12 INFINITE SEQUENCES AND SERIES
12.1 SEQUENCES
SUGGESTED TIME AND EMPHASIS 1 class Essential material
POINTS TO STRESS
1. The basic definition of a sequence; the difference between the sequences {an} and the functional value f (n).
2. The meanings of the terms “convergence” and “the limit of a sequence”.
3. The notion of recursive sequences (including the use of induction and the Monotonic Sequence Theorem to establish convergence).
QUIZ QUESTIONS
• Text Question: Could there be a sequence {an} = {f (n)} such that limx→∞ f (x) exists, but limn→∞an does not? Could lim
n→∞an exist, but not lim
x→∞ f (x)?
Answer: No to the first question. Yes to the second; an example is f (x) = sin (2πx).
• Drill Question: Can you give an example of a sequence {an} that is monotonic and bounded above and below, but lim
n→∞an does not exist?
Answer: No such sequence exists, by the Monotonic Sequence Theorem.
MATERIALS FOR LECTURE
• Point out that if an = f (n) for some function f , and if limx→∞ f(x) = L, then limn→∞an = L. Thus, an
converges if f has a horizontal asymptote as x → ∞. Note that the converse is not true. For example, take an = sin nπ and f (x) = sin (πx).
• Discuss monotonicity and the Monotonic Sequence Theorem. Perhaps apply the theorem to show that the sequence {0.1, 0.12, 0.123, 0.1234, . . . , 0.123456789, 0.12345678910, 0.1234567891011, . . .}
converges. (The limit of this sequence is called the Champernowne Constant.) One interesting fact about the Champernowne Constant is that its decimal expansion clearly contains every possible finite sequence of numbers. For example, the sequence 3483721589712 will appear somewhere, because of the counting nature of the constant. So if one were to take any book and convert it to a number using the code A = 1, B = 2, etc., that book appears somewhere in the Champernowne Constant. (This book could be already written, or even a book that has not been written yet, such as one that reveals any person’s life story — including their future!)
• Consider an = (cos n)n
ln(n + 1) and ask students how they might determine the convergence or divergence of this sequence. Then remind them that−1 ≤ (cos n)n ≤ 1 for all n and hence the Squeeze Theorem can be used to show that the limit is 0.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
WORKSHOP/DISCUSSION
• Determine the convergence or divergence of the following sequences {an} = f (n) by first looking at f (x). Make sure to write out the first few terms of the sequences for each case, to emphasize their discrete nature.
1. an = n 1+ n2 2. bn = ln(1 + 2en)
n
3. cn = (n + 1)1/2− n1/2 4. dn = 1+ n cos (2πn)
n
• Do a non-obvious example that uses the Squeeze Theorem to establish convergence, such as an = sin n+ cos n
n2/3 .
• Compute the limit of a recursive sequence such as a1 = 2, an = 4 − 1
an−1, after first either proving convergence (using induction and the Monotonic Sequence Theorem) or giving a numerical argument for convergence.
GROUP WORK 1: Practice with Convergence
After the students have warmed up by doing one or two of the problems as a class, have them start working on the others, checking one another’s work by plotting the sequences on a graph. If a group finishes early, give them Group Work 2, the Random Decimal, which makes a nice sequel.
Answers:
1. Converges to 0 2. Diverges 3. Converges to 0 4. Converges to−1 5. Converges to−1 6. Converges to 43 7. Diverges 8. Converges to 0
GROUP WORK 2: The Random Decimal
This works as an addition to Group Work 1. It can also stand alone. (Groups of four or five work best for this problem) The students may not be familiar with the idea of concatenation, so you may want to do an example for them if they seem to be having trouble understanding the idea.
Answers: (Answers to the first two problems will vary.) 1. a1= 0.5358
2. a1= 0.5358 a2= 0.53589793 a3= 0.535897932384 a4= 0.5358979323846264 a5= 0.53589793238462643383
3. The sequence is always increasing, and has an upper bound (1 will always be an upper bound, for example; 0.6 is a better upper bound in this case.) Therefore, by the Monotone Convergence Theorem, this sequence does converge. It can be proven that if the numbers generated are truly random, then this number will be irrational.
SECTION 12.1 SEQUENCES
GROUP WORK 3: Recursive Roots
This problem gives the tools to show that if x > 0, then
x+ x+
x+√
x+ · · · exists and equals 1+√
1+ 4x
2 . The students may find Questions 2 and 3 difficult. One hint to give them is that for a and b> 0, a < b ⇒ a2 < b2. In a less rigorous class, it may be acceptable for students to notice that there is a trend (the terms are increasing, and approaching 1.6 < 2) but they should also realize that noticing a trend isn’t the same thing as proving that the trend will continue forever. This exercise can be done with less rigor by having the students skip Questions 2 and 3 entirely.
Answers: (Answers to the first two problems will vary.) 1. a1= 0
a2=√
1+ 0 = 1 a3=
1+√
1+ 0 =√
2≈ 1.4142 a4=
1+ 1+√
1+ 0 ≈ 1.5538 a5=
1+
1+ 1+√
1+ 0 ≈ 1.5981
2. The easiest way to prove this is by induction. We want to show that if an < 2, then√
1+ an < 2. The base case is trivial (0 < 2). The induction step: If an < 2, then√
1+ an <√
1+ 2 < 2. If the students haven’t learned mathematical induction, this argument can be put into less formal language.
