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STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 1 of 49

UNIT CONTENT PAGE Nr

I REAL NUMBER SYSTEM 02

II SEQUENCES 05

III BEHAVIOUR OF MONOTONIC SEQUENCES 19

IV SERIES 28

V ALTERNATIVE SERIES 39

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 2 of 49

UNIT - I

BOUNDED SETS

that A is said to be bounded if it is both bounded above and bounded below.

Least Upper Bound and Greatest Lower Bound:

Definition: Let A be a subset of R and

u is called the least upper bound or supremum of A if i. u is an upper bound of A. ii. if then v is not an upper bound of A.

Let A is a subset of R and

is called the greatest lower bound or infimum of A if i. l is a lower bound of A. ii. if then m is not a lower bound of A.

Examples:

1. Let A = {1, 3, 5, 6}. Then glb of A = 1 and lub of A = 6

2. Let A = (0,1). Then glb of A = 0 and lub of A = 1. In this case both glb and lub do not

belong to A.

Bounded Functions:

Definition: Let

be any function. Then the range of f is a subset of R. f is said to be bounded function if its range is a bounded subset of R. Hence f is a bounded function iff there exists a real number m such that

Examples:

1. f : [0,1]

R given by f(x) = x + 2 is a bounded function where as f :

R given by

is not a bounded function.

2. f :R

R defined by f(x) =sin x is a bounded function. Since

Absolute Value:

Definition: For any real number x we defined the modulus or the absolute value of x denoted by |x| as follows . Clearly

Triangle inequality

1. Prove triangle inequality: for all

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 3 of 49

6 I

Cauchy-schwarz inequality

Theorem:1.1 If

and are real numbers, then

Or, equivalently

We will use mathematical induction as a method for the proof. First we observe that

By expanding the square we get

After rearranging it further and completing the square on the left-hand side, we get By taking the square roots of both sides, we reach which proves the inequality (2) for n = 2. Assume that inequality (2) is true for any n terms. For n + 1, we have that By comparing the right-hand side of equation (4) with the right-hand side of inequality (3) we know that Since we assume that inequality (2) is true for n terms, we have that which proves the C-S inequality.

Theorem:1.2

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 4 of 49

Given real numbers a and b such that

for every . Then

Proof:

Given

Suppose

Choose

Now,

Therefore,

, which is a contradiction to (1) Hence

Theorem: 1.3

If n is positive integer which is not a perfect square, then is irrational.

Proof:

Let n contains no square factor > 1

Suppose

is rational Then , where a and b are integers having no factor in common. implies But b2n is a multiple of n, so a2 is also a multiple of n However if a2 is a multiple of n, a itself must be a multiple of n. (since n has no square factor >1 ) where c is an integer sub in (1) Therefore b is a multiple of n, which is a contradiction to a and b have no factor in common. Hence is irrational If n has a square factor, then n = m2 k, where k> 1 and k has no square factor > 1. Then If is rational, then the numbers is also rational. Which is a contradiction to k is no square factor > 1.

Hence n has no square factor.

Problem: Prove that

is irrational.

Theorem: Prove that e is irrational.

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 5 of 49

UNIT - II

SEQUENCES

Definition. Let f : Գ їԹ be a function and let f (n) = an. Then is called the sequences in Թ determined by the function f and is denoted by (an). an is called the nth term of the sequence. The range of the function f which is a subset of Թ, is called the range of the sequence

Examples.

a) The function f : Գ їԹ given by b) The function f : ԳїԹ given by

Definition:

A sequence

is said to be bounded above if there exists a real number k such that

A sequence

is said to be bounded below if there exists a real number k such that for all n. k is called a lower bound of the sequence

A sequence

is said to be a bounded sequence if it is both bounded above and below. Note.

A sequence

is bounded if there exists a real number such that

Examples.

bounded sequence.

݈.b of the sequence.

ʹ1 is the ݈.u.b of the sequence.

5. Any constant sequence is a bounded sequence. Here 1.u.b = g. l. b = the constant term of

the sequence.

