of a sequence, an = f (n). Example For the sequence f1; 2; 3; 4; 5; 6;::: g, we can give a formula for the n th term. an = n. Example Assuming the following sequences follow the pattern shown, give a formula for the n-th term: I f1; −1; 1; −1; 1;::: g n+1 I nth term = an = (−1) . I {−1=2; 1=3; −1=4; 1=5; −1=6;::: g (−1)n I nth term = an = n+1 . Factorials are commonly used in sequences 0! = 1; 1! = 1; 2! = 2·1; 3! = 3·2·1;:::; n! = n·(n−1)·(n−2)·····1: Example Find a formula for the n th term in the following sequence ( ) 2 4 8 16 32 ; ; ; ; ;:::; a = ; 1 2 6 24 120 n 2n I nth term = an = n! . Annette Pilkington Lecture 23 : Sequences Different ways to represent a sequence Below we show 3 different ways to represent the sequences given: A. n 1 2 3 n o n n o1 n ; ; ;:::; ;::: ; ; an = : 2 3 4 n + 1 n + 1 n=1 n + 1 B. n −3 5 −7 (2n + 1) o ; ; ;:::; (−1)n ;::: ; 3 9 27 3n 1 n n (2n + 1) o n (2n + 1) (−1) ; an = (−1) : 3n n=1 3n C. n e e2 e3 en o n en o1 en ; ; ;:::; ;::: ; ; an = : 1 2 6 n! n! n=1 n! Annette Pilkington Lecture 23 : Sequences Graph of a sequence A sequence is a function from the positive integers to the real numbers, with f (n) = an. We can draw a graph of this function as a set of points in the plane. The points on the graph are (1; a1); (2; a2); (3; a3);:::; (n; an);::: 1 (−1)n 1 n 2n3−1 o Example Graph the sequences f n gn=1 and n3 . n=1 3 H-1Ln 2 n - 1 points Hn, L, n = 1...100 points Hn, L, n = 1...100 n n3 2.0000 0.10 1.9999 0.05 1.9998 20 40 60 80 100 1.9997 -0.05 1.9996 -0.10 20 40 60 80 100 We can see from these pictures that the graphs get closer to a horizontal asymptote as n ! 1, y = 0 on the left and y = 2 on the right. Algebraically (−1)n 2n3−1 this means that as n ! 1, we have n ! 0 and n3 ! 2. Annette Pilkington Lecture 23 : Sequences Limit of a sequence Definition A sequence fang has limit L if we can make the terms an as close as we like to L by taking n sufficiently large. We denote this by lim an = L or an ! L as n ! 1: n!1 If limn!1 an exists (is finite), we say the sequence converges or is convergent. Otherwise, we say the sequence diverges. 1 Graphically: If limn!1 an = L, the graph of the sequence fangn=1 has a unique horizontal asymptote y = L. Equivalent Definition A sequence fang has limit L and we write lim an = L or an ! L as n ! 1 n!1 if for every > 0 there is and integer N with the property that if n > N then jan − Lj < . Annette Pilkington Lecture 23 : Sequences Determining if a sequence is convergent. Using our previous knowledge of limits : Theorem If limx!1 f (x) = L and f (n) = an, where n is an integer, then limn!1 an = L. Example Determine if the following sequences converge or diverge: n 2n − 1 o1 n 2n3 − 1 o1 A: ; B: 2n n=1 n3 n=1 2x −1 1−2−x I A. limx!1 2x = limx!1 1 = 1: 1 n 2n−1 o I Therefore the sequence 2n converges and n=1 2n−1 limn!1 an = limn!1 2n = 1. 2x3−1 2−1=x3 I B. limx!1 x3 = limx!1 1 = 2. 