UNIT 5 SEQUENCES

Structure 5.1 Introduction

Objectives

5.2 Real Sequences Bounded Sequences Monotonic Sequences

5.3 Convergent Sequences 5.4 Criteria for the Convergence of Sequences

Cauchy Sequences

5.5 Algebra of Convergent Sequences 5.6 Summary 5.7 Answers/Hints/Solutions

5.1 INTRODUCTION

In Unjt 2, you were introduced to the structure of the real numbers. In Unit 3, some interesting properties of the system of real numbers were discussed. In addition to these properties, there are several other fascinating features of the real numbers. In this unit, we discuss one such feature. This is related to the problem of obtaining the sum of an infinite 'number of real humbew.

You kAow thnt it is easy to find the sum of a finite number of real numbers. The addition of an infinite number of real numbers, however, poses some problem. Apparently, you may conclude that it is not possible to add an infinite number of real numbers. But an infinite sum i.e. the sum of an infinite number of real numbers is not artificial. Under certain limiting processes, it is possible to give a meaning to an infinite sum of the form

Recall that this is the infinite sum of a Geometrical Progressi~n with first term 1 and 1

common ratio - 2 '

To obtain the infinite sums, we need the notion of a sequence of real numbers, and its convergence to a limit. What is, then, a sequence? What is the meaning of the convergence of a sequence? What is the criteria to determine the convergence of a sequence? We shall try to find answers for these questions. Also we shall discuss a few related concepts such as boundedness and monotoncity of sequences. We shall frequentby use these concepts in the Units 6 and 7 as well as later on in Blocks 3 and 5.

Objectives

After studying this wit, you shauld, therefore, be able to . * define a sequence and its subsequence * discuss a bounded and a monotonic sequence * find the limit of a sequenca, if it exists * verify whether a given sequence is convergent or not

,* use the criteria for the convergence of a sequence and define the Cauchy sequence. I

5*2 REAL SEQUENCES

Very often you use the word"ssquence1 in your daily life in hveral wnys. You talk of 'a . sequence of events' or 'arranging the library books in a sequence' and so on. Intuitively the

>

Sequences and Series idea of a sequence is that of a progression or succession of numbers, e.g. the first, the

second, the third and so on. For example, if you want to evaluate fi up to myiy decimal .... places, you can mange its approximate values as 1.4, 1.4 1, 1.414, 1.4142, and so on.

Thus, intuitively, a sequence of real numbers would mean a successiorn of real numbers x,, x, .........; where x, is the first element, x2 being the second element, ..... and so on. Hence, you may say that a sequence is an ordered collection of numbers. But in Mathematics, we define a sequence as a special type of function in the following way:

Recall from unit 1 that f is a function from a nonempty set A to a nonempty set B, if to each element x E A, there is assigned a unique element f(x) E B.

If a function s has its domain as the set N of natural numbers and range in R, then s is called a sequence. Let us study the following two examples:

EXAMPLE 1: Let a function s: N + R be defined as

Then s(1) = 1, s(2)=3, s(3) = 5 ,..........., and so on.

This ' function s: N+ R is called a sequence.

Let us write s(n) = s, V n E N. Then obviously,

s1 = I , s2 = 3. s3 = 5 ..... and so on.

The set {s, . s2. .......... s, ....... ) forms the range of the sequence s: N+R. The values s,, s,, ........ are called the terms of the sequence. The sequence is generally written as

(sn)nEN Or (s,),O~=, Or (s").

EXAMPLE 2: Let a function s: N+ R be defined as

1 1 1 1 Then s, = 5 ,Is2 = 5' s3 = 8' sa = ,........ and so on.

The sequence s, is given by

EXAMPLE 3: Let s: N+ R be defined by

......... In this case, sl = 1, s2 = 1, s3 = 1, where 1 is the first element of N, 2 is the second element of N, 3 is the third element of N, and so on. Thus in this case the sequence (s,,),,~ is such that sl = ~ ( 1 ) . = 1, is the first term of the sequence,' s, = s(2) = 1 is the second term of the sequence, s3 =s (3) 1 1 is the third term of the sequence, and so on.

In all these examples, we have taken the range as a subset of real numbers. Such sequences are called Real Sequences. A formal definition of a Real Sequence is as follows:

rC

DEFINITION 1: REAL SEQUENCE

A real sequence is a function s from the set N of natural numbers to the set R of real numbers whose values are denoted by Isl, s,, ....... ) o r by (s , , ) , ,~~ . o r by s(n), where s(n) = s, for n = 1, 2, 3, ....... The number s,

\ i s called the nth term of the sequence. -.

We shall use the notation (s,) throughout our discussion. Thus, the sequence in Example 1

is (2n-l), the sequence in Exaqile 2 is (&) and the sequence in Example 3 is ( 1 3 or

(1, 1, I , ...... ). It is important to distinguish between a sequence and its set of values since the validity of many results depends on whether we are working with a sequence or a set, We shall always use parentheses ( ) to denate a sequence and the braces { ] to signify a set. The sequence (s,, s,, sf, ...- ) should not be confused with the set (s,, s2, s ,,..... }. For instance I

(1, 1, 1 ...... ) is a sequence whose first term is 1, second term is 1, Bird term is 1, and so

on, wheras the set { I , 1, 1 , ...} is just the singlctou { I ) . kience to mclke the distillctio~i clear, we some:imes write this sequence as (1") or

L.ei us look at a few more exa~nplcs of Real Sequellces.

EX/BM)EE 4 : (E) The erpsessioi: ( 8 , 2, 3, 4, ...I i s a sequence. This is the sequence (n),,. , or @I=

. (ii) k sequence such as (c, c, c, C, ... ), ~l1ht:re every term is LBle same n a ~ n l ~ e r c, is called a constant sequence. This is the sequence ( c ) ~ ,. (iii) The sequence (-1)" 21as terms (-I, 1, -1, 1 , --1, 1, ... ). EXAMPLE 5 : Let a seqalsalce (sn) be given as

In this case (sn) becomes (1, I, 2, 3, 5, 8 ...., nr, n, m+n, ...). This sequence is called the Fibonacci Sequence given by an eminent Italian mathematician L. Fibonacci 11 175-1250].

It has many fascinating mc! interesting properties. Also, it ha. lot of ~pplications particiilarly in puzzles and riddles in Matllematics. In fact, Fibonacci, was illspired by Hindu-Arabic methods uf calculation. 1-11: fol111d t11i.s scquellce when he was trying to solve the following problem:

"How many pairs of rabbits can be [~roduced fr-ona a single pair in a ycer if d v t ~ y . . I hc nicll~oti 01' spccifjiir~g a

monil~ each pair bcgrt:; a new pair w h i c l ~ from the s c r o ~ ~ d month o u becomcs scqurllcc irl terms of prcvicjusly

producti~e'?" ~ I I I ) W I I tcr~ns of the sequerlce. is called a rrcorsiorr formula.

