4D7 !"#$%&'()*ŽG"˙ RkBhc Introduction to SequenceskBhc A sequence kBhcasbuSkBhcnkBhc3, is a list of numbers indexed by some index kBhcnkBhcjc. Usually when we talk about sequences we express them as a rule that tells us how to generate the kBhcnopxEkBhc thkBhc sequence element kBhcasbuSkBhcnkBhc from the index value kBhcnkBhc, starting from some kBhc n = nsbuSkBhc0kBhc)". Here are some concrete examples.kBhcasbuSkBhcnkBhc  = carF(kBhc n + 1kBhcnkBhc ; n ³ 2kBhcasbuSkBhcnkBhc  = carF[kBhc1 + (-1)opxEkBhcnkBhc2opxEkBhcnkBhc ; n ³ 0kBhcasbuSkBhcnkBhc  = carFrkBhcnopxEkBhc2kBhc  + 1kBhcnopxEkBhc3kBhc  + 1kBhc ; n ³ 1kBhc0)Once we know the rule for generating the kBhcnopxEkBhc thkBhc’‹ sequence element, we can write out the first few elements of the sequence. For example, the first few terms of the third example above arekBhc,%1, 5/9, 5/14, 17/65, 13/63, 37/217, ÉkBhcÙÒGiven just the sequence of terms written out like this it may be extremely difficult to recover the rule that generated that sequence. Fortunately, in almost every case we will encounter we will have that rule. kBhcThe limit of a sequence kBhcXQThe most fundamental question to ask about a sequence is whether or not it has a kBhc limitkBhc as the index kBhcnkBhc˙ gets large. To answer this question in an informal way, we could look at the first few terms of the sequence and then make a conjecture about where the sequence is going as the index grows. Here are the first ten terms of each of the three sequences showkBhcn above. carF$kBhc3kBhc2kBhc , carF$kBhc4kBhc3kBhc , carF$kBhc5kBhc4kBhc , carF$kBhc6kBhc5kBhc , carF$kBhc7kBhc6kBhc , carF$kBhc8kBhc7kBhc , carF$kBhc9kBhc8kBhc , carF%kBhc 10kBhc9kBhc , carF&kBhc 11kBhc 10kBhc , carF&kBhc 12kBhc 11kBhc , É kBhc 1, kBhc 0, carF$kBhc1kBhc2kBhc , 0, carF$kBhc1kBhc8kBhc , 0, carF%kBhc1kBhc 32kBhc , 0, carF&kBhc1kBhc 128kBhc,kBhc  0kBhc , É kBhc 1, carF$kBhc5kBhc9kBhc , carF%kBhc5kBhc 14kBhc , carF&kBhc 17kBhc 65kBhc , carF&kBhc 13kBhc 63kBhc , carF'kBhc 37kBhc 217kBhc , carF'kBhc 25kBhc 172kBhc , carF'kBhc 65kBhc 513kBhc , carF'kBhc 41kBhc 365kBhc , carF)kBhc 101kBhc 1001kBhc , ÉkBhc˙From examining the first few terms of a sequence we can sometimes make a conjecture about what that sequence of terms is going to do in the limit as the index grows very large. In each of the cases we are looking at here, we can make reasonable guesses abkBhc(!out what happens in the limit as kBhcnkBhc gets large.tmiLkBhc nÏ°carF(kBhc n + 1kBhcnkBhc  = 1tmiLkBhc nÏ°kBhc carF[kBhc1 + (-1)opxEkBhcnkBhc2opxEkBhcnkBhc  = 0tmiLkBhc nÏ°kBhc carFrkBhcnopxEkBhc2kBhc  + 1kBhcnopxEkBhc3kBhc  + 1kBhc  = 0kBhcA formal definitionkBhc–Since we are about to start generating theorems about the behavior of sequences, we have to establish a definition for the limit of a sequence.kBhc DefinitionkBhc! We say that the sequence kBhcasbuSkBhcnkBhc kBhcconverges to a limit kBhcLkBhc& if, given any positive number kBhcékBhc  there is a whole number kBhcNkBhc with the property that VsbA;kBhcasbuSkBhcnkBhc  - LkBhc  < ékBhc for all kBhc n ³ NkBhc.kBhcœ•Proving that a sequence converges directly from the definition is challenging. Here is a simple example. One of our examples earlier was the sequencekBhcasbuSkBhcnkBhc  = carF(kBhc n + 1kBhcnkBhc=6We conjectured that this sequence has a limit of 1 as kBhcnkBhc goes off to kBhc°kBhcB;. Here is how to go about proving that from the definition:kBhcG@Suppose that someone has given you a very small positive number kBhcékBhc. kBhc You want to show that as kBhcnkBhc2+ gets large enough, the difference between kBhcasbuSkBhcnkBhc  and kBhc L = 1kBhc is smaller than kBhcékBhc:VsbA;kBhcasbuSkBhcnkBhc  - LkBhc  = VsbACcarF(kBhc n + 1kBhcnkBhc  - 1kBhc  < éVsbAOcarFCGrapkBhc n + 1kBhc  - nkBhcnkBhc  < écarF$kBhc1kBhcnkBhc  < ékBhc n > carF$kBhc1kBhcé kBhc-&The calculation