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Textbook by: Lawrence Baggett. E-mail the author

# Sequences and Limits

Module by: Lawrence Baggett. E-mail the author

Summary: A definition of a sequence of numbers is followed by the concept of convergence and limits.

Definition 1:

A sequence of real or complex numbers is defined to be a function from the set NN of natural numbers into the setRR or C.C. Instead of referring to such a function as an assignment nf(n),nf(n), we ordinarily use the notation {an},{an},{an}1,{an}1, or {a1,a2,a3,...}.{a1,a2,a3,...}. Here, of course, anan denotes the number f(n).f(n).

REMARK We expand this definition slightly on occasion to make some of our notation more indicative. That is, we sometimes index the terms of a sequence beginning with an integer other than 1. For example, we write {an}0,{an}0,{a0,a1,...},{a0,a1,...}, or even {an}-3.{an}-3.

We give next what is the most significant definition in the whole of mathematical analysis, i.e., what it means for a sequence to converge or to have a limit.

Definition 2:

Let {an}{an} be a sequence of real numbers and let LL be a real number. The sequence {an}{an} is said to converge to L,L, or that LL is the limit of {an}{an}, if the following condition is satisfied. For every positive number ϵ,ϵ, there exists a natural number NN such that if nN,nN, then |an-L|<ϵ.|an-L|<ϵ.

In symbols, we say L=limanL=liman or

L = lim n a n . L = lim n a n .
(1)

We also may write anL.anL.

If a sequence {an}{an} of real or complex numbers converges to a number L,L, we say that the sequence {an}{an} is convergent.

We say that a sequence {an}{an} of real numbers diverges to ++ if for every positive number M,M, there exists a natural number NN such that if nN,nN, then anM.anM. Note that we do not say that such a sequence is convergent.

Similarly, we say that a sequence {an}{an} of real numbers diverges to -- if for every real number M,M, there exists a natural number NN such that if nN,nN, then anM.anM.

The definition of convergence for a sequence {zn}{zn} of complex numbers is exactly the same as for a sequence of real numbers. Thus, let {zn}{zn} be a sequence of complex numbers and let LL be a complex number. The sequence {zn}{zn} is said to converge to L,L, or that LL is the limit of {zn},{zn}, if the following condition is satisfied. For every positive number ϵ,ϵ, there exists a natural number NN such that if nN,nN, then |zn-L|<ϵ.|zn-L|<ϵ.

REMARKS The natural number NN of the preceding definition surely depends on the positive number ϵ.ϵ. If ϵ'ϵ' is a smaller positive number than ϵ,ϵ, then the corresponding N'N' very likely will need to be larger than N.N. Sometimes we will indicate this dependence by writing N(ϵ)N(ϵ) instead of simply N.N. It is always wise to remember that NN depends on ϵ.ϵ. On the other hand, the NN or N(ϵ)N(ϵ) in this definition is not unique. It should be clear that if a natural number NN satisfies this definition, then any larger natural number MM will also satisfy the definition. So, in fact, if there exists one natural number that works, then there exist infinitely many such natural numbers.

It is clear, too, from the definition that whether or not a sequence is convergent only depends on the “tail” of the sequence. Specifically, for any positive integer K,K, the numbers a1,a2,...,aKa1,a2,...,aK can take on any value whatsoever without affecting the convergence of the entire sequence. We are only concerned with anan's for nN,nN, and as soon as NN is chosen to be greater than K,K, the first part of the sequence is irrelevant.

The definition of convergence is given as a fairly complicated sentence, and there are several other ways of saying the same thing. Here are two: For every ϵ>0,ϵ>0, there exists a NN such that, whenever nN,nN,|an-L|<ϵ.|an-L|<ϵ. And, given an ϵ>0,ϵ>0, there exists a NN such that |an-L|<ϵ|an-L|<ϵ for all nn for which nN.nN. It's a good idea to think about these two sentences and convince yourself that they really do “mean” the same thing as the one defining convergence.

It is clear from this definition that we can't check whether a sequence converges or not unless we know the limit value L.L. The whole thrust of this definition has to do with estimating the quantity |an-L|.|an-L|. We will see later that there are ways to tell in advance that a sequence converges without knowing the value of the limit.

## Example 1

Let an=1/n,an=1/n, and let us show that liman=0.liman=0. Given an ϵ>0,ϵ>0, let us choose a NN such that 1/N<ϵ.1/N<ϵ. (How do we know we can find such a N?N?) Now, if nN,nN, then we have

| a n - 0 | = | 1 n | = 1 n 1 N < ϵ , | a n - 0 | = | 1 n | = 1 n 1 N < ϵ ,
(2)

which is exactly what we needed to show to conclude that 0=liman.0=liman.

