Unit – 2
Content
First we will understand the concept of sequence
Sequence A sequence <Sn> is said to tend to limit “l” , when given any positive number ‘ϵ’ , however small , we can always find a integer ‘m’ such that  Sn – l <ϵ , for every for all, n≥m , and we can define this as follows,
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.
Note a sequence which neither converges nor diverges , is called oscillatory sequence.
A sequence converges to zero Is called null.
Example1: Show that
Sol. For ϵ> 0 , we need to find an integer N such that ,

For all n≥N , since i=1 we need 1/n<ϵ , so that we choose any N>1 / ϵ>0 , for all n≥N , we find that
Here i = 1 , then
Example2: 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.
Example3: consider a sequence Sn= n² + (1)ⁿ.
Sol. Here we can see that, the sequence Sn is divergent as it has infinite limit.
Let {an} , {bn} and {cn} be real sequences such that an ≤ bn≤ cnfor all n . If
Then
Proof:
The hypotheses says that,
0 ≤ bn – an ≤cn – an , for all n
So that,
(cn – an ) = L – L = 0
And ,
(bn – an ) = 0
Thus,
bn=( bn – an + an)
= ( bn – an) + an
= 0 + L
= L
This is the proof of Sandwich theorem.
Example1: show that
= 1
Sol.
Let Sn =  1 for all n >0 , here Sn≥ 0 for all positive n , therefore
For all n>0 ,
For all n > 1
From Sandwich theorem, we get,
Example2: show that
For each constant b > 0
Sol.
Lets start with b ≥ 1 and fix an integer, n ≥b .
By sandwich theorem ,
For b such that 0< b < 1 , we have 1 / b >1 , hence
The function f is continuous at a if and only if it satisfies the following property
For every sequences, {Sn},
If
It says in order to f to be continuous , it is necessary and sufficient that any sequence {Sn} converging to a must force the sequence (f(Sn)) to converges to f(a).
Theorem1: Let f and g both are continuous at a .then f + g and f.g are continuous at a.
Proof:
Suppose there is sequence {Sn} which converges to ‘a’
Since f is continuous at a , then by the above statement,
)
Similarly,
If g is continuous at a , then
And ,
Similarly we can prove for f.g
Hence proved.
Suppose anbe a sequence then,
1. The sequence is increasing when , an<an+1
2. The sequence is decreasing , when , an >an+1
3. Whenan is increasing sequence or decreasing sequence we call It a monotonic.
4. When there is a number m such that m ≤an for every n we can say that the sequence is bounded below
5. When there is a number m such that m ≥an for every n we can say that the sequence is bounded above.
When a monotonic sequence which is bounded above and bounded below known as bounded monotonic sequence.
For the better understanding , we will do some examples
Example1: check whether the sequence is bounded monotonic sequence or not.
Sol. This is the decreasing sequence, hence we can say that this is monotonic for every n.
As we can notice here that the terms in this sequence will be either zero or negative , so that the sequence is bounded above.
Suppose m = 0 then,
This sequence is not bounded below ,
Hence we can not say that the sequence is bounded.
Example2: check whether the sequence is bounded monotonic sequence or not.
Sol. Here the value of the terms in this sequence will lie between 1 to +1 so the sequence is neither increasing and not decreasing , that means the sequence is not monotonic.
But , it is bounded above by 1 and bounded below by 1, then we can say that the sequence is bounded.
This is divergent sequence too.
Example3: check whether the sequence is bounded monotonic sequence or not.
Sol. This is the decreasing sequence , so that the this is a monotonic sequence.
All the terms are positive is this sequence so that it is bounded below by 0.
And the sequence is decreasing , that means the first term will be the largest one.
So that it is bounded above by 2 / 25. ( put n = 5)
Thus we can conclude that the sequence is bounded monotonic equence.
Infinite series If Un is a sequence , then U1 + U2 + U3 +………………+Un is called the infinite series.
It is denoted by .
Examples of infinite series
Covergent series  suppose n→∞ , Sn→ a finite limit ‘s’ , then the seiresSn is said to be convergent .
We can denote it as,
Divergent series whenSn tends to infinity then the series is said to be divergent.
Properties of infinite series –
1. The convergence and divergence of an infinite series is unchanged addition od 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 nonzero fixed number then converges to ks.
4. Let converges to ‘l’ and converges to ‘m’.
Example1: 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.
