SEQUENCES AND SERIES
Section 3.3 Monotone Sequences
7. If ðbnÞ is a bounded sequence and limðanÞ ¼0, show that limðanbnÞ ¼0. Explain why Theorem 3.2.3cannotbe used.
8. Explain why the result in equation (3) before Theorem 3.2.4cannotbe used to evaluate the limit of the sequenceðð1þ1=nÞnÞ.
9. Letyn:¼ ffiffiffiffiffiffiffiffiffiffiffi nþ1 p ffiffiffi
pn
forn2N. Show thatð ffiffiffi pn
ynÞconverges. Find the limit.
10. Determine the limits of the following sequences.
(a) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4n2þn
p 2nÞ; (b) ð ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
n2þ5n
p nÞ:
11. Determine the following limits.
(a) lim
ð3pffiffiffinÞ1=2n
; (b) lim
ðnþ1Þ1=lnðnþ1Þ . 12. If 0<a<b;determine lim anþ1þbnþ1
anþbn
: 13. Ifa>0;b>0;show that lim ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ðnþaÞðnþbÞ
p n
¼ ðaþbÞ=2.
14. Use the Squeeze Theorem 3.2.7 to determine the limits of the following,
(a) ðn1=n2Þ; (b)
ðn!Þ1=n2 : 15. Show that ifzn:¼ ðanþbnÞ1=nwhere 0<a<b, then limðznÞ ¼b.
16. Apply Theorem 3.2.11 to the following sequences, wherea,bsatisfy 0<a<1;b>1.
(a) ðanÞ; (b) ðbn=2nÞ;
(c) ðn=bnÞ; (d) ð23n=32nÞ:
17. (a) Give an example of a convergent sequenceðxnÞof positive numbers with limðxnþ1=xnÞ ¼1.
(b) Give an example of a divergent sequence with this property. (Thus, this property cannot be used as a test for convergence.)
18. LetX¼(xn) be a sequence of positive real numbers such that limðxnþ1=xnÞ ¼L>1. Show that Xis not a bounded sequence and hence is not convergent.
19. Discuss the convergence of the following sequences, wherea,bsatisfy 0<a<1;b>1.
(a) ðn2anÞ; (b) ðbn=n2Þ;
(c) ðbn=n!Þ; (d) ðn!=nnÞ:
20. LetðxnÞbe a sequence of positive real numbers such that limðx1n=nÞ ¼L<1. Show that there exists a numberrwith 0<r<1 such that 0<xn<rnfor all sufficiently largen2N. Use this to show that limðxnÞ ¼0.
21. (a) Give an example of a convergent sequence (xn) of positive numbers with limðx1=nn Þ ¼1.
(b) Give an example of a divergent sequence (xn) of positive numbers with limðx1=nn Þ ¼1.
(Thus, this property cannot be used as a test for convergence.)
22. Suppose thatðxnÞis a convergent sequence andðynÞis such that for anye>0 there existsMsuch thatjxnynj<efor allnM. Does it follow that (yn) is convergent?
23. Show that if (xn) and (yn) are convergent sequences, then the sequences (un) and (vn) defined by un:¼maxfxn;yngandvn:¼minfxn;yngare also convergent. (See Exercise 2.2.18.) 24. Show that if ðxnÞ;ðynÞ;ðznÞ are convergent sequences, then the sequence (wn) defined by
wn:¼midfxn;yn;zngis also convergent. (See Exercise 2.2.19.)
(i) We can use Definition 3.1.3 or Theorem 3.1.5 directly. This is often (but not always) difficult to do.
(ii) We can dominatejxnxjby a multiple of the terms in a sequenceðanÞknown to converge to 0, and employ Theorem 3.1.10.
(iii) We can identifyXas a sequence obtained from other sequences that are known to be convergent by taking tails, algebraic combinations, absolute values, or square roots, and employ Theorems 3.1.9, 3.2.3, 3.2.9, or 3.2.10.
