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MAL 100: Calculus

Lecture Notes

1 Sequences of real numbers

1.1 Real number system

We are familiar with natural numbers and to some extent the rational numbers. While

finding roots of algebraic equations we see that rational numbers are not enough to rep-

resent roots which are not rational numbers. For example draw the graph ofy = x2 2.We see that it cross the x-axis twice. The roots are such that their square is 2, but they

cannot be rational numbers according to the following theorem.

Theorem 1.1.1. Suppose thata0, a1,....,an(n1) are integers such thata0= 0, an= 0and thatr satisfies the equation

anxn +an1x

n1 +....+a1x+a0= 0.

Ifr = pq wherep, qare integers with no common factors andq= 0. Thenqdivides anandp dividesa0.

This theorem tells us that only rational candidates for solutions of the above equationhave the form pq where p divides a0 andqdivides an.

Proof: Since pq satisfies the equation, we have

anpn +an1p

n1q+...+a0qn = 0

i.e.,anpn =q(an1pn1 + ... + a0qn1). This meansqdividesan asp, qhave no common

factors. On the other hand we can also write

a0qn =

p(a

npn1 +a

n1pn2 +...+a

1qn1).

Thuspdivides a0. ///

Now we see that the possible rational roots ofx2 2 = 0 are1, 2. But it is easy tocheck that1, 2 does not satisfyx2 2 = 0. So the roots ofx2 2 = 0 are not rationalnumbers. This means the set of rational numbers has gaps. So the natural question

to ask is: Can we have a number system without these gaps? The answer is yes and

the complete number system with out these gaps is the real line R. We will not look

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into the development ofR as it is not easy to define the real numbers. We assume that

there is a set R, whose elements are called real numbers and R is closed with respect to

addition and multiplication. That is, given anya, b R, the sum a+b and product abalso represent real numbers. Moreover, Rhas an order structureand has no gaps inthe sense that it satisfies the Completeness Axiom(see below).

Let Sbe a non-empty subset ofR. IfScontains a largest element s0, then we call s0

the maximum ofS. IfScontains a smallest element s0, then we call s0 the minimum of

S. IfS is bounded above and Shas least upper bound, then we call it the supremum

ofS. IfSis bounded below andShas greatest lower bound, then we call it as infimum ofS.

Note that any bounded subset of Natural numbers has maximum and minimum.

Completeness Axiom: Every nonempty subset S ofR that is bounded above has a

least upper bound. In other words, sup Sexists and is a real number.

The completeness axiom does not hold for Q. That is, every non-empty subset ofQ

that is bounded above by a rational number need not have rational least upper bound.

For example{r Q : 0r, r2 2}.

Archimedean property: For each x R, there exists a natural number N = N(x)such that x < N.

Proof:Assume by contradiction that this is not true. Then there is no N

Nsuch thatx < N. i.e., x is an upper bound for N. Then, letu be the smallest such bound ofN.

i.e., u = sup IN. Thenu 1 N and u 1 < u. Som N such that u 1 < m butm+ 1 Nand u < m+ 1. This contradictsu = supN.Now it is easy to see the following corollary

Corollary: LetS={1n

:n N}. Then w = infS= 0.Proof: We note thatSis bounded below. Let >0 be an arbitrary positive real number.

By above Archimedean property, there exists n Nsuch that n > 1 . Then we have,

0

w

1

n

< .

Since is arbitrary, we have w = 0. (why?)

Corollary: If y > 0 be a real number, then there exists n = n(y) N such thatn 1y < n.Finally, we have the following density theorem

Theorem 1.1.2. Letx, y are real numbers such thatx < y . Then there exists a rational

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numberq such thatx < q < y.

Proof: W.l.g. assume that x > 0. Now let nN be such that y

x > 1

n

. Now

consider the set

S={m N : mn

> x}.ThenS is non-empty (by Archimedean property). By well -ordering ofN,Shas minimal

element saym0. Thenx < m0n . By the minimality ofm0, we see that

m01n

x. Then,

m0n

x+1n

< x + (y x) =y.

Therefore,

x 0 (given)

there exists a positive integerNsuch thatnN = |an L|< .

Notation. L= limn

an oranL.

Examples 1.2.4.

(i) It is clear that the constant sequence{c}n=1, c R, hasc as its limit.

(ii) Show that limn

1

n= 0.

Solution. Let > 0 be given. In order to show that 1/n approaches 0, we mustshow that there exists an integerN N such that for allnN,

1n 0 = 1n< .

But1/n < n >1/. Thus, if we chooseN N such thatN >1/, then for allnN, 1/n < .

