TEST 1
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 [pic] The number [pic]is called the first term, [pic] is the second term, and in general [pic]is the nth term . In this section we will consider infinite sequence having infinitely many terms. We represent an infinite sequence by {[pic] } or [pic] or simply by {[pic]}.
Convergence and Divergence of a sequence:
A sequence {[pic]} is convergent if [pic] exists, otherwise the sequence is divergent.
Theorem 1. If [pic], then [pic]
Increasing and Decreasing sequence:
A sequence {[pic]} is called increasing if [pic] for all [pic]. It called decreasing if [pic] for all [pic]. 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 [pic] is an arithmetic sequence, where a is the first term and d is the common difference.
A sequence of the form [pic] 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. [pic]. The given sequence is convergent because [pic], which is finite.
2. [pic]. The given sequence is divergent because [pic]. 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. [pic]. The given sequence is divergent because [pic]does not exist.
4. [pic]. The given sequence is convergent because [pic].
5. [pic]. The given sequence is convergent because [pic], by L’H[pic]pital rule .
6. Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
a) [pic]. The given sequence is decreasing [pic]. the sequence is also bounded because [pic]
b) [pic]. The given sequence is increasing because [pic], the sequence is also bounded because [pic] for all [pic].
c) [pic]. The given sequence is decreasing because [pic], the sequence is also bounded because [pic] for all [pic].
d) [pic]. The given sequence is increasing because [pic], is not convergent, the sequence is bounded because [pic] for all [pic].
Answer key:
8. 4, 4/3, 4, 4/3, 4, ….. 12. [pic] 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 [pic]
Arithmetic and geometric series
A series of the form [pic] 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 [pic], the nth term.
A series of the form [pic] 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 [pic].
Convergent series Given a series [pic]. Suppose [pic]be the partial sum of n terms of the infinite series then if [pic]is a convergent sequence and [pic]exists as a real number, the series written as [pic]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 [pic] is convergent if [pic] and its sum is given by [pic] . The geometric series is divergent if [pic]
The test of divergence
If [pic] does not exist or if [pic] then the series is divergent.
Examples
1. Find at least 10 partial sums of the given series. Is it convergent or divergent? Expalin.
a) [pic]
[pic]
[pic]The given series is a geometric series with [pic], and it is convergent, its sum is given by [pic]
b) [pic]
[pic]
[pic]
This series is a harmonic series with [pic]. By the divergence test it is divergent series.
2. Determine whether the series is convergent or divergent. Find sum if convergent.
a) [pic]. The given series is not a geometric series. Also [pic]divergent test fails. From example 7, page # 717 we know that [pic] is a divergent series, since the sequence of partial sums [pic]is divergent. So [pic] is also divergent.
b) [pic]. We have [pic]are geometric series with [pic], thus convergent and[pic].
c) [pic]. We have [pic]. Observe that the second series [pic]is a convergent geometric series and [pic] . We need to test the harmonic series [pic] We use partial fraction to get [pic]. Remember the telescoping process to find [pic], when [pic]. Thus the harmonic series is also convergent. The sum of the series is [pic]
3. Find the values of x for which the series [pic]converges. Find the sum of the series for those values of x.
The given series is a geometric series will converge if [pic]. Solving the inequality we find [pic]and the sum is [pic].
Answer key
4. Diverges 14. Converges to 5/3 16. Diverges 18. Converges, sum = 3.41
44. 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, [pic]] and let [pic]then the series [pic]is convergent iff the improper integral [pic]is convergent. Otherwise it will be divergent.
The p – series
The p – series [pic]is convergent if p > 1. Otherwise it is divergent.
Reminder estimates for integral test
Suppose [pic], f is a continuous positive decreasing function for [pic]and [pic]is convergent. If [pic], where [pic], then [pic].