(iv) We can ‘‘squeeze’’Xbetween two sequences that converge to the same limit and use Theorem 3.2.7.
(v) We can use the ‘‘ratio test’’ of Theorem 3.2.11.
Except for (iii), all of these methods require that we already know (or at least suspect) the value of the limit, and we then verify that our suspicion is correct.
There are many instances, however, in which there is no obvious candidate for the limit of a sequence, even though a preliminary analysis may suggest that convergence is likely.
In this and the next two sections, we shall establish results that can be used to show a sequence is convergent even though the value of the limit is not known. The method we introduce in this section is more restricted in scope than the methods we give in the next two, but it is much easier to employ. It applies to sequences that are monotone in the following sense.
3.3.1 Definition LetX¼(xn) be a sequence of real numbers. We say thatXisincreasing if it satisfies the inequalities
x1x2 xnxnþ1 : We say thatX isdecreasingif it satisfies the inequalities
x1x2 xnxnþ1 : We say thatX ismonotoneif it is either increasing or decreasing.
The following sequences are increasing:
ð1;2;3;4;. . .;n;. . .Þ; ð1;2;2;3;3;3;. . .Þ;
ða;a2;a3;. . .;an;. . .Þ if a>1: The following sequences are decreasing:
ð1;1=2;1=3;. . .;1=n;. . .Þ; ð1;1=2;1=22;. . .;1=2n1;. . .Þ;
ðb;b2;b3;. . .;bn;. . .Þ if 0<b<1: The following sequences are not monotone:
þ1;1;þ1;. . .;ð1Þnþ1;. . . ;
1;þ2;3;. . .;ð1Þnn. . . The following sequences are not monotone, but they are ‘‘ultimately’’ monotone:
ð7;6;2;1;2;3;4;. . .Þ; ð2;0;1;1=2;1=3;1=4;. . .Þ:
3.3.2 Monotone Convergence Theorem A monotone sequence of real numbers is convergent if and only if it is bounded. Further:
3.3 MONOTONE SEQUENCES 71
(a) If X ¼(xn)is a bounded increasing sequence, then limðxnÞ ¼supfxn:n2Ng:
(b) If Y ¼(yn)is a bounded decreasing sequence, then limðynÞ ¼inffyn:n2Ng:
Proof. It was seen in Theorem 3.2.2 that a convergent sequence must be bounded.
Conversely, letXbe a bounded monotone sequence. ThenXis either increasing or decreasing.
(a) We first treat the case whereX¼(xn) is a bounded, increasing sequence. SinceXis bounded, there exists a real numberMsuch thatxnMfor alln2N. According to the Completeness Property 2.3.6, the supremum x¼supfxn:n2Ng exists inR; we will show thatx¼limðxnÞ.
Ife>0 is given, thenxeis not an upper bound of the setfxn:n2Ng, and hence there existsxKsuch thatxe<xK. The fact thatXis an increasing sequence implies that xK xn whenevernK, so that
xe<xK xnx<xþe for all nK:
Therefore we have
jxnxj<e for all nK:
Since e>0 is arbitrary, we conclude that (xn) converges tox.
(b) IfY ¼(yn) is a bounded decreasing sequence, then it is clear thatX:¼ Y ¼ ðynÞ is a bounded increasing sequence. It was shown in part (a) that limX¼ supfyn:n2Ng, Now limX¼ limY and also, by Exercise 2.4.4(b), we have
supfyn:n2Ng ¼ inffyn:n2Ng:
Therefore limY¼ limX¼inffyn:n2Ng: Q.E.D.
The Monotone Convergence Theorem establishes the existence of the limit of a bounded monotone sequence. It also gives us a way of calculating the limit of the sequence providedwe can evaluate the supremum in case (a), or the infimum in case (b). Sometimes it is difficult to evaluate this supremum (or infimum), but once we know that it exists, it is often possible to evaluate the limit by other methods.