Now consider the scqucnce ( sl, ) gii'cn by ql =- ( - 1 )" n2. For111 a secguence ( t, 1 w!~ost:

terms are the positive terms of the sequence ( s,, ). Then the telnls of (s l l ) are (--I, 4, -4,

16, -25, 36, -49, 64; ...I slid the te:tnc of i t , ) are (4,16, 36, 64, ...I. Cbviocsly ( i , ) 1s

obpr' a ' by selecting, in ordcr, an infinite number of the dcrllls of (s"). 111 such a case, w t say hat (t,) is a subsequence of the sequerlce (sn ).

Let ( s n ) be a sequealcc. Let n,, n,, n3, ... he natorul numbers such that n, < n2 I I ~ <: ... Then a sequerice

(s,,,, Sn2, S,, ...I = (s,, 3 = ( t, I,., or ( t, 1, is called a subsequence of ( sn ).

In other wordb, a subsequence of a sequence is a sequence obtained by omining some teniis of the original sequence and not disturbing the relative positions of the remaining terms. For instance, (s,, s,, s,, ...) and (s;, s,, so, ... ) are subsequences of the sequence (s1, s2, s,, s4, ...). But (s2, sI, s4, s3.,. ) is not a subsequence of (s,, s,, s,, s,, ...). The sequence (-1)" has two subsequence namely (1, 1, 1, ...) and (-1, -1, -1, ...).

Note that n, < n, .: n, < ....... < nk-, .: n, n,,, .: .,, defines an infinite subset of N, namely, {n,, n,, n,, ...I. Conversely, you can say that every infinite subset of N can be described by

Thus, a subsequence of ( s n ) is a sequerlce obtairled by selecting, in order, an infinite subset (: the terns of the sequence. Now by the following exercise. -- - EXERCISE 1 Which of the fallowing sequences are subsequences of the sequcnce (1, 2, 3, 4, .,.j?

i) ( l , O , I , 0 , 1 , 0 ,...) ii) ( I , 3, 6, 10, 15, ... ) iii) ( 1 , 1 , 1 , 2 , 1 , 3 , 1 , 4 , 1 , 5 , . . . ) .

----

Sequences &Series Since a sequence is a function from N to R and N has the natural order, it makes sense to talk about a sequence being bounded and a sequence being monotonic. Recall from Unit 4, the definitions of bounded and monotonic functions.

DEFINITION 3: BOUNDED SEQUENCE A sequence s: N -+ R is said to be bounded if its range is bounded in R. In other words, a sequence ( s,, ) is bounded if there exists a number K > 0 such that

) s n ( < K f o r n = 1 , 2 , 3 ,... For instance, the sequence in Example 1 is not bounded whereas the sequence in Example 3 is bounded. What about the sequence in Example 2?. Is it bounded or not? Verify it yourself. Again, the sequence in Exercise 1 (i) is bounded, while the sequence in Exercise I (ii) is not bounded. What about the sequence in Exercise 1 (iii) 7

Just as, jn Unit 4, we defined a function which is bounded below only or bounded above only, you can similarly define a sequence which is bounded below only or bounded dbove only or both bounded below and bounded above i.e, bounded.

Let us consider the sequence

Since, for k' 2 2,

We have,

Thus, the sequance (sn) is bounded above by 2. Next, if we consider a sequence (sn), where

(s,Z - 2), then, in view of identity (s,+J2 - 2 = -

4s; , it follows easily that s n ' r 6 V n 2 2 i.e.,

the sequence (sn) is bounded below by 42.

Now, try the following exercise.

EXERCISE 2 Define a sequence which is bounded below ohly or bounded above only. Give a n example for each. Verify whether an Arithmetical Sequence ( a, a+d, a+2d, ... ), d # 0 is bounded or bounded below only or bounded above only.

EXERCISE 3 Examine which of the sequence in Exercise 1 are bounded and unbounded.

Again, recall the notion of monotonic functions as discussed in Unit 4. Since a sequence is a special type of function; therefore, we can say something about a monotonic sequence. First, let us study the following examples:

1 2 - 3 4 1 2 Consider the sequence (-Z-, --;j; ,...). Here s, = $1) = 7, s, = s (2) =

3 $3) = r, and so on. You can see that s, < s, < s, c s, < ..., that is, s(1) < s(2) <

s(3) < ... . In other words, the sequence preserves the order in N. Such a sequence is called a monotonically incresing sequence.

1 1 1 Again, consider the sequence ( 1, - 9 - 9 - 9 ...). Here, the inequalities are reversed 2 3 4 1 1 1 1 > - > -j- > > ...; that is, 2

s(1) > $2) > s(3) > s(4) > ...

In this case the sequence reverses the order in N. Such a sequence is called a monotonically decreasing sequence.

1 Also look at the sequence (1, - a, +, - 7 ...-).

You can see that tl~is sequence neither preserves nor reverses the order in PI. Such a sequence is neither monotonically increasing nor monotonically decreasing.

A sequence which is either monotonically increasing or monotonically decreasing is called a monotonic sequence. We have the formal definition as follows:

DEFINITION 4: MONOTONIC SEQUENCE A sequence ( s,, s,, ..... ) is called a nionotonic sequence if either s, 5 s, 5 s, s , .., or s,> s, 2 s, ,... . In the first case, the sequence is called a nlonoto~~ically increasing sequence, wllile in the second case it is called a monotonically decreasing sequence.

s 2 + 2 Once again, let us look at the sequence (4) : s, = 1, sn,, = % , n 2 1. Since

and, so, in view of the fact that sn 2- 6, for n 2 2, it follows easily that sn - sntl 2 0 i.e., sntl I. sn, foi all n r 1. Hence tlie sequence (sn) is a mo~lotonically decreasing sequence.

I However, if you look at the sequence sn = (1 + +r, then we call show that it is an

monoto~iiciilly increasing sequence. Let us see how. By Binomial Theorem,

In the analogous formula for sntl, every term on the r.h.s. of last equality (except the first-term) increases in value and an additional tenn will appear on the right. Therefore, s,+~ > s,,, 'd n 21.

1 Thus (sn), where sn = (1 4- -) , is an strictly increasing sequence. In fact, this is bounded n as well.

For, observe that

(by (*) above)

As you will see below (Theorem 4), any such sequence has to be convergent. The limit

1 of the convergent sdquence sn = ( 1 i- ---)n is an important number denoted by e. Recall

I that the number e is same as the value of the exponential function e: for x = I I

Now try the following exercises.

EXERCISE 4 I I Which of the following sequences are monotonic? 1 i) ( sinn); ii) ( tann);

iv) (2n+ (-1)")).

EXERCISE 5

i), Show that a subsequence of a monotonic sequence is also monotonic.

ii) Do there exist sequences which are both monotonically increasing and monotonically decreasing?

Sequences

Sequences & Series

Since the number m depends upon the clloice of c > 0. Therefore, someti~nes the number m is dcnoted as mq.

