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Existence of Certain Fundamental Limits

Module by: Lawrence Baggett. E-mail the author

Summary: Nondecreasing, nonincreasing, eventually nondecreasing, and eventually nonincreasing sequences are defined. Convergence is then established for these sequences, and some practice exercises are included.

We have, in the preceding exercises, seen that certain specific sequences converge. It's time to develop some general theory, something that will apply to lots of sequences, and something that will help us actually evaluate limits of certain sequences.

Definition 1:

A sequence {an}{an} of real numbers is called nondecreasing if anan+1anan+1 for all n,n, and it is called nonincreasing if anan+1anan+1 for all n.n. It is called strictly increasing if an<an+1an<an+1 for all n,n, and strictly decreasing if an>an+1an>an+1 for all n.n.

A sequence {an}{an} of real numbers is called eventually nondecreasing if there exists a natural number NN such that anan+1anan+1 for all nN,nN, and it is called eventually nonincreasing if there exists a natural number NN such that anan+1anan+1 for all nN.nN. We make analogous definitions of “eventually strictly increasing” and “eventually strictly decreasing.”

It is ordinarily very difficult to tell whether a given sequence converges or not; and even if we know in theory that a sequence converges, it is still frequently difficult to tell what the limit is. The next theorem is therefore very useful. It is also very fundamental, for it makes explicit use of the existence of a least upper bound.

Theorem 1

Let {an}{an} be a nondecreasing sequence of real numbers. Suppose that the set SS of elements of the sequence {an}{an} is bounded above. Then the sequence {an}{an} is convergent, and the limit LL is given byL=supS=supan.L=supS=supan.

Analogously, if {an}{an} is a nonincreasing sequence that is bounded below, then {an}{an} converges to infan.infan.

Proof

We prove the first statement. The second is done analogously, and we leave it to an exercise. Write LL for the supremum supan.supan. Let ϵϵ be a positive number. By Theorem 1.5, there exists an integer NN such that aN>L-ϵ,aN>L-ϵ, which implies that L-aN<ϵ.L-aN<ϵ. Since {an}{an} is nondecreasing, we then have that anaN>L-ϵanaN>L-ϵ for all nN.nN. Since LL is an upper bound for the entire sequence, we know that LanLan for every n,n, and so we have that

| L - a n | = L - a n L - a N < ϵ | L - a n | = L - a n L - a N < ϵ
(1)

for all nN.nN. This completes the proof of the first assertion.

Exercise 1

  1. Prove the second assertion of the preceding theorem.
  2. Show that Theorem 1 holds for sequences that are eventually nondecreasing or eventually nonincreasing. (Re-read the remark following the definition of the limit of a sequence.)

The next exercise again demonstrates the “denseness” of the rational and irrational numbers in the set RR of all real numbers.

Exercise 2

  1. Let xx be a real number. Prove that there exists a sequence {rn}{rn} of rational numbers such that x=limrn.x=limrn. In fact, show that the sequence {rn}{rn} can be chosen to be nondecreasing. HINT: For example, for each n,n, use (Reference) to choose a rational number rnrn between x-1/nx-1/n and x.x.
  2. Let xx be a real number. Prove that there exists a sequence {r'n}{r'n} of irrational numbers such that x=limrn'.x=limrn'.
  3. Let z=x+iyz=x+iy be a complex number. Prove that there exists a sequence {αn}={βn+iγn}{αn}={βn+iγn} of complex numbers that converges to z,z, such that each βnβn and each γnγn is a rational number.

Exercise 3

Suppose {an}{an} and {bn}{bn} are two convergent sequences, and suppose that liman=aliman=a and limbn=b.limbn=b. Prove that the sequence {an+bn}{an+bn} is convergent and that

lim ( a n + b n ) = a + b . lim ( a n + b n ) = a + b .
(2)

HINT: Use an ϵ/2ϵ/2 argument. That is, choose a natural number N1N1 so that |an-a|<ϵ/2|an-a|<ϵ/2 for all nN1,nN1, and choose a natural number N2N2 so that |bn-b|<ϵ/2|bn-b|<ϵ/2 for all nN2.nN2. Then let NN be the larger of the two numbers N1N1 and N2.N2.