3.3.3 Examples (a) limð1=pffiffiffinÞ ¼0.
It is possible to handle this sequence by using Theorem 3.2.10; however, we shall use the Monotone Convergence Theorem. Clearly 0 is a lower bound for the set f1=pffiffiffiffiffiffin: n2Ng, and it is not difficult to show that 0 is the infimum of the setf1= ffiffiffi
pn
:n2Ng; hence 0¼limð1= ffiffiffi
pn Þ.
On the other hand, once we know thatX:¼ ð1= ffiffiffi pn
Þis bounded and decreasing, we know that it converges to some real number x. Since X¼ ð1= ffiffiffi
pn
Þconverges to x, it follows from Theorem 3.2.3 that XX¼ ð1=nÞ converges to x2. Therefore x2¼0, whencex¼0.
(b) Let hn:¼1þ1=2þ1=3þ þ1=n forn2N.
Sincehnþ1¼hnþ1=ðnþ1Þ>hn, we see thatðhnÞis an increasing sequence. By the Monotone Convergence Theorem 3.3.2, the question of whether the sequence is convergent or not is reduced to the question of whether the sequence is bounded or not. Attempts to use direct numerical calculations to arrive at a conjecture concerning the possible boundedness of the sequence ðhnÞ lead to inconclusive frustration. A computer run will reveal the approximate values hn11:4 for n¼50;000, and hn 12:1 for n¼100;000. Such
numerical facts may lead the casual observer to conclude that the sequence is bounded.
However, the sequence is in fact divergent, which is established by noting that h2n ¼1þ1
2þ 1 3þ1
4
þ þ 1
2n1þ1þ þ 1 2n
>1þ1 2þ 1
4þ1 4
þ þ 1
2nþ þ 1 2n
¼1þ1 2þ1
2þ þ1 2
¼1þn 2:
SinceðhnÞis unbounded, Theorem 3.2.2 implies that it is divergent. (This proves that the infinite series known as the harmonic series diverges. See Example 3.7.6(b) in Section 3.7.)
The termshnincrease extremely slowly. For example, it can be shown that to achieve hn>50 would entail approximately 5.21021 additions, and a normal computer per- forming 400 million additions a second would require more than 400,000 years to perform the calculation (there are 31,536,000 seconds in a year). A supercomputer that can perform more than a trillion additions a second would take more than 164 years to reach that modest goal. And the IBM Roadrunner supercomputer at a speed of a quadrillion operations per
second would take over a year and a half. &
Sequences that are defined inductively must be treated differently. If such a sequence is known to converge, then the value of the limit can sometimes be determined by using the inductive relation.
For example, suppose that convergence has been established for the sequenceðxnÞ defined by
x1¼2; xnþ1¼2þ 1
xn; n2N:
If we letx¼limðxnÞ, then we also havex¼limðxnþ1Þsince the 1-tailðxnþ1Þconverges to the same limit. Further, we see that xn2, so that x6¼0 and xn6¼0 for all n2N. Therefore, we may apply the limit theorems for sequences to obtain
x¼limðxnþ1Þ ¼2þ 1
limðxnÞ¼2þ1 x:
Thus, the limitxis a solution of the quadratic equationx22x1¼0, and sincexmust be positive, we find that the limit of the sequence isx¼1þ ffiffiffi
p2 .
Of course, the issue of convergence must not be ignored or casually assumed. For example, if we assumed the sequenceðynÞdefined byy1 :¼1;ynþ1 :¼2ynþ1 is conver- gent with limity, then we would obtain y¼2yþ1, so thaty¼ 1. Of course, this is absurd.
In the following examples, we employ this method of evaluating limits, but only after carefully establishing convergence using the Monotone Convergence Theorem. Additional examples of this type will be given in Section 3.5.
3.3.4 Examples (a) Let Y¼(yn) be defined inductively byy1:¼1; ynþ1:¼14ð2ynþ3Þ forn1. We shall show that limY ¼3=2.
3.3 MONOTONE SEQUENCES 73
Direct calculation shows thaty2¼5=4. Hence we havey1<y2<2. We show, by Induction, thatyn<2 for alln2N. Indeed, this is true forn¼1; 2. Ifyk<2 holds for somek2N, then
ykþ1¼14ð2ykþ3Þ<14ð4þ3Þ ¼74<2; so thatykþ1 <2. Therefore yn<2 for alln2N.