3. We now want to show an <√
1+ anIt suffices to show that(an)2 < 1 + an, or(an)2− an− 1 < 0. The quadratic formula, or a graph, can show that this is true if 0< an < 1+2√5 ≈ 1.618. (Actually, this is true for 1−
√5
2 < an < 1+2√5.) We can use an induction argument like the one in the previous part to show that if an < 1+2√5, then an+1< 1+2√5.
4. a=
1+ 1+
1+√
1+ · · · =
1+
1+ 1+
1+√
1+ · · ·. Therefore a =√ 1+ a.
5. Since a =√
1+ a, we have a2 = 1 + a and a = 1+2√5. 6. 2, 3
GROUP WORK 4: Euler’s Constant Revisited
This activity revisits Group Work 4 in Section 5.4 (which also appears as Group Work 2 in Section 7.2*).
Answers:
1. It is the sum of positive terms.
2. Each time n is incremented, more positive area is added to the total.
3. The Monotone Convergence Theorem gives the result.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
HOMEWORK PROBLEMS
Core Exercises: 2, 3, 8, 9, 19, 28, 36, 50, 55, 59, 62
Sample Assignment: 2, 3, 6, 8, 9, 13, 15, 19, 21, 28, 36, 43, 50, 51, 55, 59, 62, 73, 75
Exercise D A N G
2 ×
3 ×
6 ×
8 ×
9 ×
13 ×
15 × ×
19 ×
21 ×
28 ×
36 ×
43 ×
50 × ×
51 × ×
55 × ×
59 ×
62 ×
73 ×
75 ×
GROUP WORK 1, SECTION 12.1 Practice with Convergence
Do the following sequences{an} converge or diverge? Justify your answers.
1. an = en
3n 5. an = (−1)n+ n
(−1)n− n
2. an = (−1)n√
n 6. an = ln
e4n 3n
3. an = (−1)n 1
√n 7. an = (−1)ncosπ
2 (n + 1)
4. an = (−1)2n+1 8. an = (−1)nsinπ
2 (2n + 1)
GROUP WORK 2, SECTION 12.1 The Random Decimal
1. Have each person in your group think of a random integer from 0 through 9. Let a1 be 0.wxyz where w, x, y and z are your numbers. For example, if you came up with 2, 4, 1, and 8, then you would write a1= 0.2418.
a1 = ________
2. Have each person in your group think of a new integer, and add those integers to the end of a1to form a2. For example, if you already had a1 = 0.2418, you might come up with a2 = 0.24185299. Continue the process to form a3, a4 and a5.
a2 = ________________
a3 = ________________________
a4 = ________________________________
a5 = ________________________________________
3. If you continued this process infinitely many times, you would have an infinite sequence{an}. Does this sequence converge, diverge, or is it impossible to tell? Why?
GROUP WORK 3, SECTION 12.1 Recursive Roots
We want to find the value of
1+
1+
1+√ 1+ ...
1. Consider the recursive sequence a0 = 0, an+1=√
1+ an. Compute the next five terms a1, a2, a3, a4, and a5.
2. Show that an < 2 for all n.
3. Show that an+1> an.
Recursive Roots
4. Since {an} is increasing and bounded above by 2, the Monotone Sequence Theorem says that {an} converges. If lim
n→∞an = a, show that a =√ 1+ a.
5. What is the value of
1+ 1+
1+√
1+ · · ·?
6. Using similar reasoning, try to compute
2+ 2+
2+√
2+ · · · and
6+ 6+
6+√
6+ · · ·.
GROUP WORK 4, SECTION 12.1 Euler’s Constant Revisited
Recall the following picture:
and the sequenceγn = 1 +12 +13 + · · · + 1n − ln (n + 1) = 1 +12 +13 + ... +1n −n+1
1 1
t dt.
1. By a previous group work (Group Work 4 in Section 7.4), we know thatγn ≤ 1. Explain why 0 ≤ γn for all n.
2. Using the picture above, explain whyγn is monotone increasing.
3. Why can we conclude that lim
n→∞γn = γ exists?
LABORATORY PROJECT Logistic Sequences
This project is useful as an in-class extended exercise, if computers are available, or as an out-of-class project.
Students are asked to numerically analyze the logistic difference equation, the discrete variant of the logistic differential equation, examining its long-term behavior. Problems 1 and 2 provide a basic analysis of the situation and can be covered in a shorter period. Problems 3 and 4 lead the students to discovering chaotic behavior.
In their report, students should include a paragraph about the similarities and differences between the behavior of the difference equation and the logistic growth differential equation from Section 10.4.
12.2 SERIES
TRANSPARENCY AVAILABLE
#27 (Figures 2 and 3)
SUGGESTED TIME AND EMPHASIS 2 classes Essential material
POINTS TO STRESS
1. The basic concept of a series. The difference between the underlying sequence and the sequence of partial sums.
2. The relationship between lim
n→∞an, and the convergence/divergence of ∞
n=1an. 3. The analysis and applications of geometric series.
4. The Test for Divergence.
QUIZ QUESTIONS
• Text Question: Is the following always true, sometimes true, or always false? If the series ∞
n=1an converges, and the series ∞
n=1bn converges, then their sum converges.