Examples
1. Determine using integral test whether the series convergent or divergent
a) [pic]is a p – series with p = 4 > 1, which is convergent. Now we will verify using integral test. [pic]converges, thus the series converges.
b) [pic] is a p – series with p = 0.85 < 1, which is divergent. Now we will verify using integral test. [pic]does not exist.
c) [pic] is not a p – series. We will test convergence using integral test. [pic]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: [pic])
d) [pic] is not a p – series. We will test the convergence using integral test. [pic]is finite and exists. It is convergent.
e) [pic] is not a p – series. We will test the convergence using integral test. [pic] does not exist. It is divergent.
2. Find p so that [pic]. For convergence [pic]must exist. The integral will exist if [pic]
3. Approximate the sum of [pic] 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.
[pic]
Now [pic], which is the at most size of the error.
For the required accuracy we need to have [pic]. We need 37 terms.
The comparison test and the limit convergence test (Page # 730)
The comparison test
Suppose that [pic]and [pic]are series with positive terms.
i) If [pic]is convergent and [pic] , then is [pic]also convergent.
ii) If [pic]is divergent and[pic] , then is [pic]also divergent.
The limit comparison test
Suppose that [pic]and [pic]are series with positive terms. If [pic], finite and positive, then either both series converges or both diverges.
Examples
1. Test the convergence of the series
a) [pic]. We use comparison test, consider [pic], which is a convergent p – series with p = 2. And also observe that [pic]. So the given series is convergent.
b) [pic]. We use comparison test, consider [pic], which is a divergent p – series with p = 1/2. And also observe that [pic]. So the given series is divergent.
To check the inequality one can verify the result
[pic]
is true.
Or, we may use limit test: [pic], as [pic] is a divergent series, and the given series is also divergent.
Test the following series:
c) [pic]is convergent d) [pic]is convergent e) [pic]is divergent
f) [pic]is divergent g) [pic]is divergent h) [pic]is convergent (Hint: consider [pic] a convergent geometric series with [pic], [pic])
i) [pic]is convergent j) [pic] 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 [pic] is convergent if i) [pic] and ii) [pic], for all n and [pic]
Estimation: If [pic]is the sum of an alternating series, with i) [pic] and
ii) [pic] then [pic]
Examples
1. Test the convergence of the series
a) [pic]. We have [pic], and [pic] . So the given series is divergent by the alternating series.
b) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
c) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
d) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
e) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
f) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
g) [pic]. We have [pic], and [pic] . So the given series is divergent by the alternating series.
h) [pic]. We have [pic], and [pic], [pic] . So the given series is convergent by the alternating series.
The Absolute convergence: Ratio and root test (Page # 740)
1. A series [pic]is called absolutely convergent if the series of absolute values [pic] is convergent.
2. A series [pic]is called conditionally convergent if the series is convergent but not absolutely convergent
3. Theorem If a series [pic]is absolutely convergent then the series is convergent.
The ratio test:
i) If [pic], then the series [pic]is absolutely convergent and therefore convergent.
ii) If [pic], or [pic] then the series [pic]is divergent
iii) If [pic], then the ratio test is inconclusive.
The root test:
i) If [pic], then the series [pic]is absolutely convergent and therefore convergent.
ii) If [pic], or [pic] then the series [pic]is divergent
iii) If [pic], then the root test is inconclusive.
Examples
1. Test the convergence of the series
a) [pic]. We have [pic] . So the given series is absolutely convergent by the ratio test.
b) [pic]. We have [pic] . 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) [pic] b) [pic] c) [pic] d) [pic]
e) [pic]
3. Show that following series are conditionally convergent:
a) [pic]. We have [pic], by ratio test the it is inconclusive. But by absolute convergence test [pic] is a p - series with p =1/4 < 1, divergent, on the other hand by alternating series test it is convergent, since [pic]. So the given series is conditionally convergent.
b) [pic]. We have [pic], by ratio test the it is inconclusive. But by limit comparision test (Section 11.4) [pic] is divergent since [pic]is a divergent p - series with p =1. On the other hand by alternating series test it is convergent, since [pic]. So the given series is conditionally convergent.
