Lecture note (last updated on October 29, 2007)

Dr Firoz

Section 8.1 Sequences

Definition: A sequence can be written as a list of numbers in a definite order like The number is called the first term, is the second term, and in general is the nth term . In this section we will consider infinite sequence having infinitely many terms. We represent an infinite sequence by { } or or simply by { }.

Convergence and Divergence of a sequence: A sequence { } is convergent if exists, otherwise the sequence is divergent.

Theorem 1. If , then

Increasing and Decreasing sequence:A sequence { } is called increasing if for all . It called decreasing if for all

. It is called monotonic if it is either increasing or decreasing.

Theorem Every bounded, monotonic sequence is convergent.

Arithmetic and geometric sequences

A sequence of the form is an arithmetic sequence, where a is the first term and d is the common difference.

A sequence of the form is a geometric sequence, where a is the first term and r is the common ratio.

Examples Determine whether the sequence converges or diverges, if converges find the limit

1. . The given sequence is convergent because , which is finite.

2. . The given sequence is divergent because . The

sequence converges to 1 when n is even, on the other hand it converges to -1 when n is odd. The sequence does not converge to single finite number.

3. . The given sequence is divergent because does not exist.

4. . The given sequence is convergent because .

MAT 266 FALL 2007

5. . The given sequence is convergent because , by L’H pital

rule .

6. Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?

a) . The given sequence is decreasing . the sequence is also bounded

because

b) . The given sequence is increasing because , the sequence is also

bounded because for all .

c) . The given sequence is decreasing because , the sequence is

also bounded because for all .

d) . The given sequence is increasing because , is not convergent, the

sequence is bounded because for all .

Answer key:

8. 4, 4/3, 4, 4/3, 4, ….. 12. 16. converges to 1/3 18. converges to 1

22. diverges 24. converges to 1 28. converges to 1 54. decreasing, bounded by 1/5 56. increasing, bounded by 2/3 60. increasing, bounded below by 2

Section 8.2 Series

An infinite series can be written as

Arithmetic and geometric series

A series of the form is an arithmetic series, where a is the first term and d is the common difference. The partial sum of n terms of an arithmetic series

is given by , the nth term.

A series of the form is a geometric series, where a is the first term and r is the common ratio. The partial sum of n terms of a geometric series is given by

.

Convergent series Given a series . Suppose

be the partial sum of n terms of the infinite series then if is a convergent

sequence and exists as a real number, the series written as

is also a convergent series. The number s is called the sum of the

series. Otherwise the series is divergent.

Convergent Geometric series

The geometric series is convergent if and its

sum is given by . The geometric series is divergent if

The test of divergence

If does not exist or if then the series is divergent.

Examples

1. Find at least 10 partial sums of the given series. Is it convergent or divergent? Expalin.

a)

The given series is a geometric series with , and it is convergent, its sum is given by

b)

This series is a harmonic series with . By the divergence test it is divergent series.

2. Determine whether the series is convergent or divergent. Find sum if convergent.

a) . The given series is not a geometric series. Also divergent test fails. From example

7, page # 717 we know that is a divergent series, since the sequence of partial sums is

divergent. So is also divergent.

b) . We have are geometric series with , thus

convergent and .

c) . We have . Observe that the

second series is a convergent geometric series and . We need to test

the harmonic series We use partial fraction to get . Remember the

telescoping process to find , when . Thus the harmonic

series is also convergent. The sum of the series is

3. Find the values of x for which the series converges. Find the sum of the series for those

values of x. The given series is a geometric series will converge if . Solving the

inequality we find and the sum is .

Answer key4. Diverges 14. Converges to 5/3 16. Diverges 18. Converges, sum = 3.4144. Converges, sum = -2/(x+1) Section 8.3 The integral test and Comparison tests

The integral test

Suppose f is continuous, positive decreasing function on [1, ] and let then the series is

convergent iff the improper integral is convergent. Otherwise it will be divergent.

The p – series

The p – series is convergent if p > 1. Otherwise it is divergent.

Reminder estimates for integral test

Suppose , f is a continuous positive decreasing function for and is convergent. If

, where , then .

Examples

1. Determine using integral test whether the series convergent or divergent

a) is a p – series with p = 4 > 1, which is convergent. Now we will verify using integral test.

converges, thus the series converges.

b) is a p – series with p = 0.85 < 1, which is divergent. Now we will verify using integral test.

does not exist.

c) is not a p – series. We will test convergence using integral test.

does not exist. One may observe that the given series is not a

decreasing series as well. The series is divergent. (One can use divergent test: )

d) is not a p – series. We will test the convergence using integral test.

is finite and exists. It is convergent.

e) is not a p – series. We will test the convergence using integral test.

does not exist. It is divergent.

2. Find p so that . For convergence

must exist. The integral will exist if

3. Approximate the sum of by using the sum of first 10 terms. Estimate the error involved in this

approximation. How many terms are required to ensure that the sum is accurate to within 0.0005.

Now , which is the at most size of the error.

For the required accuracy we need to have . We need 37

terms.

The comparison test and the limit convergence test (Page # 730)

The comparison test

Suppose that and are series with positive terms.

i) If is convergent and , then is also convergent.

ii) If is divergent and , then is also divergent.

The limit comparison test

Suppose that and are series with positive terms. If , finite and positive,

then either both series converges or both diverges. Examples

1. Test the convergence of the series

a) . We use comparison test, consider , which is a convergent p – series

with p = 2. And also observe that . So the given series is convergent.

b) . We use comparison test, consider , which is a divergent p –

series with p = 1/2. And also observe that . So the given series is divergent.To check the inequality one can verify the result

is true.

Or, we may use limit test: , as is a divergent series, and the given series is

also divergent.

Test the following series:

c) is convergent d) is convergent e) is divergent

f) is divergent g) is divergent h) is convergent (Hint:

consider a convergent geometric series with , )

i) is convergent j) is convergent

Section 8.4 Other convergence tests (Page # 437)

An alternating series is a series whose terms are alternately positive and negative. An alternating

series is convergent if i) and ii) , for all n and

Estimation: If is the sum of an alternating series, with i) and

ii) then

Examples

1. Test the convergence of the series

a) . We have , and . So the given series is divergent by

the alternating series.

b) . We have , and , . So the given series is

convergent by the alternating series.

c) . We have , and , . So the given series is

convergent by the alternating series.

d) . We have , and , . So the given

series is convergent by the alternating series.

e) . We have , and , . So the given series is

convergent by the alternating series.

f) . We have , and ,

. So the given series is convergent by the alternating series.

g) . We have , and . So the given series is divergent by the

alternating series.

h) . We have , and , . So the given series is

convergent by the alternating series.

The Absolute convergence: Ratio and root test (Page # 740)

1. A series is called absolutely convergent if the series of absolute values is

convergent.

2. A series is called conditionally convergent if the series is convergent but not

absolutely convergent

3. Theorem If a series is absolutely convergent then the series is convergent.

The ratio test:

i) If , then the series is absolutely convergent and therefore

convergent.

ii) If , or then the series is divergent

iii) If , then the ratio test is inconclusive.

The root test:

i) If , then the series is absolutely convergent and therefore

convergent.

ii) If , or then the series is divergent

iii) If , then the root test is inconclusive.

Examples

1. Test the convergence of the series

a) . We have . So the given series is absolutely convergent by

the ratio test.

b) . We have . So the given series is absolutely

convergent by the root test.

2. Apply ratio test to verify that the given series are absolutely convergent and thereby convergent.

a) b) c) d)

e)

3. Show that following series are conditionally convergent:

a) . We have , by ratio test the it is inconclusive. But by absolute

convergence test is a p - series with p =1/4 < 1, divergent, on the other hand by

alternating series test it is convergent, since . So the given series is

conditionally convergent.

b) . We have , by ratio test the it is inconclusive. But by limit

comparision test (Section 11.4) is divergent since is a

divergent p - series with p =1. On the other hand by alternating series test it is convergent, since

. So the given series is conditionally convergent.

Strategy for Testing Series We have learnt the following:

1. is a divergent p – series, converges when p > 1 and diverges when .

2. or is a geometric series, converges when , diverges when .

3. If the series diverges (Divergent test)

4. Series with factorials use ratio test

Test the convergence off the following series:

1. , by alternating series test the series is convergent

2. , by alternating series test the series is convergent

.

3. , by limit comparison test > 0 , the series is divergent

since is a divergent p – series with p =1.

4. , by ratio test > 0 , the series is

divergent.

Homework problem answers (Even only)

2. D 4. C 6. C 8. D 10. C 12. C14. D 20. C 32. C

Section 8.5 The Power Series (Page # 447)

A power series is a series of the form where x is a variable and

are called the coefficients of the series. The sum of a power series is a function

whose domain is the set of all x for which the series is convergent.

A power series of the form is power series in (x –

a) or a power series centered at a or a power series about a.

Theorem: The power series may have three

possibilities: 1) the series converges for x = a2) the series converges for all x3) There exists R > 0 such that the series converges if |x – a | < R, diverges for

|x – a | > RThe number R is called the radius of convergence.

There four possibilities of interval of convergence a) I = (a – R, a + R), b) ) I = (a – R, a + R], c) I = [a – R, a + R), d) ) I = [a – R, a + R]

Examples

1. Find the radius of convergence and interval of convergence of the power series

a) . The power series is convergent if (by absolute convergent test)

, The radius of convergence R = 1 and the

possible interval of convergence is (4, 6). We need to check the convergence separately for x = 4 and x =6.

When x= 4 we have our series , by alternating series test it is convergent.

When x= 6 we have our series , Harmonic series which is divergent. Therefore

the interval of convergence is I = [4, 6)

b) . The alternating power series is convergent if (by absolute convergent test)

, The radius of convergence R = 1 and the possible interval

of convergence is (-1, 1). We need to check the convergence separately for x = -1 and x =1

When x= -1 we have our series , a harmonic series, divergent.

When x= 1 we have our series , convergent by alternating series test. The interval of

convergence is I = (-1, 1]

c) . The power series is convergent if (by Ratio test)

, The radius of convergence R = 1 and the possible interval

of convergence is (-1, 1). We need to check the convergence separately for x = -1 and x =1.

When x= -1 we have our series , the alternating series is divergent.

When x= 1 we have our series , also diverges, the interval of convergence is I = (-1, 1)

d) Check that has radius of convergence R = 1 and interval of convergence is

I = (4, 6)

e) The series has radius of convergence R = 1, and interval of convergence is

given by I = (-2, -1] (Hint: check with problem number 21, section 11.3 and problem number 22, section 11.6)

Homework problem answers (Even only)

4. C, I =(-1, 1] 6. C on (-1, 10 8. C on {0} 12. C on [-5, 5]14. C on whole real line 16. C on (4, 6) 20. Con [-1/3, 5/3] 22. C on [3, 5] 26. C on (-2, -1]

Section 8.6 Representation of a function as a Power Series (Page # 452)

Representations of functions as power series

In this section we have the following representations: We use radius of convergence = R, and interval of convergence is I.

1. , R = 1,

2. , R = 1

3. ,

4. , R = 1,

5. , R = 1

6.

7.

Differentiation and integration of power series:

The sum of a power series is a function

whose domain is

the interval of convergence of the series.

Theorem The power series has radius of convergence R > 0, then the function is

differentiable and therefore continuous on the interval (a – R, a + R).

Observe that

1)

2)

3.

4.

Examples

1. Find a power series representation of the function and determine interval of convergence.

a)

= , R = 1, and I =( -1, 1)

b)

= , R = 1, and I =( - , )

c)

We have that the interval of convergence of is and is

. Thus the interval of convergence of the given series is .

d)

. Thus the interval of convergence of

the given series is .

e)

We consider

Differentiating on both sides we find

, Thus the interval of convergence of the

given series is .

2. a) Find a power series representation of ,what is the radius of convergence of ? b) Find the power series representation of c) Find the power series representation of

We have

Integrating on both sides w. r. to x, we find

using the initial condition x = 0, we find C = 0. Then

b) Using result from a) we can write

c) Using result from a) we can write

3. Evaluate as a power series and find radius of convergence.

a)

We have

Now

b)

We have

Now