Answer: Always true
• Drill Question: Does the series 0.1 + 0.01 + 0.001 + 0.0001 + 0.00001 + · · · converge? If so, find its sum.
Answer: 19
MATERIALS FOR LECTURE
• Present an intuitive approach to the definition of a series. Explain that the way we add an infinite number of terms is to keep adding more and more of them to a running total, in a systematic way, to create partial sums. If the limit of the partial sums exists, we say the series converges to that limit. Show the distinction between the nth term an, the nth partial sum sn, and the connection between them (sn−1+ an = sn).
• Discuss Theorems 6 and 7 and Note 2, explaining how the converse to Theorem 6 is not true in general.
Use Theorem 7 to explain why ∞
n=1cos(1/n) diverges.
• Derive the formula for the sum of a geometric series. Illustrate why this type of series diverges for |r| > 1, and why it also diverges for r = ±1. Provide details about the geometric justification found in Figure 1.
• Represent a geometric series visually. For example, a geometric view of the equation ∞
n=11/2n = 1 is given below.
1
21_
0 3_4 7_8 1516_
321_
161_
81_
41_
21_
An alternative geometric view is given in Group Work 3, Problem 2. If this group work is not assigned, the figure should be shown to the students at this time.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
• Sometimes we can express a mathematical constant as the sum of a series. A classic example is e = ∞
k=1
1 k!. Summations forπ are historically important, and a nice simple research project might be for students to find a few of the more unusual ones. In 1985, David and Gregory Chudnovsky used the series
1 π = ∞
n=0
2n
n
16n
3
42n+ 5 16 where
2n n
= (2n)!
n! n!, to compute π to more than 4 billion decimal places. Each term of this series produces an additional 14 correct digits. (If the students have access to aCAS, this is a fun formula to play with.)
WORKSHOP/DISCUSSION
• Foreshadow the Integral Test. Show that ∞
n=11/n diverges by using an integral as a lower bound, as illustrated by the following figure:
• Ask the students to write down an example of a series ∞
n=1ansuch that the terms of the series go to zero, yet the series diverges. Since the students have seen the harmonic series in both the lecture and in the text, this should be an easy question for them to answer. But there is something to be gained in their hearing an abstract question and thinking of a concrete example that they know.
• Introduce the idea that for any two real numbers A and B, the statement A = B is the same as saying that for any integer N , |A − B| < 1/N. Now use this idea to show that 0.9999 . . . = 0.9 = 1, since
1− 0.9 <
1− 0.99999 . . . 99
N nines
= 0.00000 . . . 000
N−1 zeros
1 = 10−N = 1
10N. Then use the usual approach to define 0.9 as ∞
n=19/10n and show directly that 0.9 = 1. Generalize this result by pointing out that any repeating decimal (0.3, 0.412, 0.24621) can be written as a geometric series, and can thus be written as a fraction using the formula for a geometric series. Demonstrate with 0.412 = 1000412 1
1−1/1000
= 412999.
• Check the convergence/divergence of the following series:
∞ n=1
1
kn, k> 1 ∞
n=1
4· 5n− 5 · 4n 6n
∞ n=1(−1)n ∞
n=1sin
n
n+ 1
∞
n=1(−1)2n ∞
n=1
5
n(n + 1)−
−1 2
n
SECTION 12.2 SERIES
Answers:
∞ n=1
1
kn, k> 1 is geometric and converges to 1 k− 1. ∞
n=1
4· 5n− 5 · 4n
6n is a sum of two geometric series and converges to 10.
∞
n=1(−1)n diverges by the Test for Divergence.
∞ n=1sin
n
n+ 1
diverges by the Test for Divergence.
∞
n=1(−1)2ndiverges by the Test for Divergence.
∞ n=1
5
n(n + 1)−
−1 2
n
is the sum of a geometric series and 5 times the series from Example 6, and converges to 163.
• Using a diagram similar to Figure 2 in Section 12.3, show that ln n < n
k=11/k < 1 + ln n. Make sure the students know that ∞
k=11/k goes to infinity. Now ask them this question: “We know that the harmonic series diverges. Assume that in the year 4000B.C., you started adding up the terms of the harmonic series, at the rate of, say, one term per second. We know that the sum gets arbitrarily large, but approximately how big would your partial sum be as of right now?” (If you wish the students to discover some of these concepts for themselves, Have them explore Group Work 2: The Harmonic Series.)
• Define the “middle third” Cantor set for the students. (Let C be the set of points obtained by taking the interval [0, 1], throwing out the middle third to obtain
0,13
∪2
3, 1
, throwing out the middle third of each remaining interval to obtain
0,19
∪
29,13
∪
23,79
∪
89, 1
, and repeating this process ad infinitum).
Point out that there are infinitely many points left after this process. (If a point winds up as the endpoint of an interval, it never gets removed, and new intervals are created with every step). Now calculate the total length of the sections that were thrown away: 13+ 2 ·19+ 4 ·271 + ... = ∞
k=0
2k
3k+1 = 1. Notice the apparent paradox: We’ve thrown away a total interval of length 1, but still infinitely many points remain. (See also Exercise 73.)
GROUP WORK 1: The Leaning Tower
This exercise, an expansion of Problems Plus #12, should take about 45–50 minutes.
Group the students, and give them materials to stack. Packs of small notepads, CD jewel boxes, or wooden blocks all make good materials for stacking. Their goal is to make a stack with the top block one length away from the bottom, without having the stack fall over. (See the diagram below.)
