Unit - 2

Sequences and Subsequences

2.1 Sequences and Subsequences

Sequence –

A function f: N , where S is a non-empty set, is called sequence, for each nϵN.

The sequence is written as f(1) , f(2) , f(3) , f(4)……….f(n).

Any sequence f(n) can be denoted as <f(n)> or {f(n)} or (f(n)).

Suppose f(n) =

Then it can be written as -    and can be denoted as <>or  {} or ()

is the n’th term of the sequence.

Example:  suppose we have a sequence – 1 , 4 , 9 , 16 ,……….. And its n’th term is

This sequence can be written as  -<>

Example: its n’th term will be .and can be written as <>

Types of sequences –

1. Finite sequence- A sequence which has finite number of terms is called finite sequence.

2. Infinite sequence-  A sequence which is not finite , called infinite sequence.

Sub-sequences-

Definition:

Given a sequence {}, consider a sequence {} of positive integers such that < < . . . Then the sequence {} is called a subsequence of {}. If {} converges its limit is called a subsequential limit of {}.

It is clear that {} converges to p if and only if every subsequence of {} converges to p.

Subsequences and Compact Metric Spaces

Theorem:

(a) If {} is a sequence in a compact metric space X, then some subsequence of {} converges to a point of X.

(b) Every bounded sequence in contains a convergent subsequence.

Subsequences Limits

Theorem:

The sub-sequential limits of a sequence {} in a metric space X form a closed subset of X.

Note- A sequence {} converges to s iff its every subsequence converges to s.

Similarly lim every subsequence of {} tends to

Key takeaways:

1. A sequence which has finite number of terms is called finite sequence.
2. Given a sequence {}, consider a sequence {} of positive integers such that < < . . . Then the sequence {} is called a subsequence of {}.
3. Every bounded sequence in contains a convergent subsequence.

2.2 Bounded sequence, Convergent sequence, Limit of a sequence

Bounds of a sequence:

Bounded above sequence: A sequence {} is said to be bounded above if there exists a number K such that,

Bounded below sequence: A sequence {} is said to be bounded below if there exists a number K such that,

Bounded sequence: A sequence is said to be bounded when it is bounded above and bounded below. K an k are respectively the upper and the lower bounds of the sequence.

Note-

1. Every convergent sequence is bounded.
2. A sequence can not converges to more than one limit.
3. Every convergent sequence is bounded and has a unique limit.

Convergent sequence- A sequence Sn is said to be convergent when it tends to a finite limit. That means the limit of a sequence Sn will be always finite in case of convergent sequence.

Divergent sequence- when a sequence tends to ±∞ then it is called divergent sequence.

Oscillatory sequence- When a sequence neither converges nor diverges then it is called oscillatory sequence.

Note- A sequence which neither converges nor diverges, is called oscillatory sequence.

A sequence converges to zero is called null.

Example-1:  consider a sequence 2, 3/2 , 4/3 , 5/4, …….. HereSn =  1 + 1/n

Sol. As we can see that the sequence Sn is convergent and has limit 1.

According to def.

Example-2: consider a sequence Sn=  n² + (-1)ⁿ.

Sol. Here we can see that, the sequence Sn is divergent  as it has infinite limit.

Example: suppose , here the sequence is said to be oscillate. Because it is between -2 and 2.

Limit of a Sequence- A sequence <> is said to tend to limit “l” , when given any positive number ‘ϵ’ , however small , we can always find a integer ‘m’ such that | – l| <ϵ , for every for all, n≥m , and we can define this as follows,

Example: If    , then the limit of will be,

  = =  = ½

Hence the limit of the sequence is 1/2.

Some important limits to remember-

1.

2.

3.

4.

Key takeaways:

1. A sequence is said to be bounded when it is bounded above and bounded below.
2. A sequence Sn is said to be convergent when it tends to a finite limit.
3. When a sequence tends to ±∞ then it is called divergent sequence.
4. A sequence converges to zero is called null.