We state a theorem (without proof! which we shall use i11 the next section:

THEOREM I :'Every sequence Ilas a monotonic subsequence. 1 For example, the sequence (T) has a subsequence

which is monotonic, is it monotonically increasing or decreasi~g? 7

5.3 CONVERGENT SEQUENCES - The basic tool of analysis is the notion of a limit and the simplest f o m ~ of a limit is that of the lirnit of a sequence. A real number s is called the limit of a sequei~ce ( sll ) if a large number of values of (s,,) are close to s.

1 For example, consider the sequence ti;-). It is intuitively clear that the tenns of the

sequence "approach" the number 0 as n becomes larger and larger. In other words, we can say that the n"' term of the sequence is 'as close to 0.' as we desire f ~ r

1 "sufficiently large n". See Fig. 1 for the limit of the sequence ($.

1 Alternatively, we can say that the sequence (+ approaches the limit 0 if

be made as small as possible for larger and larger values of n. Note that such a behaviour is not true of the seqrlence (n). Check why?

Thus, we have the following definition of the liniit of a sequence.

DEFINITION 5 : LIMIT OF A SEQUENCE A real number s is said to be a limit of a real sequence (sil) if, for every E > 0, there exists a number rn EN such thqt

Isn - SI < E , for all n > m.

The condition that a sequence (sn) has a limit s is often expressed by saying that the sequence (s,,) tends to a limit s. We say also that a sequence (sn) is convergent if there exists a number s (called the limit of the sequence) such that Isn - sl can be made "as small as possible" for sufficiently large values of n. Note, however, our intuition is not

1 sharp enough to tell if the sequence (n sin T) converges, we, therefore, need a precise definition of the convergence of a sequence, and also soine criteria to determine whether a given sequence converges or not. We shall first define converges of sequence in this section, and later on take up the criteria of convergence of seqiiences in the next section.

DEFINITION 6 : CONVEKGENCE OF A SEQUENCE A real sequence (sn) is said to converge to a real number s (called the limit of the sequence) if, for a given E > 0, there exists a positive integer m such that

IS,,- I] < E, for all n > rn.

We express the above fact in .several ways. We say that

(i) The sequence (sn) converges to a real number s;.or (ii) lim sn = s; or

n-tm

(iii) S,+S asn+oo.

The number s is called the limit of the sequence (sll). The convergence of (s,,) to s can be reviewed in the language of neighbourhood (Unit 3) as follows:

We say that $4; s,, = s if and only if except for finite number of terms, whole of the

sequence is in every neighbourhood of s.

GeC)n~etrir,al]y, sn -> s if, given E > 0, we are able to cut off an initial segment of the sequence, is,, s,, ... sm} such that, every member of the 'tail' sm+,, sn,+ ...a. is in the interval ]s-.E. SSE [. The initial segment that must be cut off depends on ~ . i . e . how close to s are the tail elements.

Figure 2 below illustrate the limit concept graphically. Consider the horizontal strip of width 2~ generated by the lines y = s - E and y = s+e. A given term sn of the sequence ( sn ) lies inside this strip exactly if the inequality Isn - sJ < E hold good. n u s , for a llumber s to be the litnit of the scquellce (s,,), we must be able to specify a point m on the x-axis such that, for all n lying to the right of nl, the corresponding term s,, lie wilhin the horizontal strip.

n>m

Fig. 2

Thus, if (s,,) is a sequence having the number s as a lirni!, we write /&sn= s

(or simply t&sn = s) and say the limit of sn is s as n tends t:. Iariinity or the sequence

(sn) converges to the limit s. Let us look at some examples.

1 EXAMPLE 6: Discuss tile convergence of the sequence (?).

SOLUTION : Intuitively, you can see that !5-$ - 0.

Let us use Definition 6 to justify our intuitive understanding. For this, cons'd i er an arbitrary E > 0 and let us by p find a number rn (depending upon E) such that

As you know that quite often (for example in the proof of the trigotlometric identities), we use the if method i.e. we initially work backward from our desired conclusion, but in the formal proof we must nuke sure that our steps ai-e reversible. In the present example, we want to know how big n must be so that

So, we will operate on this inequality algebraically and try to solve it for n. Thus, we

1 start with -- < E. n2

By multiplying both sides by n2 and dividing both sides by e, we find that we want

If our steps are reversible, then we can see that

1 This suggests that we should take 111 - -..? . Tllus, we proceed as follows: &

1 Let E > 0 and take m = -- . Then 6

Sequenccs & Series

I which proves that the sequence (-,) converges to the limit 0. n

A similar problem is given in the next exercise, which you should solve to check your understanding and computational skill.

EXERCISE 6 1 Use Definition 6 to prove that the sequence ($ converges to the limit 0.

EXAMPLE 7: Test the convergence of the following sequences i) (s,,), where sn = n;

ii) (I, -1, 1, -1, ... ). SOLUTION : (i) Look at the sequence sn = (n). This sequence can not converge. For, if

1 possible, suppose the sequence converges to a limit s. Taking E = , all but finitely

1 1 many elements of s,, should lie in the interval 1s - -, s + - [. But this is clearly not 2 2 so. Thus the sequence (n) does not converge.

1 ii) Suppose the sequence [ (-1)" converges to some number s. Taking E = 7, all

1 , I but finitely many elements of this sequence rrlust lie in the interval 1s - 7, s + -2- [. I 1 That means both 1 and -1 are in 1s - 7 , s It 7 [. But that is impossible because

1 1 the distance between 1 and -I is 2, while the length of the interval Is- 3, s + -2- [ is 1.

Thus the sequence (1, -1, 1, -1, ... ) does not converge.

In above example, we have come across two sequences which are not convergent. Such sequences are called divergent sequences.

Thus a sequence is said to be divergent if it is not convergent. Now we take up the question: Can a sequence converge to two different limits? The answer to this question is no and we prove this in the following form.

THEOREM 2: If a sequence is convergents, then its limit is unique.

PROOF : Suppose that a sequence ( sn) has two distinct limits s and sf. Then s # sf. Let S - s'

us assume, without loss of generality, that s . sf. Take E = - 3 .

Sincr: sn+ s, all but iinitcly many elements of the sequence are in the interval IS-E, S+E [. For the same reason, all but finitely many elements of the seqence are in tlle interval ]st-E, s t + E [. Look at the Fig. 3.

The two intervals ]sf- E, S'+E [ and Is-&, s t & [ are disjoint (why?), and a tail of the sequence cannot be contained in both the intervals. Hence, it is not possible for the sequence (sn) to converge to two different limits. This proves the theorem.

Having defined convergence of sequence, the question that remains to be settled is : How to test the convergence of a ;+:en sequence? Let us, first, obtain necessary conditions for the convergence of a sequence. This condition is provided by the following theorem.