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(iii) Consider the sequence{(1)n+1}n=1. It is intuitively clear that this sequence doesnot have a limit or it does not approach to any real number. We now prove this

by definition. Assume to the contrary, that there exists an L R such that thesequence{(1)n+1}n=1 converges to L. Then for= 12 , there exists anN N suchthat

|(1)n+1 L|< 12

, nN. (1.1)Forn even, (1.1) says

| 1 L|< 12

, nN. (1.2)while forn odd, (1.1) says

|1 L|0 be given. As limn

an= L, there exists N1 Nsuch that

nN1 = |an L|< . (1.4)

Similarly as limn cn= L, there exists N2 N

nN2 = |cn L|< . (1.5)

LetN= max{N1, N2}. Then,L < an (from (1.4)) andcnL + ( from (1.5)). Thus

L < anbncnL+.

Thus|bn L|< for allnN. Hence the proof. ///

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Examples 1.2.7.

(i) Consider the sequencecos nn

n=1

. Then1

n cos n

n 1

n. Hence by Sandwich

theorem limn

cos n

n = 0.

(ii) As0 12n 1n and 1n 0 asn , 12n also converges to 0 by Sandwich theorem.

(iii) (1)n 1n0 as1

n (1)n 1

n 1

n for alln1.

(iv) Ifb >0, then limn

n

b= 1.

Solution. First assume thatb >1. Letan= b1n 1.Asb >1, an> 0 for alln N.

Further,b= (1 +an)

n =n

k=0

nCkakn1 +nan.

Then0an b 1n

. Thusan0, i.e., b 1n 1 asn .Now ifb 0 then limnnx

(1+x)n = 0.

(iii) Ifp >0, then limn

log(n)np

= 0.

Solution. (i) Letan= n1n 1. Then0an1 for alln N. Further,

n= (1 +an)n =

nk=0

nCkakn nC2a2n=

n(n 1)2

a2n.

Thus 0

an

2(n1)

(n

2). As 2(n1)

0 as n

, by Sandwich theorem,

an0, i.e., n 1n 1 asn .(ii) Letk be an integer such thatk > x, k >0. Then forn >2k,

(1 +p)n > nCkpk =

n!

k!(n k)!pk =

pk

k!

k

i=1

[n i+ 1] > nk

2kpk

k!.

Hence,

0< nx

(1 +x)n 2k).

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Asx k nj. Then the sequence{ani}

i=1 is calleda subsequence of{an}.Theorem 1.2.10. If the sequence of real numbers{an}1 , is convergent to L, then anysubsequence of{an} is also convergent to L.Proof. Let{ni}i=1 be a sequence of positive integers such that{ani}i=1 forms a subse-quence of{an}. Let > 0 be given. As{an} converges to L, there exists N N suchthat

|an L|< ,nN.

Choose M N such that niN for iM. Then|ani L|< ,iM.

Hence the proof. ///

Definition 1.2.11. (Bounded sequence): A sequence{an}is said to be bounded above,if there existsM R such thatanM for allnN. Similarly, we say that a sequence{an} is bounded below, if there exists N R such that an N for all n N. Thus asequence{an} is said to be bounded if it is both bounded above and below.

Lemma 1.2.12. Every convergent sequence is bounded.

Proof. Let{an}be a convergent sequence and L= limn

an. Let = 1. Then there exists

N Nsuch that|an L|< 1 for allnN. Further,

|an|=|an L+L| |an L| +L

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1.3 Operations on convergent sequences

Theorem 1.3.1. Let

{an

}1 and

{bn

}1 be two sequences such that lim

n

an = L and

limn

bn= M. Then

(i) limn

(an+bn) =L+M.

(ii) limn

(can) =cL, c R..

(iii) limn

(anbn) =LM.

(iv) limn

anbn

=

L

M ifM= 0.

Proof. (i) Let >0. Since an converges to L there exists N1 Nsuch that

|an L|< /2nN1.

Also, as bn converges to M there exists N2 Nsuch that

|bn M|< /2nN2.

Thus

|(an+bn) (L+M)| |an L| + |bN M|< nN= max{N1.N2}.

(ii) Easy to prove. Hence left as an exercise to the students.

(iii) Let >0. Sincean is a convergent sequence, it is bounded by M1 (say). Also asanconverges toL there existsN1 Nsuch that

|an L|< /2M nN1.

Similarly as bn converges to M there exists N2 Nsuch that

|bn M|< /2M1nN2.