2.3 Limit Theorems, Monotone Sequences

Limit theorems:

Def: A sequence X = of real numbers is said to be bounded if there exists a real number M > 0 such that | for all

Thus, the sequence (xn) is bounded if and only if the set of its values is a bounded subset of R.

Theorem: A convergent sequence of real numbers is bounded

Proof:

Suppose that lim = x and let e := 1. Then there exists a natural number

K = K(1) such that | < 1 for all n K. If we apply the Triangle Inequality with

n K we obtain

If we set

Then it follows that | for all

Theorem: If X =() is a convergent sequence of real numbers and if   0 for all n N, then x = lim()

Proof:

Suppose the conclusion is not true and that x < 0; then e:= -x is positive. Since X converges to x, there is a natural number K such that

In particular, we have < x + e = x + (-x) = 0. But this contradicts the hypothesis that  0 for all n N. Therefore, this contradiction implies that x 0.

Theorem If X =() and Y =()  are convergent sequences of real numbers and

If for all n N, then

Proof:

Let so that Z := and 0 for all n N. It follows

So that

Note- The next result asserts that if all the terms of a convergent sequence satisfy an inequality of the form , then the limit of the sequence satisfies the same inequality. Thus if the sequence is convergent, one may ‘‘pass to the limit’’ in an inequality of this type.

Theorem If X =() is a convergent sequence and if for all n 2 N, then

Proof:

Let Y be the constant sequence (b, b, b,…). Above theorem  implies that . Similarly one shows that a lim X.

Monotonic sequence:

A sequence is said to be monotonic increasing if and monotonic decreasing if .

It is said to be monotonic if it is either increasing or decreasing.

A sequence is strictly increasing if and strictly decreasing if .

Here note that the monotonic sequences never oscillate. They either converge of diverge.

Note- A monotonic increasing sequence converges to its least upper bound and a monotonic decreasing to the greatest lower bound.

Theorem: Every monotonic increasing sequence which is not bounded above, diverges to +

Proof:

Let be a monotonic increasing sequence, not bounded above. Let G be any number however large.

Since the sequence is unbounded and monotonic increasing, a positive integer m such that

Hence,

Example: Show that the sequence where

Sol:

Here

Now

Hence the sequence is monotonic increasing.

Again

Which means-

The sequence is bounded.

Hence the sequence, being bounded and monotonic increasing, is convergent.

Key takeaways:

1. A sequence X = of real numbers is said to be bounded if there exists a real number M > 0 such that | for all
2. A convergent sequence of real numbers is bounded
3. A monotonic increasing sequence converges to its least upper bound and a monotonic decreasing to the greatest lower bound.
4. Every monotonic increasing sequence which is not bounded above, diverges to +

2.4 Divergence Criteria, Bolzano Weierstrass theorem for Sequences

Divergence Criteria:

If a sequence X = ( of real numbers has either of the following properties, then X is divergent.

(i) X has two convergent subsequences X’ = and whose limits are not equal.

(ii) X is unbounded.

Example:

1. The sequence X = is divergent.
2. The sequence (1, ½, 3, 1/4.,….) is divergent.
3. The sequence S := (sin n) is divergent

Bolzano-Weierstrass theorem:

Statement- Every sequence has a limit point.

Proof:

Let be a bounded sequence and S = { be its range.

Since the sequence is bounded, therefore its range set S is also bounded.

There are two possibilities:

1. S is finite
2. S is infinite

Now take first case- If S is finite then there must exist at least one member , such that for an infinite number of values of n. This means that every neighbourhood, contains for an infinite number of values of n.

Thus is a point of the sequence

Case-2: When S is infinite, since it is bounded. It has by Bolzano-Weierstrass theorem, at least one limit point (say

Again, since is a limit point of S therefore every neighbourhood contains an infinity of members of S.

Which means for an infinity of values of n. Hence is a limit point of the sequence.

Note- the converse of the theorem is not true, for there do exist unbounded sequences having only one real limit point.

Key takeaways:

1. If a sequence X = ( of real numbers has either of the following properties, then X is divergent.
2. Every sequence has a limit point.