We now show, by Induction, thatyn<ynþ1for alln2N. The truth of this assertion has been verified for n ¼ 1. Now suppose that yk<ykþ1 for some k; then 2ykþ3
<2ykþ1þ3, whence it follows that
ykþ1¼14ð2ykþ3Þ<14ð2ykþ1þ3Þ ¼ykþ2:
Thusyk<ykþ1 implies thatykþ1<ykþ2. Thereforeyn<ynþ1 for alln2N.
We have shown that the sequenceY ¼ ðynÞis increasing and bounded above by 2. It follows from the Monotone Convergence Theorem thatY converges to a limit that is at most 2. In this case it is not so easy to evaluate limðynÞby calculating supfyn:n2Ng. However, there is another way to evaluate its limit. Sinceynþ1¼14ð2ynþ3Þfor alln2N, thenth term in the 1-tailY1ofYhas a simple algebraic relation to thenth term ofY. Since, by Theorem 3.1.9, we havey:¼limY1¼limY, it therefore follows from Theorem 3.2.3 (why?) that
y¼14ð2yþ3Þ;
from which it follows that y¼3=2.
(b) LetZ¼(zn) be the sequence of real numbers defined byz1:¼1; znþ1:¼ ffiffiffiffiffiffiffi 2zn
p for n2N. We will show that limðznÞ ¼2.
Note thatz1¼1 andz2¼ ffiffiffi p2
; hence 1z1<z2<2. We claim that the sequenceZ is increasing and bounded above by 2. To show this we will show, by Induction, that 1zn<znþ1<2 for alln2N. This fact has been verified forn¼1. Suppose that it is true forn¼k; then 22zk<2zkþ1<4, whence it follows (why?) that
1< ffiffiffi p2
zkþ1¼ ffiffiffiffiffiffiffi 2zk
p <zkþ2¼ ffiffiffiffiffiffiffiffiffiffiffi 2zkþ1
p < ffiffiffi p4
¼2:
[In this last step we have used Example 2.1.13(a).] Hence the validity of the inequality 1zk<zkþ1<2 implies the validity of 1zkþ1<zkþ2<2. Therefore 1zn <
znþ1<2 for alln2N.
Since Z ¼ ðznÞ is a bounded increasing sequence, it follows from the Monotone Convergence Theorem that it converges to a number z:¼supfzng. It may be shown directly that supfzng ¼2, so thatz¼2. Alternatively we may use the method employed in part (a). The relationznþ1¼ ffiffiffiffiffiffiffi
2zn
p gives a relation between thenth term of the 1-tailZ1ofZ and the nth term of Z. By Theorem 3.1.9, we have limZ1¼z¼limZ. Moreover, by Theorems 3.2.3 and 3.2.10, it follows that the limitzmust satisfy the relation
z¼ ffiffiffiffiffi 2z p :
Hencezmust satisfy the equationz2¼2z, which has the rootsz¼0; 2. Since the terms of z¼ ðznÞ all satisfy 1zn2, it follows from Theorem 3.2.6 that we must have
1z2. Thereforez¼2. &
The Calculation of Square Roots
We now give an application of the Monotone Convergence Theorem to the calculation of square roots of positive numbers.
3.3.5 Example Let a > 0; we will construct a sequence (sn) of real numbers that converges to ffiffiffi
pa .
Lets1>0 be arbitrary and definesnþ1:¼12ðsnþa=snÞforn2N. We now show that the sequence (sn) converges to ffiffiffi
pa
. (This process for calculating square roots was known in Mesopotamia before 1500B.C.)
We first show that s2n a for n2. Since sn satisfies the quadratic equation s2n2snþ1snþa¼0, this equation has a real root. Hence the discriminant 4s2nþ14a must be nonnegative; that is,s2nþ1afor n1.