THEOREM 3: Every convergent sequence is boutadcd. Sequences

PROOF : Let (sn ) be a sequence which converges to s. Then, for E > 0, there is number rn E N such that 1 s,, -s / < 6 , for n > m.

With E = 1, we obtain m in N so that

Using triangle inequality (see Unit 3), we see that

Thus, n > r n = . I s , , / < l s / + 1. ChooseK=max { (sit I, ls,l, Is,I, ..... /sml}. Then,

we bave Isn 1 5 K, for all n s N. Hence ( s,, ) is bounded.

From this theorem, i t follows that if a sequence is convergent, then it is always bounded. Is the cowerse true? That is to say that if a sequence is bounded, then, is it convergent? The answer is No. For example the sequence (-1, 1, -1, 1, -1, I.....) is bounded but not convergent, Thus boundedness is a necessary condition for the convergence of a sequence. It is not a sufficient condition, However, for a monotonic sequence, boundedness is both necessary and sufficient, We prove this in the next theorem:

THEOREM 4: A tnonotonic sequence is convergent if and only if it is bounded.

PROOF : We already know that every convergent sequence is bounded. Thus, it is enough to prwe that a rn~notonic and bounded sequrnce is convergent. We shall prove this assertion for a monotonically increasing sequehce. Let ( sn) be a tnonotonically increasing sequence which is bounded above.

Let S denote the 6et (sn : n E N) = {s,, s ,,.,, ), where s, 5 s, S s, ... Since ( s n ) is bounded aboye, therefore, there exists an upper bound for the sequence ( s,, s,, s, ... ). Thus S hrls the lessf upper bound, say u. We slailn that the sequence (sll) converges to u,

For, let F > 0 be some real number. Since u - E is not an upper bound for the set S, therefore, u - E < U, This rneans that there exists some integer m such that sm > u - E. Also, since the sequence (s,,) is an iacreasing, we have

s,,, < s,,, for all n m.

But, sn < p, for all n. Therefore,

3 IS,, - UJ < e,

which shows that the sequel~ce (s,) converges to the limit u.

In case sf a monotonically decreasing sequence, the proof is similar to the one given above. 'So, we leave this as an easy exercise for you.

-- Sequei,ci.s & Sel its EXEBCISE 7

Let (sll) bc a monotonically decrensi~ag sequcnce which is bounded Seloav. Show that the sequence converges to its greatest lower bound.

EXERCISE 8 i) Suppose that the sequence (s,:) converges to s and sn 5 A, for every n. Show that

s 5 A. ii) Supposc the sequence (sn) converges to s and s,, 2 a, for every n. Show that s 2 a.

EXERCISE 9 i) Suppose that the sequence (st) ccnverges to s. Show that every subsequence of (sll)

also converges to s. ii) If a sequence docs not converge, then no subsequence of it caal converge. Is the

statement true? Justify your answer. ---- - - In fact, it is not difficult to see t!lat a sequlenca (fill) converges to s if and only if all its s~absequcnce converges to the mme limit s. However, if a sequence is such that its subsequences converges to different limits, then the sequence will not convergent. For example. the sequepce (s,,), where sn = (- I)", is ona s ~ ~ c h sequence. Because, here (- 1, -. 1, - 2 ,...) and (1, 1, 1 ,...) are its two subsequences converging to - 1 and 1, respecti\leiy. But, chc: sequence ((- 1)") itself is not convergent.

EXERCISE 10 Sappo'ie the sequence (sn) converges to s. Show that the sequence (!s,,() converges to Isl. Give an example to show that the converse of this is not true.

We now state a result (without proof) in tlie follouling theorem which is also referred to as rhe Bulzano-Weierstreas Theorem.

THEOREM 5 : Every bounded sequence of real numbers c o ~ ~ t a i n s a convtrgent subsequence.

We have already talked about the divagent sequences.

You have encountrred two examples of divergent sequences in Example 7, namely (1, 2, 3, 4 ,... ) and (1, -1, 1, -1, ...). The sequencc (1, 2, 3, 4, ....) is said to be divergent sequence because its terms become "hvo big", whereas (I, -1, 1, -1, ....) is divergent because it has no limit point.

We now give a formal definition of a divergent sequence.

DEFINITION 7: DIVERGENT SEQUENCE . A seqllence is said to be divergent if it is not convergent. A sequence (sn) diverges to +a if, given any real number-e > 0, there is a positivc integer rn such that sn E, fo r all n > m.

In this case, we also write lim sn - co or sn -+ m as 11 -+ co. Such a sequence is said to be divergent sequence. You can similarly define the divergence of a sequence (sn) to -m.

We car1 write this as 1% sn = -m.

Let us consider the sequence

Suppose ( sn ) is convergent and l i~n sn = s. It is clear, from the definition of n+m

convergence, that if sn -4 s, then s," -, s, as well.

But, for (s,,) as given above,

and so, we would have, Sequences

This contradiction establishes that ( sn ) does not converge, .

EXAMPLE 8: Show that the sequence ( dn ) diverges to +m.

'SOLUTION : Let E > 0 be given. Then > E for all n - > 2. This shows that thc sequence 6 + + w as n -+ w . Hence the sequence ( dn ) diverges to +m.

you can easily see that if a sequence ( s n ) diverges to +& or -m, then the sequence is unbounded. Some diverges to +m; while sqme other diverges to -a.

'

'TO end this section, let us look at a very important sequence, which will be frequently used in the larer sections:

EXAMPLE 9 : i ) If 0 5: x < I, then ( xu) converges to 0.

ii) If x = 1, then ( x" ) converges to I .

iii) Lf x 1, then ( xn ) diverges to +m.

SOLUTION: i) If x = 0, then tlie sequence (xn) is the constant sequence. (0, 0. 0 .....). Hence it converges to zero. Let 0 < x < I . Then

x"l = X. xn C xn,

implies that the sequence ( xn ) is a monotonic decreasing sequence, which is bounded below by 0. Hence the sequence converges to its grzatest lower bound. Therefore, it is enough to show that 0 is the greatest lower boultd of (XI, x2, x3, ,..).

clearly, 0 is a lower bound of the sequence. Therefore, it is enough to show tl'iat if r > 0, then r is not a lower bound of the sequence. Let r > 0. We wish to show that, for some n E N, xn < 1.. That is, for some n E N, nlogx < log r. (Recall from Unit 4 that .

1

I log x is an increasing function). This is equivalent to finding n EN such that n > - log r i log x '

Surely there are infinitely many such n's. This' sl?ows Lhat, when 0 < x < 1, liln xn = 0. Thus, when 0 5 x < 1, tlie sequence ( xn ), converges to 0.

ii) If x = 1, the sequence ( x") is die constant sequence (1, I , 1, ...) and thus converges 1 to 1. I

iii) Let x > 1. Then, silice xn*' > xn the sequence (xn) is a lnonotonic increasing sequence. To show that the sequence diverges to +w, it is sufficient to show .that the sequence is unbounded.