2.5 Cauchy sequence, Cauchy’s Convergence Criterion

Cauchy Sequence:

A sequence {} in a metric space (X, d) is said to be a Cauchy sequence if for every > 0 there is an integer N such that d(, ) < for all n, m N.

Definition:

Let E be a non-empty subset of a metric space (X, d), and let S = {d(p, q) : p, q E}. The diameter of E is sup S.

If {} is a sequence in X and if En consists of the points , , . . ., it is clear that {} is a Cauchy sequence if and only if

Cauchy Sequences and Convergent Sequences

Theorem:

(a) In any metric space X, every convergent sequence is a Cauchy sequence.

(b) If X is a compact metric space and if {} is a Cauchy sequence in X then {} converges to some point of X.

(c) In every Cauchy sequence converges.

Cauchy’s Principle of Uniform Convergence:

The necessary and sufficient condition for a sequence of functions () defined on A to converge uniformly on A is that for every > 0, there exists a positive integer m such that

Proof: Condition is necessary. It is given that () is uniformly convergent on A.

Let – f uniformly on A. Then given  > 0, there exists a positive integer m such that

and

triangular inequality)

  for n>km and

This proves the necessary part. Now we prove the sufficient part.

Condition is sufficient: It is given that for every   > 0, there exists a positive integer m such that

|(x) – (x)| <   for n > k  m and for all x in A. But by Cauchy’s principle of convergence of sequence of real numbers, for each fixed point x of A, the sequence of numbers ((x)) converges.

In other words, () is pointwise convergent say to f on A. Now for each   > 0, there exists a positive integer m such that

|(x) – (x)| < for  n>k

Fix k and let n ® ¥. Then (x)  ® f(x) and we get

This is true for k m and for all x in A. This shows that (f) is uniformly convergent to f on A, which proves the sufficient part.

Example: If , { are two Cauchy’s sequences, then the sequences { and { if ( are also Cauchy’s sequences.

Example: Show that the sequence where

Cannot converge.

Sol:

Lets take and p = n = m In Cauchy’s general principal of convergence, so that

Thus

Which is the contradiction, hence the sequence can not converge.

Key takeaways:

A sequence {} in a metric space (X, d) is said to be a Cauchy sequence if for every > 0 there is an integer N such that d(, ) < for all n, m N.

2.6 Infinite series, convergence and divergence of infinite series

Series

Infinite series- If  is a sequence, then is called the infinite series.

It is denoted by .

Examples of infinite series-

Convergent series - Suppose n→∞, Sn→ a finite limit ‘s’, then the series Sn is said to be convergent.

We can denote it as,

Divergent series- When Sn tends to infinity then the series is said to be divergent.

Oscillatory series- When Sn does not tends to a unique limit (finite or infinite) , then it is called Oscillatory series.

Properties of infinite series –

1. The convergence and divergence of an infinite series is unchanged addition or deletion of a finite number of term from it.

2. If positive terms of convergent series change their sign, then the series will be convergent.

3. Let   converges to s , let k be a non-zero fixed number  then converges to ks.

4. Let converges to ‘l’ and converges to ‘m’.

Example-1: check whether the series is convergent or divergent. Find its value in case of convergent.

Sol.  As we know that,

Sn = 

Therefore,

Sn = 

Now find out the limit of the sequence,

  = ∞

Here the value of the limit is infinity, so that the series is divergent as sequence diverges.

Example-2: check whether the series  is convergent or divergent. Find its value in case of convergent.

Sol. The general formula for this series is given by,

Sn =  = )

We get,

) = 3/2

Hence the series is convergent and its values is 3/2.

Example-3: check whether the series  is convergent or divergent.

Sol. The general formula can be written as,

We get on applying limits,

) = 3/4

This is the convergent series and its value is 3 / 4

Example-4: check whether the following series is convergent or divergent. If convergent, find its value.

