Infinite Sequences and Series

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1 CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, Symbolically the terms of a sequece are represeted with idexed letters: a 1, a 2, a 3, a 4, a 5, a 6, a 7,..., a,... Sometimes we start a sequece with a 0 (idex zero) istead of a 1. Notatio: the sequece a 1, a 2, a 3,... is also deoted by {a } or {a } =1. Some sequeces ca be defied with a formula, for istace the sequece 1, 3, 5, 7,... of odd positive itegers ca be defied with the formula a = 2 1. A recursive defiitio cosists of defiig the ext term of a sequece as a fuctio of previous terms. For istace the Fiboacci sequece starts with f 1 = 1, f 2 = 1, ad the each subsequet term is the sum of the two previous oes: f = f 1 +f 2 ; hece the sequece is: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, Limits. The it of a sequece is the value to which its terms approach idefiitely as becomes large. We write that the it of a sequece a is L i the followig way: For istace a = L or a L as. 1 = 0, 82

Notatio: the sequece a 1, a 2, a 3,... is also deoted by {a } or {a } =1. Some sequeces ca be defied with a formula, for istace the sequece 1, 3, 5, 7,.

2 4.1. SEQUENCES 83 etc. + 1 = 1, If a sequece has a (fiite) it the it is said to be coverget, otherwise it is diverget. If the sequece becomes arbitrarily large the we write For istace a =. 2 = Theorem. Let f be a fuctio defied i [1, ]. If x f(x) = L ad a = f() for iteger 1 the a = L (i.e., we ca replace the it of a sequece with that of a fuctio.) l Example: Fid. l x Aswer: Accordig to the theorem that it equals x x, where x represets a real (rather tha iteger) variable. But ow we ca use L Hôpital s Rule: hece l x x x = (l x) x (x) Example: Fid r (r > 0). l = 0. 1/x = x 1 = 0, Aswer: This it is the same as that of the expoetial fuctio r x, hece 0 if 0 < r < 1 r = 1 if r = 1 if r > 1

If x f(x) = L ad a = f() for iteger 1 the a = L (i.e., we ca replace the it of a sequece with that of a fuctio.) l Example: Fid.

3 4.1. SEQUENCES Operatios with Limits. If a a ad b b the: (a + b ) a + b. (a b ) a b. ca ca for ay costat c. a b ab. a b a b if b 0. (a ) p a p if p > 0 ad a > 0 for every. Example: Fid Aswer: We divide by 2 o top ad bottom ad operate with its iside the expressio: = = = Squeeze Theorem. If a b c for every 0 ad a = c = L, the b = L. Cosequece: If a = 0 the a = 0. Example: Fid cos. Aswer: We have 1 cos 1, ad 1 0 as, hece by the squeeze theorem Other defiitios. cos = Icreasig, Decreasig, Mootoic. A sequece is icreasig if a +1 > a for every. It is decreasig if a +1 < a for every. It is called mootoic if it is either icreasig or decreasig. Example: Prove that the sequece a = + 1 is decreasig.

If a b c for every 0 ad a = c = L, the b = L. Cosequece: If a = 0 the a = 0. Example: Fid cos. Aswer: We have 1 cos 1, ad 1 0 as, hece by the squeeze theorem 4.1.5. Other defiitios. cos = 0. 4.1.5.1. Icreasig, Decreasig, Mootoic.

4 4.1. SEQUENCES 85 Aswer: a +1 a = a +1 < a for all positive. = 1 ( + 1) < 0, hece Bouded. A sequece is bouded above if there is a umber M such that a M for all. It is bouded below if there is a umber m such that m a for all. It is called just bouded if it is bouded above ad below. Example: Prove that the sequece a = + 1 is bouded. Aswer: It is i fact bouded below because all its terms are positive: a > 0. To prove that it is bouded above ote that a = + 1 = sice 1/ 1 for all positive iteger Mootoic Sequece Theorem. Every bouded mootoic sequece is coverget. For istace, we proved that a = + 1 is bouded ad mootoic, so it must be coverget (i fact +1 1 as ). Next example shows that sometimes i order to fid a it you may eed to make sure that the its exists first. Example: Prove that the followig sequece has a it. Fid it: 2, 2 + 2, ,... Aswer: The sequece ca be defied recursively as a 1 = 2, a +1 = 2 + a for 1. First we will prove by iductio that 0 < a < 2, so the sequece is bouded. We start (base of iductio) by oticig that 0 < a 1 = 2 < 2. Next the iductio step. Assume (iductio hypothesis) that for a give value of it is true that 0 < a < 2. From here we must prove that the same is true for the ext value of, i.e. that 0 < a +1 < 2. I fact (a +1 ) 2 = 2 + (a ) < = 4, hece 0 < a +1 < 4 = 2, q.e.d. So by the iductio priciple all terms of the sequece verify that 0 < a < 2.

Aswer: It is i fact bouded below because all its terms are positive: a > 0. To prove that it is bouded above ote that a = + 1 = 1 + 1 2. sice 1/ 1 for all positive iteger. 4.1.6.

5 Now we prove that a is icreasig: hece a +1 > a SEQUENCES 86 (a +1 ) 2 = 2 + a > a + a = 2a > a a = (a ) 2, Fially, sice the give sequece is bouded ad icreasig, by the mootoic sequece theorem it has a it L. We ca fid it by takig its i the recursive relatio: a +1 = 2 + a. Sice a L ad a +1 L we have: L = 2 + L L 2 = 2 + L L 2 L 2 = 0. That equatio has two solutios, 1 ad 2, but sice the sequece is positive the it caot be egative, hece L = 2. Note that the trick works oly whe we kow for sure that the it exists. For istace if we try to use the same trick with the Fiboacci sequece 1, 1, 2, 3, 5, 8, 13,... (f 1 = 1, f 2 = 1, f = f 1 + f 2 ), callig L the it we get from the recursive relatio that L = L+L, hece L = 0, so we deduce f = 0. But this is wrog, i fact the Fiboacci sequece is diverget.

We ca fid it by takig its i the recursive relatio: a +1 = 2 + a. Sice a L ad a +1 L we have: L = 2 + L L 2 = 2 + L L 2 L 2 = 0.

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008 I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces