1.1 Convergence and divergence | unit 1 infinite series

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M I

Unit - 1

Infinite series

1.1 Convergence and divergence

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 an 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.

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 tend 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 terms 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 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.

Key takeaways-

A function f: N

, where S is a non-empty set, is called sequence, for each nϵN.

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

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.

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

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

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

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

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

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

Let

converges to ‘l’ and

converges to ‘m’.

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

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

If two series

and

are convergent, then

is also convergent.

1.2 Comparison, D' Alembert's ratio 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.

Here we have

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,

= 1/2

Which is finite and not zero.

By comparison test and converge or diverge together.

But,

Is divergent. (p = ½)

So that is divergent.

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,

Therexists and

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’Alembert 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.

3. Suppose we have two series of positive terms and then,

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

1.3 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) represents 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.

1.4 Raabe’s, Logrithmic

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,

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,

are in A.P nth term

So that,

By ratio test converges if |x|<1 and diverges if |x|>1, but if |x| = 1 the true. test fails,

Then,

and

By Raabe’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 ratio 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 ratio test fails, then

By Rabee’s test is convergent, hence the given series is convergent when x<4 and divergent If x >=4.

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.

Key takeaways-

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.

2. 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.

1.5 Cauchy 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:

positive number

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.

1.6 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-

Each term should be numerically less than its preceding 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 preceding 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 Leibnitz’s test the series is convergent.

Key takeaways-

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

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

Each term should be numerically less than its preceding term

ii.

1.7 Absolute and conditional convergence

Absolute 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-

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. ||<||

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.

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. ||< ||

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,

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.

Key takeaways-

A series

is said to be absolutely convergent if the series

is convergent.

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.

1.8 Applications of Differentiation : Taylor's and Maclaurin's 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: 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-

Example: Expand sin x in powers of

Sol. Let f(x) = sin x

Then,

By using Taylor’s theorem-

+ ……. (1)

Here f(x) = sin x and a = π/2

f’(x) = cosx, f’’(x) = - sin x, f’’’(x) = - cos x and so on.

Putting x = π/2, we get

f(x) = sin x = = 1

f’(x) = cos x = = 0

f’’(x) = -sin x = = -1

f’’’(x) = -cos x = = 0

from equation (1) put a = and substitute these values, we get-

+ …….

= ………………………..

Example: Find the Taylor’s series for the function f(x) = 3x² - 6x + 5 in ascending powers of x-3.

Sol. We know that the Taylor’s theorem for the function f(x) in ascending powers of (x – a) is,

f(x) = f(a) + (x – a) f’(a) + + ………. …………. (1)

here,

f(x) = 3x² - 6x + 5 and a = 3. So that f(a) = 14

Now, we will find out the derivatives,

f’(x) = 6x – 6 and f’ (3) = 12

f’’(x) = 6 and f’’ (3) = 6

f’’’(x) = 0 and f’’’(x) = 14

put these values in Taylor’s theorem, we get

3x² - 6x + 5 = 14 + 12(x-3) + 6 +…………. Ans.

Example: Expand the function in ascending powers of (x – 1) by using Taylor’s theorem.

Sol. We know that the Taylor’s theorem for the function f(x) in ascending powers of (x – a) is,

f(x) = f(a) + (x – a) f’(a) + + ………. …………. (1)

here f(x) = and a = 1

now,

f’(x) = and f’(a) = e

f’’(x) = and f’’(a) = e

put these values in equation (1),

…………….

Take ‘e’ as common,

…………….}

Which is the required expansion.

Example: Expand in powers of x using Taylor’s theorem,

Solution:

Here

i.e.,

Here

h = 2

By Taylors series

… (1)

By equation (1)

Example: By using Maclaurin’s series expand tan x.

Sol.

Let-

Put these values in Maclaurin’s series we get-

Example: Expand by using Maclaurin’s series.

Sol.

Let

Put these values in Maclaurin’s series-

Key takeaways-

Taylor’s Theorem-

+ …….+ + ……..

2. Maclaurin’s Theorem-

+ …….

1.9 Asymptotes, Curvature Asymptotes

Definition-

“An asymptote of a curve is straight line at a finite distance from the origin, to which a tangent to the curve tends as the point of contact recedes to infinity.”

Or in other words-

A straight line is said to be an asymptote to a curve, if as a point P moves to infinity along the curve, the perpendicular distance of P from the straight line tends to zero.

Vertical asymptote-

A line x = a is called a vertical asymptote to the curve of the function f if or as x approaches a from either side. In vertical asymptotes, we have used limits to describe the behavior of f (x) as x approaches a.

Horizontal asymptote-

A line y = L is said to be a horizontal asymptote to the curve of the function f if i.e.,

Note-

To find the asymptotes parallel to x-axis, equate to zero the coefficient of the highest power of x in the equation, provided this is not nearly the constant.

To find the asymptotes parallel to y-axis, equate to zero the coefficient of the highest power of y in the equation, provided this is not nearly the constant.

Example: Find the asymptotes of the following curve-

Sol.

Here the highest power of x is and the coefficient is

We can find the asymptote parallel to x-axis as-

And the highest power of y is and its coefficient is .

The asymptotes parallel to y-axis, are given by-

Therefore, the asymptotes are x = 0, x = 1, y = 0 and y = 1.

Example: Find the vertical asymptote of .

Sol.

has vertical asymptotes when ,

thus x = 4 and x = -1 are the vertical asymptotes of

Asymptotes of polar curves-

It can be shown that

Example: Find the horizontal asymptote of the function given below-

Sol.

In order to find the horizontal asymptotes, we find-

Hence the horizontal asymptote is the line y = 2.

Example: Find the asymptotes parallel to x-axis and y-axis of the following curve-

Sol.

We can write the given equation as-

Asymptote parallel to x-axis-

Now by putting the coefficient of equals to zero, we get

Which gives-

Asymptote parallel to y-axis-

By putting the coefficient of equals to zero, we get

Hence are the asymptotes parallel to y-axis.

Slant asymptote-

Definition:

A line y = mx + c (m is non-zero) is an oblique asymptote or slant asymptote to the graph of the function f if

Example: Find the slant asymptote of

Sol.

Here we will find m and c,

Hence the slant asymptote is y = x

Key takeaways-

A straight line is said to be an asymptote to a curve, if as a point P moves to infinity along the curve, the perpendicular distance of P from the straight line tends to zero.

A line y = L is said to be a horizontal asymptote to the curve of the function f if therefore,

To find the asymptotes parallel to x-axis, equate to zero the coefficient of the highest power of x in the equation, provided this is not nearly the constant.

To find the asymptotes parallel to y-axis, equate to zero the coefficient of the highest power of y in the equation, provided this is not nearly the constant.

References:

E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.

P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.

S. Ross, “A First Course in Probability”, Pearson Education India, 2002.

W. Feller, “An Introduction to Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.

N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010

Higher engineering mathematics, HK Dass

BV ramana, higher engineering mathematics

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