Sol. n’th term of the series will be,

General properties of series

The general properties of series are-

1. The nature of a series does not change by multiplication of all terms by a constant k.

2. The nature of a series does not change by adding or deleting of a finite number of terms.

3. If two series and are convergent, then is also convergent.

Exampple-1: Prove that the following series is convergent and find its sum.

Sol. Here,

And

Hence the series is convergent and the limit is ½.

Example-2: Test the convergence of the series-

Sol. Here we can see that the given series is in geometric progression

As its first term is 1 and common ratio is ½.

Then we know that the sum of n terms of a geometric progression is-

Hence the limit will be-

So that the series is convergent.

Key takeaways-

1. Suppose n→∞, Sn→ a finite limit ‘s’, then the series Sn is said to be convergent.
2. When Sn tends to infinity then the series is said to be divergent.
3. The convergence and divergence of an infinite series is unchanged addition or deletion of a finite number of term from it.
4. If positive terms of convergent series change their sign, then the series will be convergent
5. Let converges to ‘l’ and converges to ‘m’.
6. The nature of a series does not change by multiplication of all terms by a constant k.
7. The nature of a series does not change by adding or deleting of a finite number of terms.
8. If two series and are convergent, then is also convergent.

2.7 Cauchy Criterion

Definition A sequence X = ( of real numbers is said to be a Cauchy sequence if for every > 0 there exists a natural number H( such that for all natural numbers n, m , the terms ; , satisfy | < .

The significance of the concept of Cauchy sequence lies in the main theorem of this section, which asserts that a sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Example: The sequence (1/n) is a Cauchy sequence.

Sol:

If > 0 is given, we choose a natural number H = H() such that H > 2/. Then if

m; n H, we have 1/n  1/H /2 and similarly 1/m < /2. Therefore, it follows that

If m, n H, then

Since > 0 is arbitrary, we conclude that (1/n) is a Cauchy sequence.

Lemma: Lemma If X = ( is a convergent sequence of real numbers, then X is a Cauchy sequence.

Proof:

If x := lim X, then given > 0 there is a natural number K( such that if n K( then | < =2. Thus, if H(: = K( and if , then we have

Since > 0 is arbitrary, it follows that () is a Cauchy sequence.

Lemma: A Cauchy sequence of real numbers is bounded.

Proof:

Let X = ( be a Cauchy sequence and let := 1. If H := H(1) and n H, then | < 1. Hence, by the Triangle Inequality, we have | for all

n H. If we set

Then it follows that |M for all n N.

Cauchy Convergence Criterion A sequence of real numbers is convergent if and only if it is a Cauchy sequence

We have seen, above in Lemma, that a convergent sequence is a Cauchy sequence.

Conversely, let X = ( be a Cauchy sequence; we will show that X is convergent to some real number. First we observe from Lemma that the sequence X is bounded.

Therefore, by the Bolzano-Weierstrass Theorem, there is a subsequence X’ = ( of X that converges to some real number x.

We shall complete the proof by showing that X converges to .

Since X = (  is a Cauchy sequence, given > 0 there is a natural number H( /2) such that if n, m H( then

Since the subsequence X’= (

Converges to , there is a natural number K

Belonging to the set {n1; n2; . . .} such that

Since K , it follows from (1) with m = K that

Therefore, if n , we have

Since > 0 is arbitrary, we infer that lim( = . Therefore the sequence X is convergent.

2.8 Tests for convergence: Comparison test

Positive term series-

If all the terms in an infinite series are positive after few negative terms , then the series said to be a positive term series.

Suppose,

-22-65+ 55 +69 99+125+………….is a positive term series.

If we remove these negative terms, then the nature of the series does not change.

Comparison test-

Statement-

Suppose we have two series of positive terms and then,

, where k is a finite number, then both series converges or diverges together.

Proof- we know that by the definition of limits, there exist a positive number epsilon(ε)

Which is very small. Such that

According to definition(comparison test)

||<ε for n>m, that means

k-ε<    for n>m

Ignoring the first m terms of the series,

We get

k-ε<    for n>m  for all n  ………………..(1)

There will be two cases-

Case-1: is convergent, then

() = r (say), where r is finite number

From (1),

()<() =

Therefore  is also convergent.

