UNIT 1

SEQUENCES AND SERIES

1.1           Sequences

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.

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-

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, …….. Here Sn =  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.

1.2           Series

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

It is denoted by .

Examples of infinite series-

Covergent series -  suppose n→∞ , Sn → a finite limit  ‘s’  , then the seires 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 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 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,

]  = ½

1.3           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 1/2 .

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.

1.4           Series of positive terms

A series is said to be a positive term series if it has all positive terms after few negative terms.

Like-   

The series remains unchanged by neglecting the negative terms.

Necessary condition for convergence-

If a positive term series is convergent, then then

Suppose as is convergent.

Also

But

So that

1.5           Comparison tests

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 comapare 

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.

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is divergent. ( p = ½)

So that is divergent.

1.6           tests of Convergence

The series will be

1. Convergent if the value of p is :   p>1

2. Divergent if the value of p is:    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

Example-1: 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 p-series test

Is convergent . So that  is convergent.

Example-2: Test the convergence of the given series:

Sol.  Here,

Take,

is divergent by p-series test. (p = 0 <1)

Example-3: Test the convergence of the series:

Sol.

By p-series test,

Is convergent ( p = 3>1).

1.7           D’Alembert’s 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: Proof:

Case-1:

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

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

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.

1.8           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.

1.9           Raabe’s test

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

, then

1. The series is convergent if k>1

2.the series is divergent if k<1.

3. Test fails at k = 1

Proof: let us consider the series,

Case-1: 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.

Example-1: 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, now

By Raabes’s test  converges hence the given series is convergent when |x|≤ 1 and divergent If |x| >1.

Example-2: 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.

Example-3: 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.

Example: Check for the convergence of the following series:

Where x>0.

Sol.  Here , we have the series,

Neglecting the first term , we get

We get the limit,

By ratio test, is convergent if and divergent if 

But at   , the ratio test fails,

Now we will apply Raabe’s test

So that by Raabe’s test , the series converges.

Therefore is convergent is and divergent if

1.10     Logarithmic test

Statement-  Suppose is series with positive terms such that

Then,

1. If k>1 , then the series is convergent.

2. If k<1 , then the series is divergent.

Proof: if k>1

Compare with , if k>p>1 then converges.

Taking log on both sides, we get

k>p which is true as k>p>1

Hence is convergent.

When p<1,

Similarly when p<1, then is divergent.

At p = 1 , then this test fails.

Example: Test the convergence of the following series:

Sol.  We have the series,

Here , And

Which gives,

  , the series is convergent.

If   , the series is divergent.

.

Thus the series is divergent.

1.11     Alternating series

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.

1.12     Series of positive and negative terms

1. If the series of terms

Be such that the series ||+||+||+…+||+…..

Is convergent, then the series is said to be absolutely convergent.

2. If is divergent but is convergent, then said to be conditionally convergent

1.13     Absolute convergence and Conditional convergence

Absolute convergent-

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 said to be conditionally convergent

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.

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 p-series 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.