shows that if we pick kBhcNkBhc/( to be the smallest integer larger than kBhc 1/ékBhc, then kBhc n ³ NkBhc implies that VsbA;kBhcasbuSkBhcnkBhc  - LkBhc  < ékBhc-&. This demonstrates that the sequence kBhcasbuSkBhcnkBhc*# converges and that its limit is 1.5kBhc0)Here is another very useful example. Let kBhcrkBhc81 be any positive number less than 1. The sequence4kBhcasbuSkBhcnkBhc  = ropxEkBhcn3kBhcconverges to 0 as kBhcnkBhc gets large:2VsbA;kBhcasbuSkBhcnkBhc  - LkBhc  = VsbA;kBhcropxEkBhcnkBhc  - 0kBhc  < é4kBhcropxEkBhcnkBhc  < é1kBhc n ln r < ln é0kBhc n > carF*kBhc ln ékBhc ln rkBhcSome helpful theoremskBhc˙Working with the definition and using the definition to prove that a particular sequence converges quickly becomes very challenging as soon as you move beyond the most basic examples. The most effective method to prove that a particular sequence has a limkBhc81it is to use one of the following three theorems.kBhc-&Theorem One: Combinations of SequenceskBhc  Let kBhcasbuSkBhcnkBhc  and kBhcbsbuSkBhcnkBhc$ be convergent sequences with tmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc  = A kBhc andkBhc tmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc = B ­ 0kBhc then tmiLkBhc nÏ°GrapSkBhcasbuSkBhcnkBhc  + bsbuSkBhcnkBhc  = tmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc  + tmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc = A + B! tmiLkBhc nÏ°GrapSkBhcasbuSkBhcnkBhc  - bsbuSkBhcnkBhc  = tmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc  - tmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc = A - B" tmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc  bsbuSkBhcnkBhc  = GrapGtmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc GrapGtmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc  = A B#tmiLkBhc nÏ°kBhc carFTkBhcasbuSkBhcnkBhcbsbuSkBhcnkBhc  = carFŠtmiLkBhc nÏ°kBhc  asbuSkBhcntmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc  = carF$kBhcAkBhcB$kBhcÒËThe most common way we use this theorem will be to Ôbreak downĠ sequences into simpler component parts. Provided that we can ultimately show that those component parts converge, the process is justified.%tmiLkBhc nÏ°kBhc GrapkBhc 1/2opxEkBhcnkBhc  + carFJGrapkBhc 1/3opxEkBhcnkBhc4kBhc  =& tmiLkBhc nÏ°kBhc GrapkBhc 1/2opxEkBhcnkBhc  + tmiLkBhc nÏ°kBhc carFJGrapkBhc 1/3opxEkBhcnkBhc4kBhc  ='tmiLkBhc nÏ°kBhc GrapkBhc 1/2opxEkBhcnkBhc  + carFŠtmiLkBhc nÏ°kBhc GrapkBhc 1/3opxEkBhcntmiLkBhc nÏ°kBhc4kBhc  =(kBhc 0 + carF$kBhc0kBhc4/kBhc;4In the last step we used the fact that the sequence kBhcasbuSkBhcnkBhc  = ropxEkBhcnkBhc  converges to 0 whenever kBhcrkBhc is less than 1.*kBhc#Theorem Two: Squeeze TheoremkBhc  Let kBhcasbuSkBhcnkBhc  and kBhcbsbuSkBhcnkBhc$ be convergent sequences with) tmiLkBhc nÏ°kBhc  asbuSkBhcnkBhc  = L kBhc andkBhc tmiLkBhc nÏ°kBhc  bsbuSkBhcnkBhc  = L,kBhc and suppose further that kBhccsbuSkBhcnkBhc' is a sequence and that for all kBhc n > nsbuSkBhc0kBhc we have6kBhcasbuSkBhcnkBhc  ² csbuSkBhcnkBhc  ² bsbuSkBhcn5kBhcthen the sequence kBhccsbuSkBhcnkBhc converges to kBhcLkBhc.-kBhc@9We can use the squeeze theorem to prove that the sequence7kBhcasbuSkBhcnkBhc  = carF[kBhc1 + (-1)opxEkBhcnkBhc2opxEkBhcn6kBhc.'converges to 0. It is easy to show that.kBhc 0 ² carF[kBhc1 + (-1)opxEkBhcnkBhc2opxEkBhcnkBhc  ² carF<kBhc2kBhc2opxEkBhcnkBhc  = 2 Grap0carF$kBhc1kBhc2opxEkBhcn3kBhc–Both the sequence consisting of all 0s and the sequence on the right converge to 0, so our sequence is forced to converge to 0 along with them.4kBhc(!Theorem Three: Comparison TheoremkBhc Suppose kBhc f(x)kBhc,% is a function with the property that5tmiLkBhc xÏ°kBhcf(x) = L4 kBhc and that kBhcasbuSkBhcnkBhc = f(n)kBhc. Then the sequence kBhcasbuSkBhcnkBhc converges to kBhcLkBhc.3kBhcC MkBhc whenever kBhc n > NkBhc..kBhc2+Sequences that neither converge nor diverge-kBhcC