Example2: 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.
Example3: check whether the series is convergent or divergent.
Sol. The general formula can be written as,
We get on applying limits,
This is the convergent series and its value is 3 / 4
Suppose there are infinite series having infinite number of terms such as,
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s4 = a1 + a2 + a3 + a4
s5 = a1 + a2 + a3 + a4 + a5
……………………………
……………………………..
Sn = a1 + a2 + a3 + a4 + a5 + ………..+ an =
Sn is called the sum of all series if we combine them.
Here {Sn} is the sequence
=
this is called the infinite series , that starts with i = 1
A series is called harmonic series. This series can be proved divergent by using integral test.
Progression of the series can be represented by,
Example1: show that the following series is divergent
Sol. To prove this series divergent , we follow the steps below,
As we know that will always diverge.
And if we multiply a divergent series with a constant then the resulting series will also be divergent.
Hence we can say that theeries will be divergent.
Example2: Show that the harmonic series is divergent.
Sol. First we will expand its three terms ,
= 1+ +
We get,
Here if we subtract a finite number from a divergent series , then the resulting series will not be affected by this number.
Therefore we can say that is the divergent series which is also a harmonic series.
A positive term series
∅(1) + ∅(2) +∅(3) + ∅(4)+ …………..∅(n) + ………
Where ∅(n) decreases as n increases , is convergent or divergent according as the integral is finite or infinite.
Proof: suppose, Sn = ∅(1) + ∅(2) +∅(3) + ∅(4)+ …………..∅(n)
Here n = 1,2,3,4,………n , n+1, ……..
Hence the total area under the curve ( see fig) lies between the sum of areas of all interior rectangles and sum of the area of all exterior rectangles
So that,
As n tends to infinity , then limit of Sn will be finite or infinite according as is finite or infinite.
Example1: determine the following series is convergent or divergent.
Sol. By using integral test
= ∞
By the integral test, given series is divergent as
Example2: 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.
Example3: 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.
The infinite series,
Is convergent , when
(1) convergent when p >1
(2) divergent if p <= 1
Proof;
Let p>1
Similarly,
And so on.
Adding we get,
The right hand side of the above inequality is an infinite geometric series , the sum of the geometric series is finite hence the given series is convergent.
Example1: Test the convergence of the following series:
Sol.
Here we take,
Which not zero and finite ,
So by comparison test , and both converges or diverges, but by pseries test
Is convergent .so that is convergent.
Example2: Test the convergence of the given series:
Sol. Here,
Take,
is divergent by pseries test. (p = 0 <1)
Example3: Test the convergence of the series:
Sol.
By pseries test,
Is convergent ( p = 3>1).
Let , and be two series of positive terms and let be convergent ,
Then , converges ,
1. If,
Or
2.
Or
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.
Which is finite and not zero.
By comparison test and converge or diverge together.
But,
Is divergent. ( p = ½)
So that is divergent.
and
Be the series of positive terms.
Then
1. Both series converge or diverge for –
2. The series converges if –
3. Diverges for
Example: Test the convergence of the series:
Sol. Let
For all n> 0we show that An has the same order of magnitude as 1/n².
We put accordingly
For all n>0, then
As n tends to infinity , thus
Then the series converges by limit comparison rule.
Example: Test the series for convergence.
Sol. For all n>= 0 let,
And let,
For all n>0, since
It follows,
By limit comparison test , the series is divergent
Example: Test the series
Sol. Take,
And,
For all n>0, we get
Thid is a divergent series.
Let be a series of positive terms and
Then , the series,
(1) is convergent if k < 1
(2) is divergent if k>1
Proof:
Case1:
We know that from the definition of limits ,it follows,
But,
Is the finite quantity. So is convergent.
Case2:
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
So that is divergent.
Let be a series of positive terms and
Then,
(1) if k >1 , is convergent and
(2) if k < 1 , then is divergent
(3) this test fails if k = 1.
Proof: let us consider the series,
Case1: in this case,
We choose a number ‘p’ for all k > p >1 , comparing thr series with which is divergent ,
We get will converge if after some fixed numver of terms ,
That means,
If k >p , which Is true . Hence is convergent.
We can prove the second case similarly.
Example1: Test the convergence of the following series.