Strategy for Testing Series
We have learnt the following:
1. [pic] [pic]is a divergent p – series, converges when p > 1 and diverges when [pic].
2. [pic] or [pic] is a geometric series, converges when [pic], diverges when [pic].
3. If [pic] the series diverges (Divergent test)
4. Series with factorials use ratio test
Test the convergence off the following series:
1. [pic], by alternating series test the series is convergent [pic]
2. [pic], by alternating series test the series is convergent [pic].
3. [pic], by limit comparison test [pic] > 0 , the series is divergent since [pic] is a divergent p – series with p =1.
4. [pic], by ratio test [pic] > 0 , the series is divergent.
Homework problem answers (Even only)
2. D 4. C 6. C 8. D 10. C 12. C
14. D 20. C 32. C
Section 8.5 The Power Series (Page # 447)
A power series is a series of the form [pic] where x is a variable and [pic] are called the coefficients of the series. The sum of a power series is a function[pic] whose domain is the set of all x for which the series is convergent.
A power series of the form [pic]is power series in (x – a) or a power series centered at a or a power series about a.
Theorem: The power series [pic]may have three possibilities:
1) the series converges for x = a
2) the series converges for all x
3) There exists R > 0 such that the series converges if |x – a | < R, diverges for
|x – a | > R
The 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) [pic]. The power series is convergent if (by absolute convergent test) [pic], 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 [pic], by alternating series test it is convergent.
When x= 6 we have our series [pic], Harmonic series which is divergent. Therefore the interval of convergence is I = [4, 6)
b) [pic]. The alternating power series is convergent if (by absolute convergent test) [pic], 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 [pic], a harmonic series, divergent.
When x= 1 we have our series [pic], convergent by alternating series test. The interval of convergence is I = (-1, 1]
c) [pic]. The power series is convergent if (by Ratio test) [pic], 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 [pic], the alternating series is divergent.
When x= 1 we have our series [pic], also diverges, the interval of convergence is I = (-1, 1)
d) Check that [pic] has radius of convergence R = 1 and interval of convergence is
I = (4, 6)
e) The series [pic] 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. C on [-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. [pic], R = 1, [pic]
2. [pic], R = 1
3. [pic],
4. [pic], R = 1,
5.[pic], R = 1
6. [pic]
7. [pic]
Differentiation and integration of power series:
The sum of a power series is a function [pic] whose domain is the interval of convergence of the series.
Theorem The power series [pic] has radius of convergence R > 0, then the function is differentiable and therefore continuous on the interval (a – R, a + R).
Observe that
1) [pic]
2) [pic]
3. [pic]
4. [pic]
Examples
1. Find a power series representation of the function and determine interval of convergence.
a) [pic]
= [pic], R = 1, and I =( -1, 1)
b) [pic]
= [pic], R =[pic] 1, and I =( -[pic], [pic])
c) [pic]
[pic]
We have that the interval of convergence of [pic] is [pic] and [pic]is [pic]. Thus the interval of convergence of the given series is [pic].
d) [pic]
[pic]. Thus the interval of convergence of the given series is [pic].
e) [pic]
We consider
[pic]
Differentiating on both sides we find
[pic]
[pic], Thus the interval of convergence of the given series is [pic].
2. a) Find a power series representation of [pic],what is the radius of convergence of ? b) Find the power series representation of [pic]
c) Find the power series representation of [pic]
We have [pic]
Integrating on both sides w. r. to x, we find
[pic]
using the initial condition x = 0, we find C = 0. Then [pic]
b) Using result from a) we can write [pic]
c) Using result from a) we can write [pic]
3. Evaluate as a power series and find radius of convergence.
a) [pic]
We have [pic]
Now [pic]
b) [pic]
We have [pic]
Now [pic]
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MAT 266
FALL 2007
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