Give them time to work. When a group achieves the goal, have them try to get two lengths out. After they have been working for a while, give them the hint that it is easiest to build onto the bottom, not the top. In other words, take a balanced stack, transfer it to a new bottom piece, and then slide it as far as possible.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
When there are 25 minutes left, collect the blocks and model the situation on the board. The main thing to get across is the idea of center of mass. The center of mass will be the place where half of the mass is to the left, and half is to the right. (We don’t care that much about the vertical component; it will be n/2 units up.) The stack balances if the center of mass is over the table, otherwise it falls.
Have them try to solve the general problem: Given n things to stack, what is the farthest that they can go?
You may want to do n= 1 and n = 2 on the board to give them a start:
n = 1
Total weight on the right= 12n = 12 extension= 12
n= 2
Total weight on the right= 12n = 1 = 12 + 2x ⇔ x = 14 extension= 12 +14
After they have tried, if they did not succeed, write out the solution for n= 3 and n = 4 as follows:
n = 3: Total weight on the right = 32 = 1 + 3x ⇔ x = 16; extension= 12 +14 +16. n= 4: Total weight on the right = 2 = 32 + 4x ⇔ x = 18; extension= 12 +14 +16 +18. Therefore, if n = k, we have an extension of 12
1+12 +13 + ... +k1
.∗(See below.) At this point they should recognize the harmonic series. So the answer to
the question “What is the farthest that the stack can extend, given as many objects as desired?” is tied to the question “What is the sum of the harmonic series?” which they have already seen to be infinity. Emphasize how slowly it goes to infinity (perhaps by putting the figure at right on a transparency and noting how small the increments are at the bottom of the stack).
∗Note that, since at some point the left edges of the blocks will begin to overhang, the expression12
1+12+13+ · · · +1k
is actually a lower bound on the possible extension ofk objects.
GROUP WORK 2: The Harmonic Series
This activity was suggested by the Teachers Guide toAPCalculus published by the College Board. In addition to allowing the students to discover the divergence of the harmonic series for themselves, the last question will allow them to make an intuitive guess that will be confirmed or refuted by what they learn in the next section.
Answers:
1. s1 = 1, s2 = 1.5, s3 ≈ 1.8333, s4 ≈ 2.08333, s5 ≈ 2.28333, s6 = 2.45, s7 ≈ 2.5929, s8 ≈ 2.7179, s9≈ 2.8290, s10≈ 2.9290
SECTION 12.2 SERIES
2., 4.
0 1 2 3 4
2 4 6 8 10 n
s
3. The partial sums appear to approach 3.
5. We know ln x goes to infinity, and the partial sum sn seems to always be larger than ln n.
6. Neither of our answers is a proof, since we are just generalizing based on the first ten partial sums. There is a proof in the text that ∞
n=11/n diverges, and one may have been given in class as well.
7. Defer revealing the answer to this question until the next section.
GROUP WORK 3: Made in the Shade
Problem 1 attempts to help the students visualize geometric series. Problem 2 gives a geometric interpretation of the fact that 12+ 14+ 18+ 161 + · · · = 1.
Answers:
1. (a) a= 12π, r = 14, A= 2π3 (b) a= 14, r = 14, A = 13 2. (a) an = 1/2n
(b) 1. Note that the students may use the geometric series formula to get the answer, but it should be pointed out that the answer is immediate from the diagram.
GROUP WORK 4: An Unusual Series and Its Sums
The initial student reaction may be “I have no idea where to start!” One option is to start the problem on the board for the students. Another approach may be to encourage the students to write down the length of the largest dotted line segment (b), then to figure out the length of the next one, which they can get using trigonometry, and keep going as long as they can. Many students still resist the idea of tackling a problem
“one step at a time” if it seems difficult.
Answers:
1. L = b + b sin θ + b sin2θ+ · · · or ∞
n=1b(sin θ)n. Because there are infinitely many terms, we need to write the answer as a series.
2. L= 1−sin θb 3. L approaches infinity.
4. Geometrically: Asθ → π2, the picture breaks down. The easiest way to see this is to have the students try to sketch what happens forθ close toπ2. The dotted lines become infinitely dense.
Using infinite sums: lim
θ→π/2
b
1− sin θ diverges.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
HOMEWORK PROBLEMS
Core Exercises: 1, 4, 14, 23, 29, 35, 42, 57, 58, 64
Sample Assignment: 1, 4, 5, 9, 14, 16, 23, 28, 29, 30, 35, 42, 47, 55, 57, 58, 59, 62, 64, 72
Exercise D A N G
1 ×
4 × ×
5 × ×
9 ×
14 ×
16 ×
23 ×
28 ×
29 ×
30 ×
Exercise D A N G
35 ×
42 ×
47 ×
55 ×
57 × ×
58 ×
59 ×
62 × ×
64 ×
72 ×
GROUP WORK 2, SECTION 12.2 The Harmonic Series
In this exercise, we look at ∞ n=1
1 n. 1. What are the first ten partial sums sn?
s1 = s6 = s2 = s7 = s3 = s8 = s4 = s9 = s5 = s10=
2. The way we will compute ∞ n=1
1
n (or prove that it diverges) is to compute the limit of its partial sums. Plot the partial sums on the following graph, as accurately as you can.