Monotonic sequence

Definition: A sequence

is said to be monotonic increasing if for all n. is said to be monotonic decreasing if for all n. is said to be strictly monotonic decreasing if for all n. is said to be monotonic if it is either monotonic increasing or monotonic decreasing.

Example.

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 6 of 49

3. The sequence

monotonic decreasing. Hence is not a monotonic sequence. 4. is a monotonic increasing sequence.

Proof:

an ʹ an+1 =

Therefore

Hence the sequence is monotonic increasing.

5. Consider the sequence (an) where

. Clearly is a monotonic increasing sequence.

Note: A monotonic increasing sequence

is bounded below and q1 is the g.݈.b of the sequence.

A monotonic decreasing sequence

is bounded above and a1 is ݈. u. b of the sequence.

Solved Problems:

Show that if

is a monotonic sequence then ( ) is also a monotonic sequence.

Solution:

Let be a monotonic increasing sequence.

Therefore

Let Now, by (1)

Therefore,

Therefore

is monotonic increasing.

The proof is similar if

is monotonic decreasing.

Convergent sequences

Definition. A sequence

is said to converge to a number ݈if given ߳ positive

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 7 of 49

integer m such that . We say that is the limit of the sequence and we write lim݊їьܽ

Note.1

ї݈ iff given ߳

Theorem. 2.1

A sequence cannot converge to two different limits.

Proof. Let (an) be a convergent sequence.

If possible let ݈1 and ݈2 be two distinct limits of (an).

Let ߳

Since (anͿї݈1 , there exists a natural number n1

Such that

Since

ї݈2, there exists a natural number n2

Such that

Let m = max {n1 , n2}

Then by (1) and (2) ׵݈1 о݈2<߳and this is true for every ߳

Hence ݈1 = ݈2

Examples

1.

Proof:

Then .Hence if we choose m to any natural number such that m then for all n

Note. If

, then m can be chosen to be any natural number greater than100.In this example the choice of m depends on the given ߳and [ 1/ ߳ that satisfies the requirements of the definition.

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 8 of 49

Proof.

Let ߳

Let the given sequence be denoted by (an). Then an = 1 for all n. ʹ ϭൟсൟϭʹ ϭൟсϬф߳ for all n ߳ where m can be chosen to be any natural number. Note. In this example, the choice of m does not depend on the given ߳ 3.

Proof. Let ߳

Now, 4.

Proof.

Let ߳> 0 ܾ

Then

Therefore,

5. The sequence

is not convergent

Proof.

Suppose the sequence

converges to ݈

Then, given ߳

чൟ;ʹ1) m ʹ݈ൟнൟ;ʹ1) m + 1 ʹ ݈ൟ < ɸ+߳ = 2߳

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 9 of 49

But i.e., 1 <߳ which is a contradiction since ߳

Theorem:2.2

Any convergent sequence is a bounded sequence.

Proof.

Let be a convergent sequence. Let

Let ߳> 0 be given. Then there exists m ߳

Now, let

Then is a bounded sequence. Note. The converse of the above theorem is not true. For example, the sequence is a bounded sequence. However it is not a convergent sequence.

Divergent sequence

Definition: A sequence

is said to diverge to ьif given any real number , there exists m ߳ Note. їьif given any real number k > 0 there exists m ߳

Examples

1.

Proof: Let k > 0 be any given real number.

Choose m to be any natural number such that m > k

2. (n2 Ϳїь

Proof: Let k > 0 be any given real number.

Then n2> k for all n > m

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 10 of 49

Definition. A sequence

is said to diverge to оь if given any real number k < 0 there exists m ߳ Note. їоь iff given any real number k < 0, there exists m ߳

A sequence

is said to be divergent if either

Theorem. 2.3

Proof.

Let

Let k < 0 be any given real number. Since

їьthere exists m ߳

Theorem. 2.4

If

Proof. Let ם

Since

їь, there exists m ߳

Hence Note. The converse of the above theorem is not true. For example, consider the sequence (an) where

An = (-1)n /n. Clearly (an)

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 11 of 49

need not converge or diverge.

Theorem:2.5

If > 0 for all n ߳

Proof.

Let k > 0 be any given real number.

Hence

Theorem:2.6

Any sequence

Proof.