1 n 2n3−1 o I Therefore the sequence n3 converges and n=1 2n3−1 limn!1 an = limn!1 n3 = 2. Annette Pilkington Lecture 23 : Sequences L'Hospital's rule We can use L'Hospital's rule to determine the limit of f (x) if we have an indeterminate form. Example Is the following sequence convergent? n n o1 2n n=1 x 1 I limx!1 2x = (by l'Hospital) limx!1 2x ln 2 = 0: n I Therefore the sequence converges and imn!1 2n = 0: Diverging to 1. limn!1 an = 1 means that for every positive number M, there is an integer N with the property if n > N; then an > M: In this case we say the sequence fang diverges to infinity. Note: If limx!1 f (x) = 1 and f (n) = an, where n is an integer, then limn!1 an = 1. Annette Pilkington Lecture 23 : Sequences Important sequence/limit n 1 Example Show that the sequence fr gn=1, r ≥ 0, converges if 0 ≤ r ≤ 1 and diverges to infinity if r > 1. n x x ln r I limn!1 r = limx!1 r = limx!1 e . 8 0 if r < 1 x ln r < I limx!1 e = 1 if r = 1 : 1 if r > 1 n 1 I Therefore the sequence fr gn=1, r ≥ 0, converges if 0 ≤ r ≤ 1 and diverges to infinity if r > 1. Annette Pilkington Lecture 23 : Sequences Rules of Limits The usual Rules of Limits apply: If fang and fbng are convergent sequences and c is any constant then lim can = c lim an n!1 n!1 lim (an + bn) = lim an + lim bn n!1 n!1 n!1 lim (anbn) = lim an · lim bn n!1 n!1 n!1 lim (an − bn) = lim an − lim bn n!1 n!1 n!1 an limn!1 an lim c = c lim = if lim bn 6= 0 n!1 n!1 bn limn!1 bn n!1 p p h i lim an = lim an if p > 0 and an > 0 n!1 n!1 In fact if limn!1 an = L and f (x) is a continuous function at L, then lim f (an) = f (L): n!1 Annette Pilkington Lecture 23 : Sequences Applying the Rules of Limits Example Determine if the following sequence converges or diverges and if it converges find the limit. r n 2n + 1 1 o1 3 − : n n n=1 q q 3 2n+1 1 3 2n+1 1 I limn!1( n − n ) = limn!1 n − limn!1 n q 3 2n+1 1 I = limn!1 n − limn!1 n q q 3 2x+1 1 3 2+1=x I = limx!1 x − limx!1 x = limx!1 1 − 0 p3 I = 2 q 1 p n 3 2n+1 1 o 3 I Therefore the sequence n − n converges to 2. n=1 Annette Pilkington Lecture 23 : Sequences When there is no f(x) / Squeeze Theorem Note We cannot always find a function f (x) with f (n) = an. The Squeeze Theorem or Sandwich Theorem can also be applied : If an ≤ bn ≤ cn for n ≥ n0 and lim an = lim cn = L; then lim bn = L: n!1 n!1 n!1 1 n 2n o I Example Find the limit of the following sequence n! ; n=1 I Requires a bit of cleverness, because we cannot replace n! by a function x!. 2n I Certainly n! > 0 for all n ≥ 1. So if we can find a sequence fcng with 2n n! ≤ cn for all n ≥ 1 and limn!1 cn = 0, then we can apply the squeeze theorem. 2n 2 2 2 2 2 I Note that n! = 1 · 2 · 3 ····· n−1 · n 2 2n 2 I Since k ≤ 1 if k ≥ 2, we have n! ≤ 2 · n for all n ≥ 2. 2 2n 2 I Since limn!1 2 · n = 0, and 0 ≤ n! ≤ 2 · n for all n ≥ 2, we can conclude 2n that limn!1 n! = 0 using the squeeze theorem. 1 n 2n o I Therefore the sequence n! converges to 0. n=1 Annette Pilkington Lecture 23 : Sequences Alternating Sequences n 0 Theorem If fang is an alternating sequence of the form (−1) an where 0 an > 0, then the alternating sequence converges if and only if limn!1 janj = 0 0 or (for the sequence described above) limn!1 an ! 0. n+1 0 (also true for sequences of form (−1) an or any sequence with infinitely many positive and negative terms) Example Determine if the following sequences converge: n 2n + 1 o1 n 2n + 1 o1 A: (−1)n ; B: (−1)n n2 n=1 n n=1 n 2n+1 I A. an = (−1) n2 . 2n+1 2x+1 (2=x)+(1=x2) I limn!1 janj = limn!1 n2 = limx!1 x2 = limx!1 1 = 0 1 n n 2n+1 o I Therefore the sequence (−1) n2 converges to 0.
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