0 1 2 3 4
2 4 6 8 10 n
s
3. The partial sums appear to be approaching a limit. What is that limit?
4. Now, on the same axes, graph y = ln x and y = 1 + ln x for x ≥ 1. (Both of these graphs, as you know, go to infinity as x gets arbitrarily large.)
The Harmonic Series
5. Using your answer to Problem 4 and your graph, explain why it is reasonable to believe that ∞ n=1
1 n goes to infinity.
6. Did either of your answers to Problems 3 and 5 constitute a proof? Does ∞ n=1
1
n converge?
7. Assume that in the year 4000B.C., you started adding up the terms of the harmonic series, at the rate of, say, one term per second. We know that the sum gets arbitrarily large, but approximately how big would your partial sum be as of now? Go ahead and make a guess, based on your best judgment and intuition.
GROUP WORK 3, SECTION 12.2 Made in the Shade
1. Compute the sum of the shaded areas for each figure.
(a)
(b)
Made in the Shade
2. Consider the figure below.
aÁ
1
1 aª
a£
a¢ a°
a§
(a) Find an general expression for the area an.
(b) What is ∞ n=1an?
GROUP WORK 4, SECTION 12.2 An Unusual Series and Its Sums
Consider the following right triangle of side length b and base angleθ.
1. Express the total length of the dotted line in the triangle in terms of b andθ. Why should your answer be given in terms of a series?.
2. Compute the sum of this series.
3. What happens asθ→ π2?
4. Justify this answer geometrically and using infinite sums.
12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
TRANSPARENCY AVAILABLE
#28 (Figures 1 and 2)
SUGGESTED TIME AND EMPHASIS 1 class Essential material
POINT TO STRESS
1. The geometry and formal statement of the Integral Test, including the conditions on f (continuous, positive, decreasing).
2. The Remainder Estimate for the Integral Test, and its use in bounding the error.
3. The use of the Remainder Estimate and partial sums to estimate the sum of a series (as in Example 6).
QUIZ QUESTIONS
• Text Question: Why can the integral∞
100 f (x) dx be used to test the convergence of ∞
n=1 f (n)?
Answer: Convergence or divergence of a series depends only on the “tail”.
• Drill Question: We know that ∞
1
1
x2 d x = 1. From this fact, we can conclude that (A)
∞ n=1
1
n2 converges (B) ∞ n=1
1
n2 = 1 (C) Both (A) and (B) (D) Neither (A) nor (B) Answer: (A)
MATERIALS FOR LECTURE
• State the Integral Test and give at least a geometric justification, as done prior to the formal statement in the text. Present examples of series that can be shown to be convergent by the Integral Test, such as
∞ n=2
1 n(ln n)2.
• Discuss the basic p-series ∞
n=1
1
np, and determine the values of p for which it converges and diverges. If time permits, similarly discuss ∞
n=1
1 n(ln n)p.
• State the remainder estimate∞
n+1 f (x) dx ≤ Rn ≤∞
n f (x) dx (Formula 2) and the corresponding sum estimate (Formula 3), together with the geometric justification given in Figures 3 and 4.
WORKSHOP/DISCUSSION
• Present examples of convergent and divergent series determined by the Integral Test, such as ∞
n=1
1 n2+ 2 and
∞ n=1
1 n0.9.
• Using the series ∞
n=1
1
n2, find values of n for which the remainder Rn < 0.01, and then values of n for which Rn < 0.001. Then do the same for ∞
n=1
1 n4.
SECTION 12.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
• Using the remainder estimate for the Integral Test, answer this question (posed at the end of Group Exercise 2 in Section 12.2): If you had started adding up the harmonic series at a rate of one term per second, starting in 4000B.C., what would the partial sum be today? (Not all series that go to infinity do so quickly!) This question anticipates Exercise 35.
Answer: As of the date and time of this writing, taking into account leap years, skipped leap years, leap seconds, and the change from the Gregorian calendar, the partial sum is approximately 25.967340.
• Check the following for convergence or divergence (or not enough information given):
1. ∞
n=2
1
n ln n 2. ∞
n=2
1
n(ln n)2 3. ∞
n=1e−xcos x Answers: All converge
GROUP WORK 1: The Integral Test
A straightforward series is presented, followed by one that seems to prove the Integral Test false, to make sure the students understand the importance of the conditions of the test.
Answers:
1.
∞
1
x d x x2+ 1 = ∞ 2. It is true that ∞
1 x sin(πx) dx is divergent, which may lead the students to think that the Integral Test says the series is divergent. This is not the case, as explained in Problem 4.
3. 0+ 0 + 0 + 0 + 0 + · · · . The series converges to 0.
4. The fact that∞
1 x sin(πx) dx is divergent is not relevant, because the Integral Test applies only to positive decreasing functions.
GROUP WORK 2: Unusual Sums
Assign each group a different problem to work on first. Have the students do as many as they can. Leave time for each group to present the solution to their assigned problem.
Answers:
1. 32 2. Diverges to∞ 3. 1 4. Diverges to−∞
5. (a) They are the same. (b) 0< p < 1 HOMEWORK PROBLEMS
Core Exercises: 2, 5, 9, 11, 21, 27
Sample Assignment: 2, 4, 5, 8, 9, 11, 14, 20, 21, 25, 27, 34, 37, 39, 40
Exercise D A N G
2 × ×
4 ×
5 ×
8 ×
9 ×
11 ×
14 ×
20 ×
Exercise D A N G
21 ×
25 ×
27 ×
34 ×
37 ×
39 ×
40 × × ×
GROUP WORK 1, SECTION 12.3 The Integral Test
1. Use the Integral Test to show that ∞ n=1
x
x2+ 1 diverges.