Let M > 0 be any number. Then

x" > M a logx" > logM (why?)

i a n I O ~ X > IogM

16g M a n > -

log x I which shows that the seqlience (x") is unbounded and hence diverges to +po. j

I Try some exercises now.

i I EXERCISE 11

! i) Show that, if -1 < x < 0, then ( x " converges to 0,

ii) Discuss the nature of the sequence ( x n) when x = -1, and x < -1.

1 EXERCISE 12

Suppose (s t ) ) and ( t n ) are convergent sequences such that sn I tn, for each n. Prove 1 17

! I *

Scqucnees ti Serier that Iitn sn I liln t,,. I1 -r. I, *

- --- ---"-I-

- Wc ~onclucle tills section by drsc1rs5ins a very 1111portant erta~l-lplc of a convergent sequence.

EXAMPI I.- 10 : I.et s., = nk. Show that i i l i l sn = I , t ha t is. !jr?; 11'= I . "-,.I

1 - SOLU'FION: Writr: sn = tin = I + xn. Then x,, 0, tor n , 1

Ey Bino~n~sl 'Theorem. we Iiave

--- 2 : 2

'This Iniplies ~l la t O 4,: x,: .: -. , fcl~ 11 2 2. Hence 0 i >i,, <: y , , For n 2 2. n-. 1

Lei c 0 bc given. *Then

7 Let m E N such that m :a --T -1 1. 'l'heii. x,) < 6, h r all n , rn.

E-

Helice 1 % - I = i X" 1 C: e , fur ill1 ir :. n(. That is, ( s,, ) coilverges to l

5.4 CRITERIA FOR THE CONVERGENCE OF SEQUENCES

In section 5.3, we have delincd a sequence ! sn ) lo be convergent if we arc able to find a nu~nber s such that sll -S s as n + ~m. 111 soti~e cases, it may be easy enough to guess the existence of such a nu~nber s. Hut. quite often, i t is not easy to do so. For exa~iiple,

I ' . consider thc saque~cc sinn). Does this scquenee converge? We cannot say

I1

anytliing unless we know the limit of this sequence. What we really need is a criterion for the convergencr: of' a sequence ( sn ) which uses only the terrns of the glven sequelicc - and 170t to search for a possible limit of the sequence frum among the infinite set R. 'That is precisely what we intend to do in this section I n fact, we shall obtain a necessary

\ ' > * . 'I r r ' i~~y le Incqrral~ty as and sufficient condition for the coavcrgence of a sequence in this section. .IIVLI\~C~ 111 [;IIIL 3 We state, first, a necessary condition for a sequence to converge. You will see later that

the condition is also sufficient for a sequence to converge but under certain aclditio~ial res~r~ctions

THEOREM 6: If ' a sequence ( s,, ) is convcrgcnt, then for a given E > 0, there esists m E N S I I C ~ I that Isn - sL[ C: E, whenever n > li 1n.

PROOF: Let the sequence ( s n ) converge to a nunalsar s (say). Let E 0 be given.

Since sl, --+ s as n -+ cc , tlierc exists a natiiral number m such that

Thus isn - ski <'E. for aB n, k > 111 (n > k).

The theorem says that if the terms of the sequence (sen) get close to some number, then they are already close to each other. Motivated by the above theorem, we have the following definition due to A.L. Cauchy, an eminent French mathematician.

DEFINITION 9: CAUCHY SEQUENCE A sequence ( s " ) of real numbers is called a Cauchy sejuence or a fundamental sequence. if, for each E > 0, there exists a natural number m such that

, Isn - s,, I < E, for ,all n > k > m.

i n Theorem 3, we have seen that every convergent sequence is bounded. And, in Theorem 6 , it is shown that every convergent sequence is Cauclly. In the next two results we will prove that class of convergent sequences is same as the class of cauchy sequences and is properly contained in the class of bounded sequences. This inclusion is proper because there exists bounded sequence which are not monotone.

We state and pr,ove the following tileorem.

THEOREM 7: Evcry cauchy sequence is bounded.

PROOF: Let ( srl ) be a cauchy sequence. Then, by definition, it follc. .::s that for E = 1. there is a positive integer m sucll that

Isn - q( < 1 whenever n > k > m.

In particular, Isn - sm+,l < 1, for all n > m. In other words (see Unit 3)

I Sn t = Its,, - snl*,) + SII1+, I Isn - srn+,l -1- I Sn,+, I < I Sm+l I + 1- n > m.

Thus if

A = { I s , I, I s2 I , ..., I snl I, I sm+l i + 11,

then j'sn I I: A,' for all n E N.

This proves that the Sequence (sn) is bounded if it is Cauchy.

THEOREM 8: Every cauchy sequence is convergent.

PROOF': Let ( 5 ) be a cauchy sequence. Then by Theore111 7, it follows that ( sn ) is bounded. Using Bulzano-Weirstrass Theorem ('Theorem S), we can conclude that the sequence ( s n ) must have a convergent subsequence, say ( snr ). Suppose that

.snr + p as r -+ w.

Our claim is that sn -+ p as 11 + w.

Let E > 0 be given. Then there exists a natural number q such that, for any r > q, w e have

Again since ( s,, ) is a cauchy sequence, therefore, tht:e cxists a natural number k such . that for n > k > rn, we have

E lSn-SkI< 7.

Now, for an n > m, clloose r so large thal n, I. nl and r > q. 'Tben

Sequences

Sequences % Series E I s n r - P I ' T s

E is satisfied. Also 1 sn - s , ,~ / is satisfied with k = n,. TIILIS, tbr n > ~ n ,

(Triangle Inequality)

I s,, - P I = I Sn - Sn, -' Sn, - P I

5 Is,, - snIrl + ISt,[ - PI

E 1 < - - + - E = E . 2 2

Thus, given an E > 0, we have found an m such that, for any n > In, I s,, - p I < E.

This shows that s,, -> p as n -+ a, which concludes the proof.

Theorems 6 and 8 are solneti~nes combined into one theorem wl;.cli is popularly known as Cauchy's General Principle of Convergence o r Cauchy's criteria for the convergence of sequence.

THEOREM 9: Cauchy's General Principle of Convel.gence A necessary and sufficient condition for the convergence of a sequence'( st ' ) is that, for every E > 0, there exists a positive integer rn suc l~ that

( sn - s,, ( < E, whenever n > k > m.

Thus a sequence of real numbers converges if and only if it is a cauclly sequence

Let us illustrate this by the following example:

I EXAMPLE 11 : S l ~ o w that the sequence -;;- is a cauchy sequence. What about ( nl)?

SOLUTION : Let E > 0 be given. Then by taking m = III (E) > $ , we see that

whenever n, k 2 m. Thus, fLr n > k > rn,

In other words,

1 whenever n > In and k >m, which shows that the sequence (x) is convergent.

The sequ ce (n2), however, is not a caucliy sequence.