## Example 2

Let an=(2n+1)/(1-3n),an=(2n+1)/(1-3n), and let L=-2/3.L=-2/3. Let us show that L=liman.L=liman. Indeed, if ϵ>0ϵ>0 is given, we must find a N,N, such that if nNnN then |an+(2/3)|<ϵ.|an+(2/3)|<ϵ. Let us examine the quantity |an+2/3|.|an+2/3|. Maybe we can make some estimates on it, in such a way that it becomes clear how to find the natural number N.N.

| a n + ( 2 / 3 ) | = | 2 n + 1 1 - 3 n + 2 3 | = | 6 n + 3 + 2 - 6 n 3 - 9 n | = | 5 3 - 9 n | = 5 9 n - 3 = 5 6 n + 3 n - 3 5 6 n < 1 n , | a n + ( 2 / 3 ) | = | 2 n + 1 1 - 3 n + 2 3 | = | 6 n + 3 + 2 - 6 n 3 - 9 n | = | 5 3 - 9 n | = 5 9 n - 3 = 5 6 n + 3 n - 3 5 6 n < 1 n ,
(3)

for all n1.n1. Therefore, if NN is an integer for which N>1/ϵ,N>1/ϵ, then

| a n + 2 / 3 | < 1 / n 1 / N < ϵ , | a n + 2 / 3 | < 1 / n 1 / N < ϵ ,
(4)

whenever nN,nN, as desired. (How do we know that there exists a NN which is larger than the number 1/ϵ?1/ϵ?)

## Example 3

Let an=1/n,an=1/n, and let us show that liman=0.liman=0. Given an ϵ>0,ϵ>0, we must find an integer NN that satisfies the requirements of the definition. It's a little trickier this time to choose this N.N. Consider the positive number ϵ2.ϵ2. We know, from Exercise 1.16, that there exists a natural number NN such that 1/N<ϵ2.1/N<ϵ2. Now, if nN,nN, then

| a n - 0 | = 1 n 1 N = 1 N < ϵ 2 = ϵ , | a n - 0 | = 1 n 1 N = 1 N < ϵ 2 = ϵ ,
(5)

which shows that 0=lim1/n.0=lim1/n.

REMARK A good way to attack a limit problem is to immediately examine the quantity |an-L|,|an-L|, which is what we did in Example 2 above. This is the quantity we eventually wish to show is less than ϵϵ when nN,nN, and determining which NN to use is always the hard part. Ordinarily, some algebraic manipulations can be performed on the expression |an-L||an-L| that can help us figure out exactly how to choose N.N. Just know that this process takes some getting used to, so practice!

## Exercise 1

1. Using the basic definition, prove that lim3/(2n+7)=0.lim3/(2n+7)=0.
2. Using the basic definition, prove that lim1/n2=0.lim1/n2=0.
3. Using the basic definition, prove that lim(n2+1)/(n2+100n)=1.lim(n2+1)/(n2+100n)=1. HINT: Use the idea from the remark above; i.e., examine the quantity |an-L|.|an-L|.
4. Again, using the basic definition, prove that
limn+n2in-n2i=-1.limn+n2in-n2i=-1.
(6)
Remember the definition of the absolute value of a complex number.
5. Using the basic definition, prove that
limn3+n2i1-n3i=i.limn3+n2i1-n3i=i.
(7)
6. Let an=(-1)n.an=(-1)n. Prove that 1 is not the limit of the sequence {an}.{an}. HINT: Suppose the sequence {an}{an} does converge to 1. Use ϵ=1,ϵ=1, let NN be the corresponding integer that exists in the definition, satisfying |an-1|<1|an-1|<1 for all nN,nN, and then examine the quantity |an-1||an-1| for various nn's to get a contradiction.

## Exercise 2

1. Let {an}{an} be a sequence of (real or complex) numbers, and let LL be a number. Prove that L=limanL=liman if and only if for every positive integer kk there exists an integer N,N, such that if nNnN then |an-L|<1/k.|an-L|<1/k.
2. Let {cn}{cn} be a sequence of complex numbers, and suppose that cnL.cnL. If cn=an+bnicn=an+bni and L=a+bi,L=a+bi, show that a=limana=liman and b=limbn.b=limbn. Conversely, if a=limana=liman and b=limbn,b=limbn, show that a+bi=lim(an+bni).a+bi=lim(an+bni). That is, a sequence {cn=an+bni}{cn=an+bni} of complex numbers converges if and only if the sequence {an}{an} of the real parts converges and the sequence {bn}{bn} of the imaginary parts converges. HINT: You need to show that, given some hypotheses, certain quantities are less than ϵ.ϵ. Part (c) of (Reference) should be of help.

## Exercise 3

1. Prove that a constant sequence (ancanc) converges to c.c.
2. Prove that the sequence {2n2+11-3n}{2n2+11-3n} diverges to -.-.
3. Prove that the sequence {(-1)n}{(-1)n} does not converge to any number L.L. HINT: Argue by contradiction. Suppose it does converge to a number L.L. Use ϵ=1/2,ϵ=1/2, let NN be the corresponding integer that exists in the definition, and then examine |an-an+1||an-an+1| for nN.nN. Use the following useful add and subtract trick:
|an-an+1|=|an-L+L-an+1||an-L|+|L-an+1|.|an-an+1|=|an-L+L-an+1||an-L|+|L-an+1|.
(8)

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