2. What does the Integral Test say about the series ∞ n=1
n sin(πn)?
3. Write out the first five terms of ∞ n=1
n sin(πn). Does the series converge or diverge?
4. Do your answers to Problems 2 and 3 contradict each other? Explain.
GROUP WORK 2, SECTION 12.3 Unusual Sums
In each of the following problems, determine if the sum converges, diverges, or if there is not enough infor- mation to tell:
1. ∞
n=1
n+1
n
1 x5/3 d x
2. ∞
n=1
n+1
n x2/3d x
3.
∞ n=1
1/n
1/(n+1)x2/3d x
4. ∞
n=1
1/n
1/(n+1)
1 x5/3d x
Unusual Sums
5. (a) Suppose p> 0. What is the relationship between ∞
n=1
1/n
1/(n+1)
1
xp d x and1 0
1 xp d x?
(b) Find the values of p for which ∞
n=1
1/n
1/(n+1)
1
xp d x converges.
12.4 THE COMPARISON TESTS
SUGGESTED TIME AND EMPHASIS 1 class Essential material
POINT TO STRESS
1. The Comparison Test, stressing the fact that we need only consider the “tail” of the series to determine convergence or divergence.
2. The Limit Comparison Test
QUIZ QUESTIONS
• Text Question: When using the Comparison Test, why do we need to check the conditions an ≤ bn or an ≥ bn only for n≥ N, where N is some fixed integer (as stated in the text)?
Answer: For any fixed integer N , the sum of the first N terms of the series is finite. When checking for convergence, we are concerned only with the infinite “tail” of the series.
• Drill Question: If the improper integral
∞
5
d x
xp converges, then which of the following series must converge?
(A) ∞ n=1
1
np+1 (B) ∞ n=5
1
np+1 (C) ∞ n=1
1
np−1 (D) ∞ n=5
1
np−1 (E) BothAandB (F) BothCandD
Answer: (E)
MATERIALS FOR LECTURE
• State the Comparison Test, using a graph for informal justification. For example, illustrate how ∞
n=3
1 n2− 5 can’t be shown convergent by comparison with ∞
n=2
1
n2, but can be shown convergent by comparison with ∞
n=2
2
n2, since for n ≥ 4, we have n2− 5 ≥ n2−n2 2 = n2
2 . Discuss the fact that we need consider only the
“tail” of the series to determine convergence or divergence.
• Give some examples of using the Limit Comparison Test, such as ∞
n=1
n2+ 2n + 3
3n4+ 7n3+ 11n2+ 13n + 17 [compare to 1/
3n2
] and ∞
n=1
2n+ 17
n2ln n+ 5 [compare to 2/ (n ln n)].
• Discuss why the series 100
n=12n+ ∞
n=1011/2n converges by comparison to ∞
n=11/2n. Point out that
100
n=12n is a very large number, namely 2,535,301,200,456,458,802,993,406,410,750.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
WORKSHOP/DISCUSSION
• Check the following series for convergence or divergence (or not enough information given):
1. ∞
n=1
1
n3/2− 1/2 2. ∞
n=1
1 ln n+ n 3. ∞
n=1an2, where ∞
n=1an converges and 0< an < 1 for all n Answers:
1. Converges. Compare to ∞ n=1
2
n3/2 or use the Limit Comparison Test with ∞ n=1
1 n3/2. 2. Diverges. Compare to
∞ n=2
1
2n or use the Limit Comparison Test with ∞ n=2
1 n. 3. Converges. Compare ∞
n=1anwith ∞
n=1(an)2.
• Show that the series ∞
n=1
1
2n− 1 can be shown to converge using either the Limit Comparison Test with ∞
n=1
1
2n, or the Comparison Test with ∞ n=1
1
2n−1
(
2n1− 1 ≤ 2n−11 , since 2n− 1 ≥ 2n− 2n−1 = 2n−1forn ≥ 1
)
.• Test the series ∞
n=1
1
n2− n + 1for convergence in two ways: using the Limit Comparison Test with ∞
n=1
1 n2, and using a regular comparison with ∞
n=2
1
(n − 1)2
(
That is, n2− n + 11 ≤ 1(n − 1)2 for n≥ 2
)
.GROUP WORK 1: Practicing with the Comparison Test
This activity consists of eight comparison test problems, ranging widely in difficulty. Feel free to have the students start with a problem other than the first one. Problems 4 and 5 were foreshadowed in Group Work 2 in Section 12.3.
Answers:
1. Converges. Compare with ∞ n=1
1 n2. 2. Diverges. Compare with
∞ n=1
1
2n or use the Limit Comparison Test with ∞ n=1
1 n. 3. Converges. Compare with
∞ n=1
1 n1.5. 4. Converges. Compare with
∞ n=1
2 3
n . 5. Converges. The series is the same as
∞
1
1
x5/3d x = 3 2. 6. Converges. Compare with
∞ n=1
2 2−n
.