Case-2: is divergent, then

()→∞   …………………………..(2)

From eq. (1)

Then

()<()

From(2)

()→∞

Hence, is also divergent.

Example: Test the convergence of the following series.

Sol. We have  

First we will find  and the

And

Here, we can see that, the limit is finite and not zero,

Therefore, and converges or diverges together.

Since is of the form  where p = 2>1

So that, we can say that,

is convergent, so that will also be convergent.

Example: Test the convergence of the following series-

Sol. Here we have the series,

Now,

Now compare

We can see that the limit is finite and not zero.

Here  and converges or diverges together since,

is the form of here p = 1,

So that,

is divergent then is also divergent.

Example: Show that the following series is convergent.

Sol.

Suppose,

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is convergent. So that is also convergent.

Example: Test the series:

Sol. The series is,

Now,

Take,

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is divergent. (p = ½)

So that is divergent.

Key takeaways-

Suppose we have two series of positive terms and then,

, where k is a finite number, then both series converges or diverges together.

2.9 Limit Comparison test, Ratio Test

Limit comparison test:

Suppose that X:= ( and Y : = are strictly positive sequences and suppose that the following limit  exists in R:

1. If is convergent if and only if is convergent.
2. If is convergent if and only if is convergent.

Example: the series

It is clear that the inequality

Is valid. Since the series is convergent.

We can apply the Comparison Test to obtain the convergence of the given series.

Example:  the series

If the inequality

Were true, we could argue as in first example. However, (1) is false for all n N. The reader can probably show that the inequality

Ratio test:

Statement- suppose is a series of positive terms  such that then,

1. If k<1, the series will be convergent.

2. If k>1, then the series will be divergent.

Proof:

Case-1:

We know that from the definition of limits, it follows,

∃m>0 and l(0<l<1)∋

But,

Is the finite quantity. So is convergent.

Case-2:
 

There could be some finite terms in starting which will never satisfy the condition,

In this case we can find a number ‘m’,

∋

Ignoring the first ‘m’ terms, if we write the series as

We have , in this case

which

So that is divergent.

Example: Test the convergence of the series whose n’th term is given below-

n’th term =

Sol. We have

  and

By D’Alembert ratio test,

So that by D’Alembert ratio test, the series will be convergent.

Example: Test for the convergence of the n’th term of the series given below-

Sol. We have,

Now, by D’Almbert ratio test converges if and diverges if 

At x = 1, this test fails.

Now, when x = 1

The limit is finite and not zero.

Then by comparison test, converges or diverges together.

Since is the form of , in which

Hence diverges then will also diverge.

Therefore in the given series converges if x<1 and diverges if x≥1.

Key takeaways-

1. If is a series of positive terms  such that then,

1. If k<1, the series will be convergent.

2. If k>1, then the series will be divergent.

2.10 Cauchy’s nth root test

Let be a series of positive terms and let

Then is convergent when  l<1 and diverges when l >1.

Proof:  case-1:

Or

Since,

Is a geometric series with common ratio <1  so that the series will be convergent.

Case- 2:

By the limit concept, we can find a number,

That means

After 1st ‘r’ terms, each term is > 1

So that the series is divergent.

Example: Test the convergence of the series whose nth term is given below-

Sol.

By root test is convergent.

Example: Test the convergence of the series whose nth term is given below-

Sol.

By root test is convergent.

Example: show that the following series is convergent.

Sol.

By root test is convergent.

Example: Test the convergence of the following series:

Sol. Here, we have,

Therefore the given series is convergent.

Key takeaways-

If be a series of positive terms and let

Then is convergent when  l<1 and diverges when l >1.

2.11 Integral test

Statement- A series with positive terms f(1) + f(2) + f(3) + ……… f(n)+……..where, if we increase n, then f(n) decreases-

Converges or diverges according to the following integral-

Is finite or infinite.