Let is not bounded above

From (1) we see that an >݈ for all n.

Theorem:2.7

Any sequence

Proof is similar to that of the previous theorem

Note 1. The converse of the above theorem is not true. For example, the function f : ԳїԹ defined by f(n)= above. Also for any real number k > 0, we cannot find a natural number m such that an > k for all n >m.

Similarly f: ԳїԹ given by f(n)=

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 12 of 49

Oscillating sequence

Definition: A sequence

be an oscillating sequence. An oscillating sequence which is bounded is said to be finitely oscillating. An oscillating sequence which is unbounded is said infinitely oscillating.

Examples.

theorems). Also this sequence is not convergent . Hence ((о1)) is a finitely oscillating

sequence.

2. The function f : Գ їԹ defined by

f(n) infinitely oscillating.

The Algebra of limits

In this section we prove a few simple theorems for sequences which are very useful in calculating limits of sequences.

Theorem: 2.8

If

Proof:

Let ߳

Now |ܽ݊ + ܾ݊ оܽ ʹ ܾ|= |ܽ݊ оܾ + ܽ݊ оܾ |ч |ܽ݊ ʹ ܽ|+|ܾ݊ ʹ ܾ

Since Since

Let m = max{݊1, ݊2}

Then |ܽ݊ + ܾ݊ оܽ ʹ ܾ|< 1/2߳+1/2߳ = ߳

Note. Similarly we can prove that (ܽ

Theorem:2.9

If

Proof:

If k = 0, (݇ܽ

Then |k anо݇ a|= |݇| |ܽ݊ʹܽ

Let ߳

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 13 of 49

Theorem: 2.10

Proof.

Let ߳

Now, |ܾܽ݊݊ оܾܽ |= |ܾܽ݊݊ оܽ݊ܽ + ܾܾ݊ ʹ ܾܽ

=|ܽ݊||ܾ݊ʹ ܾ|+ |b| | ܽ

Also, since (ܽ݊ͿїĂ͕ (ܽ

Using (1) and (2) we get

Let m = max{݊1, ݊2}.

Then|ܾܽ݊݊ ʹ ܽ

(by (3),(4)and(5))

Hence (anbn)

Theorem: 2.11

If

Proof:

Let ߳

We have |1/an ʹ 1/a|=

Now, a

Since Hence

Using (1) and (2) we get

Now since (an)

Let m = max {݊1, ݊2}.

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

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Therefore (1/an)

Corollary:

Let Then

Proof:

(since If

Theorem: 2.12

If

Proof:

Let ߳

Now | |ܽnͮоͮĂͮͮчͮĂnо ܽ Hence from (1) we get ||ܽ݊ |о |ܽ|| <߳

Hence (|anͮͿї;ĂͿ͘

Theorem: 2.13

If

Proof.

׵ Ăо߳<ܽ݊< a+߳

Theorem: 2.14

If

Proof.

Also ( bn оܽ݊ Ϳїܾ

STUDY MATERIAL FOR BSC MATHEMATICS

REAL ANALYSIS - I

SEMESTER - III, ACADEMIC YEAR 2020-21

Page 15 of 49

Theorem: 2.15

If and ܽ݊ чܿ݊ чܾ݊ for all n, then (ܿ

Proof.

Let ߳

Since ї ݈ ,there exist ݊1אN such that ݈ о߳<ܽ݊<݈ + ߳ Similarly, there exist ݊2אN such that ݈ о߳<ܾ݊<݈ + ߳

Let m = max {݊1, ݊2}.

Theorem:2.16

If

Proof.

Since ܽ

Now, |

Since ( anͿїܽ тϬ͕ we obtain ܽ for all n n1

Now, let ߳

Since (an Ϳїܽ, there exist ݊2א

|an-ܽ|<߳

2 + 1)/

Let m = max {݊1, ݊2}.

Then |

Theorem: 2.17

If

Proof.

Let k > 0 be any given real number.

Since

Similarly there exists ݊2אN such that ܾ

Let m = max {݊1, ݊2}.

Then ܽ݊ + ܾ

Theorem: 2.18

If

Proof.

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