The next theorem establishes the existence of four nontrivial and important limits. This time, the proofs are more tricky. Some clever idea will have to be used before we can tell how to choose the N.N.

Theorem 2

  1. Let zCzC satisfy |z|<1,|z|<1, and define an=zn.an=zn. then the sequence {an}{an} converges to 0.0. We write limzn=0.limzn=0.
  2. Let bb be a fixed positive number greater than 1, and define an=b1/n.an=b1/n. See (Reference). Then liman=1.liman=1. Again, we write limb1/n=1.limb1/n=1.
  3. Let bb be a positive number less than 1. Then limb1/n=1.limb1/n=1.
  4. If an=n1/n,an=n1/n, then liman=limn1/n=1.liman=limn1/n=1.

Proof

We prove parts (1) and (2) and leave the rest of the proof to the exercise that follows. If z=0,z=0, claim (1) is obvious. Assume then that z0,z0, and let ϵ>0ϵ>0 be given. Let w=1/|z|,w=1/|z|, and observe that w>1.w>1. So, we may write w=1+hw=1+h for some positive h.h. (That step is the clever idea for this argument.) Then, using the Binomial Theorem, wn>nh,wn>nh, and so 1/wn<1/(nh).1/wn<1/(nh). See part (a) of (Reference). But then

| z n - 0 | = | z n | = | z | n = ( 1 / w ) n = 1 / w n < 1 / ( n h ) . | z n - 0 | = | z n | = | z | n = ( 1 / w ) n = 1 / w n < 1 / ( n h ) .
(3)

So, if NN is any natural number larger than 1/(ϵh),1/(ϵh), then

| z n - 0 | = | z n | = | z | n < 1 n h 1 N h < ϵ | z n - 0 | = | z n | = | z | n < 1 n h 1 N h < ϵ
(4)

for all nN.nN. This completes the proof of the first assertion of the theorem.

To see part (2), write an=b1/n=1+xn,an=b1/n=1+xn, i.e., xn=b1/n-1,xn=b1/n-1, and observe first that xn>0.xn>0. Indeed, since b>1,b>1, it must be that the nnth root b1/nb1/n is also >1.>1. (Why?) Therefore, xn=b1/n-1>0.xn=b1/n-1>0. (Again, writing b1/nb1/n as 1+xn1+xn is the clever idea.) Now, b=b1/nn=(1+xn)n,b=b1/nn=(1+xn)n, which, again by the Binomial Theorem, implies that b>1+nxn.b>1+nxn. So, xn<(b-1)/n,xn<(b-1)/n, and therefore

| b 1 / n - 1 | = b 1 / n - 1 = x n < b - 1 n < ϵ | b 1 / n - 1 | = b 1 / n - 1 = x n < b - 1 n < ϵ
(5)

whenever n>ϵ/(b-1),n>ϵ/(b-1), and this proves part (2).

Exercise 4

  1. Prove part (3) of the preceding theorem. HINT: For b1,b1, use the following algebraic calculation:
    |b1/n-1|=b1/n|1-(1/b)1/n||1-(1/b)1/n|,|b1/n-1|=b1/n|1-(1/b)1/n||1-(1/b)1/n|,
    (6)
    and then use part (2) as applied to the positive number 1/b.1/b.
  2. Prove part (4) of the preceding theorem. Explain why it does not follow directly from part (2). HINT: Write n1/n=1+hn.n1/n=1+hn. Observe that hn>0.hn>0. Then use the third term of the binomial theorem in the expansion n=(1+hn)n.n=(1+hn)n.
  3. Construct an alternate proof to part (2) of the preceding theorem as follows: Show that the sequence {b1/n}{b1/n} is nonincreasing and bounded below by 1. Deduce, from Theorem 1, that the sequence converges to a number L.L. Now prove that LL must be 1.1.

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