For, if n and k are any two integers ( n > k), then

'n2 - kl' = l(n-k) (n+k), > 2k1 > 1,

whatever m may be. Thus, if we choose E = I , then by above inequality

inZ - k 2 ! > E , for any m (n > k > m).

Thus the sequence (nZ) is not convergent.

The Cauchy's Criteria is sometimes described in the following way:

A sequence (s,,) is said to be a cauchy srquence, if $or any E > 0, there is a positive integer m such that

Is, - s,,1 4 E, whenever k > (I.

AS you can observe, in Theorein 9, there is no mention of iimit. Thus the advantage of the Cauchy Criteria is that we are able to test the convergence of a sequence without necessarily knowing the value of its limit. Example 10 has justified the utility of Cauchy Criteria. To further elaborate this assertion, c0nsider.a sequence ( rn ) of rational nurtlbers

which converges to 4 (e.g., sequence (8,) defined below Definition 4 in this unit). If (rn)

is treated as a sequence of real numbers, then we have a real number fi, as the limit of

the sequence ( rn ). Thus the sequence ( rn ) satisfies the definition of convergence. However, if we treat ( rn ) as a sequence of rational numbers and if our definition of convergence requires us to find a rational number which is the limit of ( rn) , then ( r n ) is

no longer convergent since fi is not a rational number. Thus lack of convergence has

arisen not due to the change of the sequence in any way, In fact, we have merely modified the context in which we are considering the sequence 'by changing the underlying field in which the scquence is being discussed. By changing the context in which the sequence is considered has no effect on whether the sequence is cauchy or not. ~ h u s : it also implies that there is no difference between convergent sequences of real numbers and cauchy sequences of real numbers. This is true because of the axiom of completeness of R, the set. of real numbers, But then this is not the case if we confine our sequences to the field of rational numbers. In view of this, the completeness of R is also described by saying that the system'of real numbers is complete if every caucl~y sequence in R has a limit in R.

5.5 ALGEBRA OF CONVERGENT SEQUENCES

In this section, we shall discuss the behaviour of convergent sequences with respect to algebraic operations like addition, niultiplicat~on, and so on. Recall that a sequence of real numbers is a function s: N -+ R. Since, the sum,, product, etc, of real-valued functions are defined, you can easily define the sum, product etc. of sequences.

DEFINITION 10: COMPOSITION OF SEQUENCES

Let (s,,) and ( t n ) be two sequences of real numbers and cc E R. Then,

i) ( sn + ( trt ) = ( Sn + tt, (slim of two sequences)

ii) ( sn') ( tQ 1 = ( sn tt, ) (product of two seqiences)

iii) a ( sn ) = (a S; ) .

(scalar multiple ut' ii sequence)

We'shall sllow that convergence and limits are preserved under above operations.

THEOREM 10: 'If lirn sn = s and fim.tn = t, then n->m !I.-)=

lirn (q, + tn ) = s + t, n-tr

PROOF: Let E > 0 be given. Then ~ 1 2 , 0. Since lirn sn = s, there is.m, E b4 such that n->n

E . ' I sn - s J < - , whenever n 2 m,. 2

Since lirn tn = t, there is m, E N such that n 4 r

E Itn - t J -, whenever n r rn,.

2

Let m = max (m,, m,). Then, for all n 2; m, we have

Thus ( s,, + tn j converges to s + t.

The next theorem is easikr to prove. I

Sequences & Scrirs THEOREM 11: I f liii s,, = s and a E R, then

lim ( a sn) a s n - r

PROOF: The theorem is obvious if u = 0. So, let us consider the case when a t 0.

Let E > 0 be given. Then > 0, because / a ( 0. Since (s,) converges to s. there 1s la I

m E N such that

E / sn - s I < -, whenever.n 2 m. la I

Thus, if n 2 m, then we have E

~ 1 s ~ - a s I = l a ( I s ~ - s ( < I c ~ I - - - = E . la I

That is, lirn ( a sn) = a s . n-+%

Now you can easily solve following exercise. .

EXERCISE 13

Let linl sn = s and !i,mP t,, = t. Let a, p E 'R. Show that lirn (.a sn + P t n ) = cc s + [Jt. 1142 n-r=

Deduce that lirn ( sn - tn ) = lirn sn .- lirn 'tn. n - l r n->z n-+x . .

Now to prove that the limit of the productGof twb convergent sequences is the prod;~ct 01' their limits, we need the following theorem:

THEOREM 12: &f lirn ( sn )' = s, then !$ s: = sz. n-+E

PROOF: Let E > 0 be given. We have to find m E N such that

1 s ; - s 2 1 ' < : ~ , f o r - a l l n > . m , t h a t i s , '

1s"- s J ( s n + S I C & , for a l l n 3 m.

Since (sn) converges, therefore it is bounded. Hence, there exists a'ieal ,nu~~lber ti' such

that 1 sn I < K, for all n. Since 1i1n s,, = s, we have Is I s K. Hence Il--IX

i I s n + s I < I s n I - I - ( s ( < 2 K fo ra l tn .

E , Since lirn sn = s, there is an m E N such that I sn - s I < -- . for all 11 2 ni.

n+r , 2K

Hence, whenever n m, we have

'This proves that l.im s: = s2. 1 1 4 5

Now we are ready to prove the following theorem:

THEOR.EM 13: Let lim s,, = s and lirn tn = t. Then lirn (s,, , tn ) = s.t. '

n+x n-rr

1 PROOF: We use the wcll-known algebraic identity ab = - [(a t- b)l - (a - b)?].

4 ' Now, by Theorem 10, we know that sn + ln -+ s 4- t as n-+m. Ti~erefore. using Tl~eorem I

12. we get

(so + tJ2 + ( ~ + t ) ~ .

-

~ 1 ~ 0 , by E 13, sn-- tn + s - t as n+m. Therefore. ligalli using Theorem 12, we have

Ifence, (sl, + tn)' - (sn - tJ2 -> ( s -1- l )' - ( S - t)'. Finally, using the algebraic, idcnl i :~ ,

we get

verify t.fiat all the steps are justif'ied. Nniice that this proof' uses no E. 'T*l~e tcuhlliqut, of ~!:;i!lg'the'iilgebraic ideritity to deal with Ihe product is c,alled polsrizi~tioa.

I'inally. we turn our attention to the qe~otient of'convery,ent seqr.ri.liccs. I:nr l.hi.r;. \rife ?sail.\ 1 I )reim: ilr7ed ttie fi.~llocv~n,b , I( (

i I'WBEOF: '1'6: p~.ove tlic ~i.leul.em, rse nccd s i i! f,(> that ;:- c;in bc rl*:iint:J. Bur: w l ~ : 3

1 : ' I : :abot,lt 7-- ! if SOIIIC g,':= 0. then --- IS 1101 ilefi;iccj. "To ovclecumc tliii; difficulty \.vt! 112:iy !> 3,)

\,Ye shall discuss tlio proof for the casc s 0 The case \ O call he discurstd 1.y' applyilig the case for s :-. O to the sequence ( --s,, )

Let c ' 0 be given. We must find rn E N such tliat

Since iim sn - s. tlierelblr tl'iere exists m, e N such that n ->I

S , This implies that sn > for all n , m,.