To see that (sn) is ultimately decreasing, we note that forn2 we have snsnþ1¼sn1
2 snþa sn
¼1
2ðs2naÞ sn 0:
Hence,snþ1sn for all n2. The Monotone Convergence Theorem implies that s:¼ limðsnÞexists. Moreover, from Theorem 3.2.3, the limitsmust satisfy the relation
s¼1 2 sþa
s
;
whence it follows (why?) thats¼a=sors2¼a. Thuss¼ ffiffiffi pa
.
For the purposes of calculation, it is often important to have an estimate ofhow rapidly the sequence (sn) converges to ffiffiffi
pa
. As above, we have ffiffiffi pa
sn for alln2, whence it follows thata=sn ffiffiffi
pa
sn. Thus we have 0sn ffiffiffi
pa
sna=sn¼ ðs2naÞ=sn for n2: Using this inequality we can calculate ffiffiffi
pa
to any desired degree of accuracy. &
Euler’s Number
We conclude this section by introducing a sequence that converges to one of the most important ‘‘transcendental’’ numbers in mathematics, second in importance only to p.
3.3.6 Example Leten:¼ ð1þ1=nÞnforn2N. We will now show that the sequenceE¼ (en) is bounded and increasing; hence it is convergent. The limit of this sequence is the famousEuler number e, whose approximate value is 2:718 281 828 459 045. . .;which is taken as the base of the ‘‘natural’’ logarithm.
If we apply the Binomial Theorem, we have en¼ 1þ1
n
n
¼1þn 11
n þnðn1Þ
2! 1
n2þnðn1Þðn2Þ
3! 1
n3 þ þnðn1Þ 21
n! 1
nn:
3.3 MONOTONE SEQUENCES 75
If we divide the powers ofninto the terms in the numerators of the binomial coefficients, we get
en¼1þ1 þ 1 2! 11
n
þ 1 3! 11
n
12 n
þ þ 1 n! 11
n
12 n
1n1 n
: Similarly we have
enþ1¼1þ1 þ 1
2! 1 1
nþ1
þ1
3! 1 1
nþ1
1 2
nþ1
þ þ1
n! 1 1 nþ1
1 2
nþ1
1n1 nþ1
þ 1
ðnþ1Þ! 1 1 nþ1
1 2
nþ1
1 n nþ1
:
Note that the expression forencontainsnþ1 terms, while that forenþ1containsnþ2 terms. Moreover, each term appearing inenis less than or equal to the corresponding term inenþ1, and enþ1 has one more positive term. Therefore we have 2e1<e2 < <
en<enþ1 < ;so that the terms ofEare increasing.
To show that the terms ofEare bounded above, we note that ifp¼1; 2;. . .;n, then ð1p=nÞ<1. Moreover 2p1p! [see 1.2.4(e)] so that 1=p!1=2p1. Therefore, ifn>
1, then we have
2<en<1þ1þ1 2þ 1
22þ þ 1 2n1: Since it can be verified that [see 1.2.4(f)]
1 2þ 1
22þ þ 1
2n1¼1 1 2n1 <1;
we deduce that 2<en<3 for alln2N. The Monotone Convergence Theorem implies that the sequence Econverges to a real number that is between 2 and 3. We define the numbereto be the limit of this sequence.
By refining our estimates we can find closer rational approximations to e, but we cannot evaluate it exactly, since e is an irrational number. However, it is possible to calculateeto as many decimal places as desired. The reader should use a calculator (or a computer) to evaluateen for ‘‘large’’ values ofn. &
Leonhard Euler
Leonhard Euler (1707–1783) was born near Basel, Switzerland. His clergyman father hoped his son would follow him into the ministry, but when Euler entered the University of Basel at age 14, he studied medicine, physics, astronomy, and mathematics as well as theology. His mathematical talent was noticed by Johann Bernoulli, who became his mentor. In 1727, Euler traveled to Russia to join Bernoulli’s son, Daniel, at the new St. Petersburg Academy. There he met and married Katharina Gsell, the daughter of a Swiss artist. During their 40-year marriage, they had 13 children, but only five survived childhood.