Sol. Neglecting the first term the series can be written as,
So that,
By ratio test converges if x<1 and diverges if x>1, but if x = 1 the the test fails,
Then ,
By Raabes’s test converges hence the given series is convergent when x≤ 1 and divergent If x >1.
Example2: Test the convergence of the series,
Sol. As we will neglect the first term, we get
By ration test is convergent when x<1 and divergent when x>1, when x= 1,
The rario test fails, then
By Rabee’s test is convergent , hence the given series is convergent when x≤ 1 and divergent If x >1.
Example3: Test the nature of the following series:
Sol.
By ration test is convergent when (x/4)<1 and divergent when x>4, when x= 4,
The rario test fails, then
By Rabee’s test is convergent , hence the given series is convergent when x<4 and divergent If x >=4.
Let be a series of positive terms and let
Then is convergent when l<1 and diverges when l >1.
Proof: case1:
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.
Definition the series,
Is alternating if each bj is positive.
Test the alternating series is convergent if {bn} is a non – increasing sequence of positive terms converging to zero.
Proof: for each integer , n ≥ 0, let
We need to show that the sequence {Sn} converges. We will do this as follows
First we have,
Note
Here
Is bounded above b0, it is non decreasing because for all n>0, we have
Which is the proof.
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 if the series has positive terms and it is convergent then this series will be absolutely convergent too.
An absolute convergent series will be convergent but the converse may not be true.
Conditional convergence:
If the series is divergent and is convergent then is aid to be conditionally convergent
Example: Test the series for absolute/conditional convergence.
Sol. The given series is an alternating series of the form,
Here,
1.
2.
And,
Hence bt Leibnitz’s test , the given series is convergent ,
But,
Is divergent by pseries test.
So that, the given series is conditionally convergent.
Example: Test for the convergence of the series:
Sol. The given series is,
By ratio test series converges, so that the series is convergent.
Example: Test for absolute convergence:
Sol. Let the series is ,
By ratio test,
is convergent , if x<1.
is absolutely convergent if x< 1.
A series
Where An are all the constant is a power series in x.
It may converges for some values of x,
We neglect first term here,
Suppose it is equals to kx,
Where,
By ratio test, the series converges, when kx<1
That means it converges
The interval ( 1 /k , 1/k) is known as the interval of convergence of the power series.
Example: Find the interval of convergence.
Sol.
We can see that, by ratio test , the given series converges when x<1
That means x = (1,1)
When x = 1
Which is convergent by pseries test.
The interval of convergence of the given series is (1,1).
A power series is a variable z is defined as a series of the form
The power series converges absolutely for all z , it converges only when z = c , or there exists a positive number ‘r’ such that the power series converges absolutely if z – c< r, and diverges if zc>r ,
Then the ‘r’ is called the radius of convergence of a power series.
Example : Find the radius of convergence for the power series,
Sol. Set,
For each n>0.
If z<=1, then a<= 3/n²
Since
Converges the given series is absolutely convergent by the comparison test. Suppose therefore z>1
Since,
For all n>0, the sequence does not converges to 0 and the series diverges .
Hence the radius of the convergence is 1.
Expand by Maclaurin’s theorem,
Log sec x
Solution:
Let f(x) = log sec x
By Maclaurin’s Expansion’s,
(1)
By equation (1)
Prove that
Solution:
Here f(x) = x cosec x
=
Now we know that
Expand upto x6
Solution:
Here
Now we know that
… (1)
… (2)
Adding (1) and (2) we get
Show that
Solution:
Here
Thus
Taylor’s Series Expansion:
a) The expansion of f(x+h) in ascending power of x is
b) The expansion of f(x+h) in ascending power of h is
c) The expansion of f(x) in ascending powers of (xa) is,
Using the above series expansion we get series expansion of f(x+h) or f(x).
Expansion of functions using standard expansions
Expand in power of (x – 3)
Solution:
Let
Here a = 3
Now by Taylor’s series expansion,
… (1)
equation (1) becomes.
Using Taylors series method expand
in powers of (x + 2)
Solution:
Here
a = 2
By Taylors series,
… (1)
Since
, , …..
Thus equation (1) becomes
Expand in ascending powers of x.
Solution:
Here
i.e.
Here h = 2
By Taylors series,
… (1)
equation (1) becomes,
Thus
Expand in powers of x using Taylor’s theorem,
Solution:
Here
i.e.
Here
h = 2
By Taylors series
… (1)
By equation (1)