2

Similarly, there exists m, E N such that

s2 E .. Isn - s I < - , whencvcr n > 111,.

2

Let m = niax (m,, m,), If 11 > rn, then we have

1 1 This proves that I,ic ,

Now, it should not hc tl;ffic~llt t b r ; \NI t o prow the

'I'HEOREM 1% I F lim sn = s :(II(J liiu t , , 'I, then 11 +T 11 .

Sequences & Scrics 2n3 + 5n 1 EXAMPLE 12: Prove that iim, (dn3 + n z ) = i .

. .

SOLUTION: You have seen earlier that liif = 0. Consequently, .

5 lirn (2 + -)

n+m n 2 . 2 - - 1 - -- 1

lim (4 + -) 4 2 '

n-+= n .

What we have proved is that, if lim sn = s and Iim t = t, then it is true that n+= n + s

lirn (sn i- tn) = s + t. In other words, convergence of the sequences (sn) and (tn) is n + r

sufficient for the convkrgence of (sn -+ tn).'It is possible for (sn + tn) to be convergent even if ( s,, ) and ( t, ) do not converge.

Similar reniarks are true for sequence (a sn ) and (sn tn). Now try the following exercises.

'EXERCISE 14

i) Given an example of divergent sequence ( sn ) and ( tn ) such that ( sn + tn ) converges.

ii) Given an example of a divergent sequence (s,,) and a codvergent sequence ( t n ) such that'( s. tn ) converges.

EXERCISE IS

Show that if (sn ) is a bounded sequence and if ( t n ) converges to 0, then (sntn) converges to 0.

We have discussed the algebra of convergent sequences. Is there an algebra of divergent sequences? The following re'sults do justify that there is algebra of divergent sequences also.

If lim sn = +. oo and lirn tn = +oo, then n-bm n-+oc

111. If ( sn ) diverges to +OD and if ( tn ) converges, then (sn +- tn) diverges to 4-m.

You can similarly try to formulate some similar results for the sequences.diverging to minus infinity.

5.6 SUMMARY

In this unit, we have initiated the study of the limiting procebs by introducing the norion . of a sequence and other related concepts. In section 5.2, we have defined a sequence, a subsequence and a few types of sequences such as bounded and monotonic sequences .

etc. We have confined our discussion to the real sequences. A real sequences is a special type of real function whose domain is the set N of natural numbers and the range is a subset of the set R of real numbers. If s: N + R is a sequence, then its values are . ,

denoted. by s,, s,, .... The sequence is generally denoted by ( s n ) where the values s,, s,, .... are known as its terms. A sequence'( tn) is called a subsequence of the sequence ( s,, ) if all terms of tn are taken in order from *those of ( sn). A sequence ( x ) is said to be bounded if there exists a real number K such that ( s,, I I K, for' every n E N. A seiuerice (s, ) is said to be monotonicall;r illcreasing if sn I snr,, for, every n E N and it is said to be monotonically decreasing if sn 2 sn+, for every n EN. The sequence ( sm) is said to be

strictly increasing if sn < sn+, (strict inequality), for every n E N and strictly decreasing if

Sn ' Sn+l' for every n EN. A sequence which is either increasing or decreasing is said to be a monotonic sequence.

Section 5.3 deals with the convergence of a sequence. When a sequence ( s n ) possesses a limit as n-+m, then it is said to be convergent. In other words, we say that a sequence (s,,) .converges to a limit s if for a given E > 0, there exists a positive integer m such that

A sequence which is not convergent is said to be a divergent sequence. This is due to the reason that ( s, ) is unbounded or because ( s n ) does not have a unique limiting value.

We have proved that a convergent sequence is always bounded. According to Bulzano- Weirstrass Theorem, every bounded sequence has a convergent subsequence. Similarly a bounded and monotonic sequence is convergent.

In Section 5.4, we have discussed Cauchy's Criteria to test the convergence of a sequence without taking the botheration of finding the limit of the sequence. Tl1i.s criteria states that a sequence ( s n ) is convergent if and only if, for an E > 0, there exists a positive integer n1 such that

1 s, - sk I < E, for a11 n, k > m ( n > k ).

Finally in section 5.5, we have discussed the algebra of convergent sequences i.e., the sum, difference, product and the quotient of two convergent sequences is a sequence which is convergent under certain necessary restrictions.

E 1 ) ii) is a subsequence o r (n) while (i) and (iii) are not subsequence of (n). .

E 2) Since d r 0, either d > -0 or d < 0. Let us consider the case d > 0. The case d < 0 , % .

is si~nilar and can be proved in an analogous way.

We have to show that ( a, a+d, a+2d, ....,) is unbounded.

Clearly it is bounded bclow by a. We show that it cannot have an upper bound.

Suppose m > 0 is any nu~liber. Then, there is a number a+nd such that a+nd > In for some n EN. TIILIS there are infinitely such ,positive integers. Hence (a, a+d, a+2d ....) is unbounded.

E 3) i) (1, 0, 1, 0, ....) is bounded, 1 is an upper bound and 0 is a lower bound.

ii) (1, 3, 6, 10, 15, ....) is not bounded above.

iii) (1, 1, 2, 1, 3, 1, 4, 1, 5, .....) is not bounded above.

E 4) i) is not monotonic.

ii) is also not monotonic.

1 1 1 iii) is monotonically decreasing since - > - > - 1+12 1+22. 1+32 > """'

- iv) The sequence is monotonically increasing.

The sequence is (1, 5, 5, 7, 7, 9, 9, .....)

E 5) i) Suppose ( sn ) is a monotonic increasing sequence. That is,

s, 5 s2 5 s3 5 ..", Let ( s,,,) be a subsequence. This means that

nI < n2 < no ... Hence s,, 5 sni I sn3 < ...

25

Sequences &Series [n other words, (s,,) is n~onotonically increasing.

ii) Yes. Suppose ( s n ) is both an increasing slid a decreasing sequence. This means that s, 2 s: 2 s, 2 .... and s , I. s, 5 s, s ..... By the law of trichotollly ( do you recall it), this can happen it and only if s, = s, = s, = ..... In other words constant sequences are the orfly sequences which are both increasing and decreasing.

E 6) Proof is similar to that of Exampie 6.

E 7 ) We have s, 2 s, 2 s, .... and we are given that the sequence is bounded. Hence, by the completeness axiom, the set is,. s,, ..,..I has the greatest lower bound. Let a be the greatest lower bound. Hence sn 2 a, for eveiy n EN.