LetN= max{N1, N2}. Then

|anbn LM|=|anbn anM+ anM LM| |an(bn M)| + |M(an L)|=|an||bn M| +M|an L|< /2 +/2 =

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(iv) In order to prove this, it is enough to prove that if lim n an = L, L= 0, thenlimn 1/an = 1/L. Without loss of generality, let us assume thatM > 0. Let

>0 be given. As{an}forms a convergent sequence, it is bounded. Choose N1 Nsuch that an> L/2 for all nN1. Also, as an converges to L, there exists N2 Nsuch that|an L|< L2/2 for all nN2. Let N= max{N1, N2}. Then

nN = 1an

1

L

=|an L||anL| < 2

L2L2

2 =. ///

Examples 1.3.2.

(i) Consider the sequence

5

n2

1

. Then limn

5

n2 = lim

n5 1

n1

n= 5 lim

n

1

n limn

1

n=

5 0 0 = 0.(ii) Consider the sequence

3n2 6n5n2 + 4

1

. Notice that 3n2 6n

5n2 + 4 =

3 6/n5 + 4/n

3/5.

(iii) Similarly one can show that limn

n 1

n

= 1.

1.4 Divergent sequence and Monotone sequences

Definition 1.4.1. Let{an}be a sequence of real numbers. We say that an approachesinfinity or diverges to infinity, if for any real numberM >0, there is a positive integer

Nsuch that

nN = anM.

Ifan approaches infinity, then we write an as n . A similar definition is given for the sequences diverging to . In this case we

writean asn .

Examples 1.4.2.

(i) The sequence{

log(1/n)}1 diverges to

. In order to prove this, for anyM >0,

we must produce aN N such that

log(1/n) eM, nN. ChooseN eM. Then, forthis choice ofN,

log(1/n)

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(ii) Consider the sequence{(1)n+1n}1 . Although this is not a convergent sequence, italso does not approach to or.

(iii) The sequence(1)n is also an example of the previous type.

Definition 1.4.3. If a sequence{an}does not converge to a value inRand also does notdiverge to or, we say that{an}oscillates.Lemma 1.4.4. Let{an}and{bn} be two sequences.

(i) If{an}and{bn} both diverges to, then the sequences{an+bn} and{anbn} alsodiverges to.

(ii) If

{an

}diverges to

and

{bn

} converges then

{an+bn

}diverges to

.

Example 1.4.5. Consider the sequence{n+ 1 n}n=1. We know that

n+ 1 andnboth converges to.But the sequence{n+ 1 n}n=1 converges to 0. To see this,

notice that, for a given >0,

n+ 1 n < if and only if1< 2 + 2n. Thus, ifNis such thatN >

1

42,then for allnN, n+ 1 n < .Thusn+ 1 nconverges

to 0. This example shows that the sequence formed by taking difference of two diverging

sequences may converge.

Definition 1.4.6. A sequence{an} of real numbers is called a nondecreasing sequence ifan

an+1 for alln

N and

{an

} is called a nonincreasing sequence ifan

an+1 for all

n N. A sequence that is nondecreasing or nonincreasing is called a monotone sequence.Examples 1.4.7.

(i) The sequences{1 1/n},{n3} are nondecreasing sequences.

(ii) The sequences{1/n},{1/n2} are nonincreasing sequences.

(iii) The sequences{(1)n},{cos(n3 )},{(1)nn},{ (1)n

n } and{n1/n} are not monotonicsequences.

Lemma 1.4.8. (i) A nondecreasing sequence which is not bounded above diverges to.

(ii) A nonincreasing sequence which is not bounded below diverges to.

Example 1.4.9. Ifb >1, then the sequence{bn}1 diverges to.Theorem 1.4.10.

(i) A nondecreasing sequence which is bounded above is convergent.

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(ii) A nonincreasing sequence which is bounded below is convergent.

Proof. (i) Let

{an

}be a nondecreasing, bounded above sequence and a = sup

nN

an. Since

the sequence is bounded, a R. We claim that a is the limit point of the sequence{an}.Indeed, let >0 be given. Sincea is not an upper bound for{an}, there existsN Nsuch that aN> a . As the sequence is nondecreasing, we have a < aN an for allnN. Also it is clear that ana for all n N. Thus,

a ana+,nN.

Hence the proof.

The proof of (ii) is similar to (i) and is left as an exercise to the students. ///

Examples 1.4.11.

(i) If0< b 0 for all n N, the sequence{bn} is bounded below. Hence, by the abovetheorem, {bn} converges. Let L = lim

nbn. Further, lim

nbn+1 = lim

nbbn =

b limn

bn =b L. Thus the sequence{bn+1} converges to b L. On the other hand,{bn+1} is a subsequence of{bn}. HenceL= b L which impliesL= 0 asb= 1.