In 1741, Euler accepted an offer from Frederick the Great to join the Berlin Academy, where he stayed for 25 years. During this period he wrote landmark books on a relatively new subject called calculus and a steady stream of papers on mathematics and science. In response to a request for instruction in science from the Princess of Anhalt-Dessau, he wrote her nearly 200 letters on science that later became famous in a book titledLetters to a German Princess. When Euler lost vision in one eye, Frederick thereafter referred to him as his mathematical ‘‘cyclops.’’
In 1766, he happily returned to Russia at the invitation of Catherine the Great. His eyesight continued to deteriorate and in 1771 he became totally blind following an eye operation.
Incredibly, his blindness made little impact on his mathematics output, for he wrote several books and over 400 papers while blind. He remained active until the day of his death.
Euler’s productivity was remarkable. He wrote textbooks on physics, algebra, calculus, real and complex analysis, and differential geometry. He also wrote hundreds of papers, many winning prizes. A current edition of his collected works consists of 74 volumes.
Exercises for Section 3.3
1. Letx1:¼8 andxnþ1:¼12xnþ2 forn2N. Show thatðxnÞis bounded and monotone. Find the limit.
2. Letx1>1 andxnþ1:¼21=xnforn2N. show thatðxnÞis bounded and monotone. Find the limit.
3. Let x12 andxnþ1:¼1þ ffiffiffiffiffiffiffiffiffiffiffiffiffi xn1
p forn2N. Show that ðxnÞ is decreasing and bounded below by 2. Find the limit.
4. Letx1:¼1 andxnþ1:¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi 2þxn
p forn2N. Show thatðxnÞconverges and find the limit.
5. Lety1:¼pffiffiffip
, wherep>0, andynþ1:¼pffiffiffiffiffiffiffiffiffiffiffiffiffipþyn
forn2N. Show thatðynÞconverges and find the limit. [Hint:One upper bound is 1þ2pffiffiffip
.]
6. Leta>0 and letz1>0:Defineznþ1:¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi aþzn
p forn2N. Show thatðznÞconverges and find the limit.
7. Letx1:¼a>0 andxnþ1:¼xnþ1=xnforn2N. Determine whetherðxnÞconverges or diverges.
8. LetðanÞbe an increasing sequence,ðbnÞbe a decreasing sequence, and assume thatanbnfor alln2N. Show that limðanÞ limðbnÞ, and thereby deduce the Nested Intervals Property 2.5.2 from the Monotone Convergence Theorem 3.3.2.
9. LetAbe an infinite subset ofRthat is bounded above and letu:¼supA. Show there exists an increasing sequenceðxnÞwithxn2Afor alln2Nsuch thatu¼limðxnÞ.
10. Establish the convergence or the divergence of the sequenceðynÞ, where yn:¼ 1
nþ1þ 1
nþ2þ þ 1
2n for n2N:
11. Letxn:¼1=12þ1=22þ þ1=n2for eachn2N. Prove thatðxnÞis increasing and bounded, and hence converges. [Hint:Note that ifk2, then 1=k21=kðk1Þ ¼1=ðk1Þ 1=k.]
12. Establish the convergence and find the limits of the following sequences.
(a)
ð1þ1=nÞnþ1 ; (b)
ð1þ1=nÞ2n ;
(c) 1þ 1
nþ1
n
; (d)
ð11=nÞn : 13. Use the method in Example 3.3.5 to calculate ffiffiffi
p2
, correct to within 4 decimals.
14. Use the method in Example 3.3.5 to calculate ffiffiffi p5
, correct to within 5 decimals.
15. Calculate the numberenin Example 3.3.6 forn¼2, 4, 8, 16.
16. Use a calculator to computeenforn¼50;n¼100, andn¼1000.
3.3 MONOTONE SEQUENCES 77