Let E > 0 be given. Then a -t E is not a lower bound of the sequence. Hence sill L= a + E , for somr: m EN. But then a + E =- sn, 2 s ,,,? , > ..... But then a - E z sill 2 sm, .... 2 a. That is, s,, lies i n the interval ]a - E, a + E [, for all n > m. In other words, lim sll = a.

n->z

E 8) i) We prove this by contradiction.

If possible, let s > A.

Take c = s-A.

Since ( sn ) converges to s, there exists m EN such that / s - s,, 1 < E , for all n In.

Hence sn > s - E = A, for all n > m, which is a contradiction to the hypothesis that: sn 5 A for all n.

ii) This is entirely analogous to Part (i). You have nlerely to reverse tlle inequalities.

E 9) i) Given that ( sl, ) converges to s.

~ e t ( s, ,~ ) = (s,,, s , , ~ , sIl3, .......) be a subsequence of ( s,, ).

This means, by definition, that n, n, < n, < ...... To show that (snr) converges to s, let E > 0 be, given.

Since ( sn ) converges to s, then exists m E N sucll that 1s - sll / < E, wilenever n 2 m.

Since n , < 11, < n, < ..... is an increasing sequence of natural numbers, there is an integer i EN such that n,, nit,, nit,, .... are all 2 m.

Hence I s - snrl < E, whenever r 2 i .

That is, ( snr) converges to s.

ii) The sL7tement is false. For instance, consider tlle sequence (1, 0, 1. 0, ...). This sequence does not converge. However (1, 1, 1, ...) is a subsequence ~d l i ch converges to 1.

E 10) Given that ( s,, ) converges to s.

To show that ( I sn 1 ) coilverges to 1 s I.

Let E > 0 be given.

Since ( Is,! - J S 1 ( E, whenever 11 > rn,

then. I I sn 1 - 1 s I I " sn - s I . [Refer to Unit 31.

Hence, 1 i s,, 1 - ( s 1 I < E, whenever n > m.

Hence ( / s, I ) converges to I s 1 . Sequences

The converse is not true, as you can see by considering the non-convergent sequence (1, -1, 1, -1 .... ).

E 11) A sequence ( s n ) is said to diverge to -w if, given any real number k < 0, there is ni E N such that sn k, for all n 2 m.

L,et -1 < x < 0. Then, 0 - 1 ?: < 1.

Let E > 0 be given. Then

( 0 - xn I = I x I".' NOW,

I x l " ~ i f n log 1x1 ' l og&.

Remembering that log 1 x I < 0 sinco 0 < ( x 1 < 1. Take

log E , , m > ---

log 1x1 log E

Hence 10 - xn I < E, for all n > - 1% 1x1 '

Hence ( x") converges to '0.

ii) If x = -1, then the sequence becomes

(-1, 1, -1, +I, ..... ), which we know, oscillates finitely.

If x < -1, then (x" oscillates infinitely.

E 12) Let.lini sn = s and l i ~ n t,, = t. Then, for E > 0, there exists integers m,, rn, E N n+T n+m

such that Isn - S I < E, M n > m,, and Itn - tl < E, V n > m,. Let m = max (m,, m,). Then, for n > m, both Isn - sJ < e and Itn - t( .< E holds.

This, in term, implies

s - sn < E and t,, - t < 'E, V n > m.

Consequently, for n > m,

s ' - t = (S - sn) + (tn - t) 4 (sn - tn)

< 2&, since sn - tn 5 0,

That is, s - t < 2 ~ , V E > 0. Thus, S - t I 0 or

E 13) . lirn (asn) = a s , by Theorem 11. n - r a

Similarly, lirn (Pt,) = Pt, by h he or em 11. n+=

lirn (asn + Ptn ) = lirn (asn ) + lirn (Pt, ), by Theorem 10. n+x n+x n+d

= a s + p t .

lirn ( -tn ) = -lim sn = -t, by Theorem 1 1. n+m n+w

Hence, lirn ( sn - tn ) = lirn sn + lim ( -tn ) = s-t. n + r n + s n+m

E 14) i) Let sn = n and tn = -n. .

Then (sn ) diverges, ( tn ) diverges, but

(sn+ t n ) is the consiant sequence (0, 0, 0, .....) which converges to 0.

27

: I Sequcnccs & Series

Then (sn) diverges, while ( sn t n ) is the constant sequence ( 1, 1, 1, .....) which converges to 1.

E 15) Let I s n ( K , f o r a l l n.

let.^ < 0 be given. E

Since ( tn ) canverges to 0; there is m E N such that

for all n > m.

Hence, for all n > m,

Hence ( sn tn ) converges to 0.

1 ii) ~ e t sn = n and tn = - n .

UNIT 6 . POSITIVE TERM SERIES

Structure 6.1 Introduction

Objectives 6.2 Infinite Series 6.3 Series of Positive Terms! 6.4 General Tests of Convergence

Co~nparison Tests

p-test 6.5 Some Special Tests of Convergence

D'Alembertls Ratio Tesl Cauchy's Root Test

Cauchy's Integral Test

Raabe's Tcst Ciauss's Test

6.6 Summary 6.7 Answers/Hints/Solutions

6.1 INTRODUCTION

In the Unit 5, you were introduced to the notion of a real sequence and its convergence to a limit. It was also stated that one of the main aims of discussing the real sequences and its convergence'was to find a method of obtaining the sum of an infinite number of real numbers. In other words, we have to give a meaning to the infinite sums of the forms

1 t 2 + 3 + 4 + 5 + .............

where the ' .................. ' is interpreted to indicate the remainiae infkite number of additions which have to be performed. The clear explanation of this concept will, then, lead us to conclude that it is possible to achieve the addition of an infinite number of real numbers, by using the limiting pracess of the real sequences.

To give a satisfactory meaning to-the summation of the infinite number of the terms of a sequence, we have to define a summation which is popularly known as an infinite series. The infinite series have been classified mainly into two categories - the positive term series and the general Series. What are, then, the positive term series and the general series? We shall try to find answers for these questions. T h e summation of an infinite series of real numbers is directly connected with the convergence of the associated real sequences. We shall, therefore, give a meaning to the term associated sequence for an infinite series and hence its convergence which will lead us ultimately. to find the sum of an infinite series.

Although the famous Greek philosopher and mathematician, Archimedes had summed up the well-known Geometric Series, yet oqer results on infinite series did not appear in Europe until the 14th century when Nicole Oresme [1330-13821 showed that the Harmonic series diverges. Since then, a lbt of work has been going on in thiS direction. There is evidence that this type of work was known in 1ndia also as early as in 1550. Indeed, even modem work has shown evidence of the discovery of a number of mathematical ideas pertaining to the infinite series in China, India and Persia much before they came to be known in Europe. In the 17th century, there seemed to be little concern for the convergence of the infinite series. But during the 18th century, two ~rench$mathematicians D' Aleinbert and Cauchy devised

+ remarkable tests for the convergence of infinite series under certain conditions which we shall discuss in this unit. Also, we shall discuss, in this unit, a few more tests for the