SECTION 12.4 THE COMPARISON TESTS
7. Converges. Compare with ∞ n=1
1 n3√3
2, or use the Limit Comparison Test with ∞ n=1
1 n3.
8. Diverges. Compare with ∞ n=1
1 n.
GROUP WORK 2: How Do I Compare?
This group work should be attempted only after the students have had an opportunity to practice using the Comparison Test on some routine problems.
Answers:
1. Converges. Compare to ∞ n=1
1 2n2.
2. Diverges. Compare to ∞ n=1
1 n.
3. Diverges. The proof can be a bit tricky. Consider the odd numbered terms only: 1 1+1
3+1 5+1
7+1 9+· · · . One can compare this series to
∞ n=1
1
2n, which diverges. The point of this problem is to get a little controversy going: encourage the students to discuss and debate this one with each other, instead of just putting down a guess.
GROUP WORK 3: Sums of Squares
The purpose of this group work is to have students realize that comparisons need only take place after a finite number of terms, and that the Limit Comparison Test is frequently useful in the cases where the “obvious”
convergent comparison series is not always larger. Be sure to first cover Exercise 40, which shows that the Limit Comparison Test holds for convergence when lim
n→∞
an
bn = 0. This group work is useful primarily for advanced students. Problem 1(d) can be done by showing that ln n≤ n1/6for sufficiently large n.
Answers:
1. (a) Intuitive justifications will vary. Correct answers must include the fact that, for large n, an < 1.
(b) The Comparison Test works directly.
2. Compare ∞
n=1anbn to ∞
n=1(cn)2, where cn = max (an, bn).
CHAPTER 12 INFINITE SEQUENCES AND SERIES
HOMEWORK PROBLEMS
Core Exercises: 1, 4, 9, 18, 24, 28, 34, 37
Sample Assignment: 1, 4, 7, 9, 14, 18, 19, 24, 28, 31, 34, 35, 37, 40, 43
Exercise D A N G
1 ×
4 ×
7 ×
9 ×
14 ×
18 ×
19 ×
24 ×
28 ×
31 ×
34 ×
35 ×
37 ×
40 ×
43 ×
GROUP WORK 1, SECTION 12.4 Practicing with the Comparison Test
For each of the following problems, determine whether the series is convergent or divergent.
1.
∞ n=1
1 n2+ 1
2.
∞ n=1
n n2+ 1
3.
∞ n=1
ln n n2+ 1
4.
∞ n=1
n+1
n
1 x5/3d x
5.
∞ n=1
n+1
n
1 x2/3d x
6.
∞ n=1
2−nan, where an = f (n) as shown.
2
0 1 2 3 n
aÁ aª a£ an
x y
7.
∞ n=1
1
√3
n9− n3+ 1 8.
∞ n=1
sin
1 n
GROUP WORK 2, SECTION 12.4 How Do I Compare?
Determine if each of the following series converges or diverges.
1. 1 2 + 1
23 + 1 2· 32 + 1
43 + 1 2· 52 + 1
63 + · · ·
2. 1 1 + 1
ln 2+ 1 3+ 1
ln 4+1 5 + 1
ln 6+ · · ·
3. 1 1 + 1
22 +1 3 + 1
42 +1 5 + 1
62 + · · ·
GROUP WORK 3, SECTION 12.4 Sums of Squares
1. Let ∞
n=1an be a convergent positive series. Now consider the series ∞
n=1an2.
(a) Why is it intuitively true that for sufficiently large n, an2≤ an? Give a reason for your answer.
(b) Using part (a), it should be possible to compare ∞ n=1an2 to
∞
n=1an to check its convergence. Use the Comparison Test to show that ∞
n=1
an2converges.
2. Show that if ∞ n=1an and
∞
n=1bnare positive series and both converge, then ∞
n=1anbn converges.
Hint: What convergent series can you compare to ∞ n=1anbn?
12.5 ALTERNATING SERIES
TRANSPARENCY AVAILABLE
#29 (Figure 2 and Section 12.6 Figure 1) SUGGESTED TIME AND EMPHASIS
12 class Recommended material POINTS TO STRESS
1. The Alternating Series Test.
2. The Alternating Series Estimation Theorem.
QUIZ QUESTIONS
• Text Question: In Example 3, how did looking at the function f (x) = x2
x3+ 1 and its derivative help us determine that the series converges?
Answer: We examined the derivative of f(x) to show that f was a decreasing function, and that therefore we were able to use the Alternating Series Test to show that ∞
n=1(−1)n+1 n2
n3+ 1 converges.
• Drill Question: Does ∞
n=1sinπn 2
x−1/2converge or diverge? Why?
Answers: The series converges by the Alternating Series Test.
MATERIALS FOR LECTURE
• Do Exercise 32. In this problem, students explore how changing the value of a parameter affects the convergence of a series, and are thus exposed to a type of reasoning used in analysis of power series.
• Some common terms that appear in alternating series are (−1)n and cos(πn). Ask the students if they can come up with other such alternating terms.
Answer: sinπ
2n
alternates, as does sec(πn) and 4n 2 −n
2
− 1.
WORKSHOP/DISCUSSION
• Write out the first five partial sums of ∞
n=1
1 n,
∞ n=1−1
n, and ∞ n=1
(−1)n+1
n and graph your answers as shown below. Make explicit that you are not graphing 1/n. Analyze the patterns.
