Module 3

Sequences and series

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.

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

p-test-

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).

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 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.

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-1: 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-2: 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

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 the series with which is divergent ,

We get will converge if after some fixed number 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 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

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.

A series of the form

Where all the ‘a’ are independent of x, is known as power series in x

Interval for convergence-

In this power series-

D’Almbert’s ratio test-

If , then by ratio test, the power series is convergent, when is less than 1.

Or in other way if |x|<1/ then the series converges and diverges for other values.

Thus the interval of convergence of power series is-

Example-1: If the series converges, then find the value of x.

Sol. Here

Then,

By D’Almbert’s ratio test the series is convergent for |x|<1 and divergent if |x|>1.

So at x = 1

The series becomes-

At x = -1

This is an alternately convergent series.

This is also convergent series, p = 2

Here, the interval of convergence is

Example-2: If the series converges, then find the value of x.

Sol. Here

Then,

By D’Almbert’s ratio test the series is convergent for ||<1 or |1-x|>1

Or

At x = 0, the series becomes- which is divergent harmonic series.

At x = 2, the series becomes-

It is an alternate series which is convergent by Leibnitz rule.

So that the series .

Taylor’s Theorem-

If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-

+ …….+ + ……..

Which is called Taylor’s theorem.

If we put x = a, we get-

+ …….+ + …….. (1)

Maclaurin’s Theorem-

If we put a = 0 and h = x then equation(1) becomes-

+ …….

Which is called Maclaurin’s theorem.

Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)

We get-

+ …….

Example-1: Express the polynomial in powers of (x-2).

Sol. Here we have,

f(x) =

Differentiating the function w.r.t.x-

f’(x) =

f’’(x) = 12x + 14

f’’’(x) = 12

f’’’’(x)=0

Now using Taylor’s theorem-

+ ……. (1)

Here we have, a = 2,

Put x = 2 in the derivatives of f(x), we get-

f(2) =

f’(2) =

f’’(2) = 12(2)+14 = 38

f’’’(2) = 12 and f’’’’(2) = 0

Now put a = 2 and substitute the above values in equation(1), we get-

Taylor’s theorem for functions of two variables-

Suppose f(x, y) be a function of two independent variables x and y. Then,

+ ……………

Maclaurin’s series is the special case of Taylor’s series-

When we put a = 0 and b = 0 (about origin) in Taylor’s series, we get-

+ ……………

Example-2: Expand f(x, y) = in powers of x and y about origin.

Sol. Here we have the function-

f(x , y) =

Here , a = 0 and b = 0 then

f(0 , 0) =

Now we will find partial derivatives of the function-

Now using Taylor’s theorem-

+………

Suppose h = x and k = y, we get

+…….

= +……….

Example-3: Find the Taylor’s expansion of about (1 , 1) up to second degree term.

Sol. We have,

At (1 , 1)

Now by using Taylor’s theorem-

……

Suppose 1 + h = x then h = x – 1

1 + k = y then k = y - 1

……

=

……..

Let’s first define the exponential functions-

A function which contains where e is the constant, called an exponential function.

Here ‘e’ is the exponent which has an approximate value 2.7183

The power series for

The value of can be defined in terms of the following power series:

If we obtain an actual value of by adding all the terms of the series, then the series is said to be convergent.

The more terms we take in the series, the closer will be the value of to its actual value.

Example: Find the value of 5, correct to five decimal places by using the power series for .

Sol. As we know that the exponential series is-

Here we get-

Now

Example: Expand as far as the term in .

Sol. We know that the power series for is-

Here we have to find-

So that-

On solving, we get-

As we know that,

And

By adding these equations-

And subtracting (2) from (1), we get-

Note-

Example-1: Verify

Sol. As we know that

And

Hence,

and

That means

The following function is a logarithmic function-

‘a’ is any value which is greater than 0, except 1

The natural logarithmic function is-

The value of e = 2.71828182….

Properties of logarithms-

1.

2.

3.

4. then x = y

Example: Simply the logarithm

Sol. We can write this as-

= 3

Example: Simplify

Sol. We know that-

So that

Sine series-

If it is required to expand the function f(x) as a sin series in 0 < x < c

Where,

Cosine series-

Example- Find a Fourier series for

;

Solution:

Here

;

Since f(x) is even function hence

It’s Fourier series is

… (1)

Where

Hence equation (1) becomes,

Example- Find half range cosine series of in the interval and hence deduce that

a)

b)

Solution:

Here

;

Hence it’s half range cosine series is,

… (1)

Where

Hence equation (1) becomes,

… (2)

Put x = 0, we get

Hence the result

Put we get,

i.e.

The Parseval’s theorem can be stated as-

Proof: we know that

Multiply this equation by f(x), we get-

Integrate from –c to +c. We get-

We get on integrating-

Which is the Parseval’s formula.

Example-1: Show that by using sine series for f(x) = 1 in 0<x<π

Sol. We know that the sine series is-

= 0 if n is even.

Then the sine series is-

So that-

Proved.

Example-2: Prove that for 0 < x <

1.

2.

Sol. 1. Half range series,

=

= 0 when n is odd

So that-

Now by Parseval’s formula-

So that-

References

1. G.B. Thomas and R.L. Finney, Calculus and Analytic geometry, 9th Edition, Pearson, Reprint, 2002.

2. Erwin kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, 2006.

3. Veerarajan T., Engineering Mathematics for first year, Tata McGraw-Hill, New Delhi, 2008.

4. Ramana B.V., Higher Engineering Mathematics, Tata McGraw Hill New Delhi, 11thReprint, 2010.

5. D. Poole, Linear Algebra: A Modern Introduction, 2nd Edition, Brooks/Cole, 2005.

6. N.P. Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint, 2008.

7. B.S. Grewal, Higher Engineering Mathematics, Khanna Publishers, 36th Edition, 2010.