Proof: in the given figure, the area under the curve x=1 to x = n+1 lies between the sum of the areas of small rectangles and the sum of the areas of large rectangles.

f(1) , f(2) , f(3) represent the height of the rectangles,

f(1) + f(2) + f(3) + ……… f(n) ≥f(2) + f(3) + ……… f(n+1)

As n , from the second inequality that  the integral has a finite value then is also finite so that is convergent.

Same as the integral is infinite, then from the first inequality that so that the series is divergent.

Example:  Test the series for its convergence.

Sol. Let,    f(x) = 

=

Here we notice that, by Cauchy’s integral test, the series is divergent.

Example: Test the series by integral test-

Sol. Here is positive and decreases when we increase n,

Now apply integral test,

Let,

X = 1, t = 5 and x = ∞ , t = ∞,

Now,

So by integral test,

The series is divergent.

Example: Test the series by integral test-

Sol. Here  decreases as n increases  and it is positive.

By using integral test,

= 

We get infinity,

So that the series is divergent.

2.12 Alternating series, Leibniz test

First we will understand about alternating series,

When the terms in a series are alternately negative , then the series is known as alternating series.

For example:

Leibnitz test for the convergence of an alternating series:

The given series will be convergent if it follows the rules given below-

1. Each term should be numerically less than its preceeding term

Example: Test the convergence of the following alternating series:

Sol.  Here in the series, we have

First condition-

So that,

|| > ||

That means, each term is not numerically less than its preceeding terms.

Now second condition-

Both conditions are not satisfied for convergence.

Hence the given series is not convergent. It is oscillatory.

Example: Test the following series for the convergence-

Sol. We have the given series, now

We see that, this is an alternating series,

Here,

Also,

By Leibinitz’s test the series is convergent.

Key takeaways-

1. When the terms in a series are alternately negative , then the series is known as alternating series.
2. The given series will be convergent if it follows the rules given below-

i. Each term should be numerically less than its preceeding term

Ii.

2.13 Absolute and Conditional convergence

A series is said to be absolutely convergent if the series is convergent.

For example- suppose the following series,

By p- series test, we can say that  is convergent.

Hence is absolutely convergent.

Note-

1. If the series has positive terms and it is convergent then this series will be absolutely convergent too.
2. An absolute convergent series will be convergent but the converse may not be true.

Example: Test for absolute convergence:

Sol. Let the series is ,

By ratio test,

is convergent, if |x|<1.

is absolutely convergent if |x|< 1.

Example: Show that the series is absolutely convergent.

Sol. We have,

|| =     and      || =

The first condition and second conditions are-

1.  ||<||

2.

Both the conditions are satisfied.

So that we can say that by Leibnitz’s rule, the series is convergent.

The series is also convergent by p-test as p = 2 > 1.

Hence the given series is absolutely convergent.

Example: Test the convergence/Divergence of the series:

Sol. Here the given series is alternately negative and positive , which is also a geometric infinite series.

1. Suppose,

S =

According to the conditions of geometric series,

Here , a = 5 , and common ratio (r) = -2/3

Thus, we know that,

So ,

Sum of the series is finite, which is 3.

So we can say that the given series is convergent.

Now.

Again sum of the positive terms,

The series is geometric, then

A = 5 and  r = 2/3 , then

Sum of the series,

Sum of the series is finite then the series is convergent.

Both conditions are satisfied, then the given series is absolutely convergent.

Key takeaways-

1. A series is said to be absolutely convergent if the series is convergent.
2. If the series has positive terms and it is convergent then this series will be absolutely convergent too
3. An absolute convergent series will be convergent but the converse may not be true.

References:

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. Advanced engineering mathematics, by HK Dass

3. Mathematical analysis by Dr. Anju panwar

4. Real analysis by Dr. Sachin Kaushal

5. Real analysis by Ajit kumar & S.Kumaresan

6. Principles of real analysis by S.C Malik

7. A BASIC COURSE IN REAL ANALYSIS- AJIT KUMAR