(ii) The sequence{(1 + 1/n)n

}

1 is convergent.

Solution. Letan = (1 + 1/n)n =

nk=0

n

k

1

n

k. Fork = 1, 2,...,n, the(k+ 1)th

term in the expansion is

n(n 1)(n 2) (n k+ 1)1 2 k

1

nk =

1

k!

1 1

n

1 2

n

1 k 1n

. (1.6)

Similarly, if we expandan+1, then we obtain(n + 2) terms in the expansion and for

k= 1, 2, 3,..., the(k+ 1)th term is

1

k!

1 1

n+ 1

1 2

n+ 1

1 k 1n+ 1

. (1.7)

It is clear that (1.7) is greater than or equal to (1.6) and hence an an+1 whichimplies that{an}is nondecreasing. Further,

an=(1 + 1/n)n =

nk=0

n

k

1

n

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Thus{an}is a bounded monotone sequence and hence convergent.

Theorem 1.4.12. Every sequence has a monotone subsequence.Proof. Pick xN1 such that xn xN1 for all n > N1. We call suchxN as peak. If weare able to pick infinitely many xNis, then{xNi} is decreasing and we are done. If thereare only finitely many xNs and let xn1 be the last peak. Then we can choose n2 such

that xn2 xn1 . Again xn2 is not a peak. So we can choose xn3 such that xn3 xn2.Proceeding this way, we get a non-decreasing sub-sequence.

The following theorem is Bolzano-Weierstrass theorem. Proof is a consequence of Theorem1.4.12

Theorem 1.4.13. Every bounded sequence has a convergent subsequence.

Theorem 1.4.14. Nested Interval theorem: LetIn= [an, bn], n1 be non-empty closed,bounded intervals such that

I1I2I3...InIn+1...

and limn

(bn an) = 0. Thenn=1In contains precisely one point

Proof. Since {an}, {bn} [a1, b1], {an}, {bn} are bounded sequences. By Bolzano-Weierstrasstheorem, there exists sub sequencesank , bnk anda, bsuch that anka, bnkb. Sinceanis increasing a1 < a2 < ......

a and b1 > b2 > ....

b. It is easy to see that a

b. Also

since 0 = lim an bn= a b, we have a = b.It is easy to show that there is no other point inn=1In.

1.5 Cauchy sequence

Definition 1.5.1. A sequence{an}is called a Cauchy sequence if for any given > 0,there existsN N such thatn, mN = |an am|< .Example 1.5.2. Let{an}be a sequence such that{an}converges to L (say). Let >0 begiven. Then there existsN N such that

|an L|< 2 nN.

Thus ifn, mN, we have

|an am| |an L| + |am L|< 2

+

2=.

Thus{an}is Cauchy.

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Lemma 1.5.3. If{an}is a Cauchy sequence, then{an}is bounded.Proof. Since

{an

}forms a Cauchy sequence, for = 1 there exists N

Nsuch that

|an am|< 1, n, mN.

In particular,

|an aN|0. Choose N1, N2 such that

n, nkN1 = |an ank |< /2

nkN2 = |ank a|< /2.Then

n, nkmax{N, N1} = |an a| |an ank | + ank a|< .Hence the claim. ///

Therefore, we have the following Criterion:

Cauchys Criterion for convergence: A sequence{an} converges if and only if forevery >0, there exists N such that

|an

am

|<

m, n

N.

Problem: Let{an}be defined as a1 = 1, an+1 = 1 + 1an . The show that{an}is Cauchy.Solution: Note that an> 1 andanan1 = an1+ 1 > 2. Then

|an+1 an|=|an1 ananan1

| 12|an an1| 1

2n1|a2 a1|,n2.

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Hence

|am an| |am am1| + |am1 am2| +....+ |an+1 an| |a2 a1|n1

1 , =1

2

So given, >0, we can choose N such that 12N1 < 2 .

Indeed the following holds,

Theorem 1.5.5. Let{an} be a sequence such that|an+1an| < |anan1| for allnN for someN and0< 0

limn a1/nn = limn

an+1an

provided the limit on the right side exists.

Proof. Let >0 be arbitrary. Suppose the second limit exists (say l), then there exists

N Nsuch thatl

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1.6 Limit superior and limit inferior

Definition 1.6.1. Let

{an

}be a bounded sequence. Then limit superior of the sequence

{an}, denoted by lim supn

an, is defined as

limsupn

an:= infkN

supnk

an.

Similarly limit inferior of the sequence{an}, denoted by lim infn

an, is defined as

liminfn

an:= supkN

infnk

an.

