Unit - 2

Sequences and Subsequences

Q1) Define sequence.

A1)

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.

Q2) What are finite and infinite sequences?

A2)

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.

Q3) Define sub-sequences.

A3)

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.

Q4) What are bounded above and bounded below sequences?

A4)

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,

Q5) Define convergent and divergent sequences.

A5)

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.

Q6) consider a sequence 2, 3/2 , 4/3 , 5/4, …….. Here Sn =  1 + 1/n

A6)

According to def.

Q7) Define limit of a sequence.

A7)

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,

Q8) Prove that A convergent sequence of real numbers is bounded

A8)

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

Q9) Prove that If X =() is a convergent sequence of real numbers and if   0 for all n N, then x = lim()

A9)

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.

Q10) Define monotonic sequence.

A10)

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.

Q11) Prove that Every monotonic increasing sequence which is not bounded above, diverges to +

A11)

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,

Q12) Show that the sequence where

A12)

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.

Q13) What is the divergence criteria?

A13)

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.

Q14) State and prove Bolzano-Weierstrass theorem.

A14)

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.

Q15) What is Cauchy’s sequence?

A15)

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

Q16) Explain the Cauchy’s sequence of uniform convergence.

A16)

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.

Q17) check whether the series  is convergent or divergent. Find its value in case of convergent.

A17)

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.

Q18) check whether the following series is convergent or divergent. If convergent, find its value.

A18)

n’th term of the series will be,

Q19) Prove that a Cauchy sequence of real numbers is bounded.

A19)

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

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

Convergent.

Q20) Test the convergence of the following series.

A20)

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.

Q21) Test for the convergence of the n’th term of the series given below-

A21)

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.

Q22) Test the convergence of the series whose nth term is given below-

A22)

By root test is convergent.

Q23) Test the convergence of the following series:

A23)

Here, we have,

Therefore the given series is convergent.

Q24) Test the series for its convergence.

A24)

Let,    f(x) = 

=

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

Q25) Test the series by integral test-

A25)

Here  decreases as n increases  and it is positive.

By using integral test,

= 

We get infinity,

So that the series is divergent.

Q26) Test the convergence of the following alternating series:

A16)

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.

Q27) Test the convergence/Divergence of the series:

A27)

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

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.