Example 1.6.2. (i) Consider the sequence

{an

}=

{0, 1, 0, 1, .....

}.Thenn= sup

{am, m

n}= 1 andn= inf{am, mn}= 0. Therefore, lim infan= 0, limsup an= 1.

(ii) Consider the sequence{an}={12 , 23 , 13 , 34 ,....}. Then for largek

1sup{am, mk} limk 1k

0< inf{am, mk} lim1k

Then by sandwich theorem, we see that lim sup an= 1 and liminfan = 0.

Lemma 1.6.3.

(i) If{an}is a bounded sequence, thenlim supn

anlim infn

an.

(ii) If{an}and{bn}are bounded sequences of real numbers and ifanbn for alln N,then

limsupn

anlim supn

bn

and

lim infn

anlim infn

bn.

(iii) Let{an}and{bn} are bounded sequences of real numbers. Then

lim supn

(an+bn)lim supn

an+ lim supn

bn

and

lim infn

(an+bn)lim infn

an+ liminfn

bn.

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Example 1.6.4. Consider the sequences{(1)n} and{(1)n+1}. Herean = (1)n andbn= (1)n+1. Also lim sup

nan= lim sup

nbn= 1. Butan+ bn= 0 for alln Nand hence

lim supn

(an+bn) = 0. Thus a strict inequality may hold in (iii) the above Lemma.

Theorem 1.6.5. If{an}is a convergent sequence, then

liminfn

an = limn

an= limsupn

an.

Proof. LetL= limn

an. Then given >0 there exists N Nsuch that

|an L|< ,nN.

EquivalentlyL < an< L + ,for all nN. Thus, ifnN,L + is an upper boundfor the set {ak|kN}. Ifk := sup{ak|kn}, then we note thatL < N L +andN+1< L+, ....,n < L+ for all nN (Asn is decreasing). Also an > L , nN = nL , nN. Therefore, lim n= L. Hence lim sup an = L. Similarly, onecan prove the lim infan= L.

///

Theorem 1.6.6. If{an}is a bounded sequence and iflim supn

an= lim infn

an= L, L R,then{an}is a convergent sequence.

Proof. Notice thatlim supn

an = limn

(sup{ak|kn})

and

lim infn

an = limn

(inf{ak|kn}).Given that L = lim sup

nan. Thus for >0, there exists N1 Nsuch that

| sup{an, an+1,...} L|< ,nN1.

This implies

an< L+,nN1 (1.8)Similarly there exists N2 Nsuch that

| inf{an, an+1,...} L|< ,nN2.

This implies

L < an,nN2 (1.9)

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LetN= max{N1, N2}. Then from (1.8)and (1.9) we get

|an L|< , nN.Thus the sequence{an}converges. ///

Examples 1.6.7. limn

1 +

1

n

n=e. Assume thate= limn

nk=0

1k! .

Solution. Letan=n

k=0

1

k! andbn=

1 +

1

n

n. Now,

bn =n

k=0

nCk

1

n

k= 2 +

n

k=2

1

k!

k1

i=1

1 i

n

an.(see(1.6)

This implies

limsupn

bnlimsupn

an= e.

Further, ifnm, then

bn =

1 +

1

n

n=

nk=0

nCk

1

n

k

mk=0

nCk

1

n

k= 2 +

mk=2

1

k!

k1

i=1

1 i

n

.

Keepingm fixed and lettingn

, we get

lim infn

bnmk=0

1

k!

which impliesanliminfn

bn. Hence

e= lim infn

anlim infn

bn.

Finally we have the following more precise version of theorem 1.6.6

Theorem 1.6.8. Let

{an

}be any sequence of nonzero real numbers. Then we have

lim inf

an+1an lim inf|an|1/n lim sup |an|1/n lim sup

an+1an .

Proof. The inequality in the middle is trivial. Now we show the right end inequality. Let

L= limsup |an+1an |. W.l.g assume L 0. Then there exists N Nsuch thatan+1an < L+ nN.

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Then for anyn > N, we can write

|an|= an

an1 an1an1 ...

aN+1aN

|aN| N. Sincelimn a

1/n = 1, we conclude that lim sup |an|1/n (L+). Since is arbitrary, we getthe result. Similarly, we can prove the first inequality.

References[1] Methods of Real Analysis, Chapter 2, R. Goldberg .

[2] Elementary Analysis: The Theory of Calculus, K. A. Ross.

[3] Calculus, G. B. Thomas and R. L. Finney, Pearson .

[4] Calculus, James Stewart, Brooks/Cole Cengage Learning.

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