Observe, by looking at the graphs, that all the partial sums sn of ∞
n=1
(−1)n+1
n appear to be between 0 and 1. This is a fact that can be proven by a more formal induction argument on the even-odd pairs of partial sums.
SECTION 12.5 ALTERNATING SERIES
• Note that in the case of ∞
n=1
(−1)n+1
n , the associated absolute value series ∞
n=1
1
n diverges, while for ∞
n=0(1.1)−ncos nπ = ∞
n=0(−1)n(1.1)−n, the associated series ∞
n=0(1.1)−n converges. Foreshadow the notion of absolute convergence.
GROUP WORK 1: The Three Conditions
Set this up by making sure the students have read the test in the text. When students apply the Alternating Series Test, it is very tempting for them to neglect to check that the sizes of the terms are decreasing. This activity allows the students to explore that condition, and its importance in the Alternating Series Test. The students may not be able to complete this activity, but there is something to be gained by having the students work on it anyway. Wrestling with this problem will help them to really internalize both the general concept of the Alternating Series Test, and its subtleties. When wrapping up this activity, make sure the students see that if there were no divergent series satisfying conditions 1 and 3, but not condition 2, then we would not need to include condition 2 as part of the test.
Answers:
1. One possible answer (there are many) is:
1− 2 + 1 − 12 + 1 −12 +14 −12 +14 −18 +14 −18 + · · ·
The terms do go to zero, but they are not strictly decreasing. The partial sums of the series go to zero:
1, −1, 0, −12,12, 0,14,−14 , 0, −18,18, 0, . . .
2. One possible answer is:
1− 12+12 −14 +13 −18 +14 −161 + 15−321 + · · ·
The positive terms form the harmonic series, and thus tend to infinity. The sum of the negative terms is
−1.
GROUP WORK 2: Exploring Infinite Series (Part 1)
In this exercise the students explore a series that is conditionally convergent, and that converges slowly. They should explore the relationship between the partial sums Sn and the actual sum S, noting that as n → ∞, Sn → S, and that for a given n we can bound the error |S − Sn|.
Answers:
1. The Alternating Series Test 2. R3= |S − S3|, Rn = |S − Sn| 3. R3 < 12, R6< √17 ≈ 0.37796447, R9 < √110 ≈ 0.31622777
CHAPTER 12 INFINITE SEQUENCES AND SERIES
HOMEWORK PROBLEMS
Core Exercises: 1, 3, 12, 21, 24, 32, 35
Sample Assignment: 1, 3, 6, 12, 15, 17, 21, 24, 29, 32, 35
Exercise D A N G
1 ×
3 ×
6 ×
12 ×
15 ×
17 ×
21 ×
24 ×
29 ×
32 ×
35 × ×
GROUP WORK 1, SECTION 12.5 The Three Conditions
Notice that the Alternating Series Test has three conditions associated with it:
1. The series must alternate
2. The terms must decrease (in absolute value) for large n 3. The nth term must go to 0
The second condition seems redundant. Can you even have a series where the first two conditions hold, but the second doesn’t?
1. Come up with a convergent series that satisfies conditions 1 and 3, but not 2.
2. Come up with a divergent series that satisfies conditions 1 and 3, but not 2.
GROUP WORK 2, SECTION 12.5 Exploring Infinite Series (Part 1)
Consider the series ∞
n=1
(−1)n+1
√n .
1. How do we know that this series converges?
2. Call the sum of this series S, and the nth partial sum Sn. Write an equation for the remainder R3. Write an equation for the remainder Rn.
3. Find upper limits on the remainders R3, R6, and R9.
12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
TRANSPARENCY AVAILABLE
#29 (Figure 1 and Section 12.5 Figure 2)
SUGGESTED TIME AND EMPHASIS 1–2 classes Essential material
POINTS TO STRESS
1. Absolute convergence and conditional convergence.
2. The Ratio Test: When it gives useful information, and when it doesn’t 3. The Root Test: When it gives useful information, and when it doesn’t
QUIZ QUESTIONS
• Text Question: Are the following statements true or false?
1. If a series is absolutely convergent, then it is convergent.
2. If a series is convergent, then it is absolutely convergent.
Answers: 1. True 2. False
• Drill Question: What can we say about the convergence of ∞
n=1an if lim
n→∞
an+1 an
= 1?
Answer: Nothing
MATERIALS FOR LECTURE
• Provide some intuition about the proof of the Ratio Test. For example, try to check the convergence of ∞
n=1
2n
n! by previous methods. First show the students that, by cleverly noting that 2n n! ≤
2 3
n
for n ≥ 7, we can use the Comparison Test to show that ∞
n=1
2n
n! converges. Alternately, if we note that an+1
an = 2 n+ 1 ≤ 2
3 for n≥ 2, we can write an+1 ≤ 23an ≤ · · · ≤2
3
n−1
a2 = 22
3
n−1
for n≥ 2, and hence the series converges by comparison with
∞ n=22
2 3
n−1 .
Show that we can generalize these observations: if we know that lim
n→∞
an+1
an = L < 1, then an+1“acts like” cLn−N, for n larger than some fixed N , and hence the series converges by comparison.
CHAPTER 12 INFINITE SEQUENCES AND SERIES
• Illustrate the power of the Ratio Test in summing recursively defined sequences. For example: let a1 = 1 and an+1 = |sin n|
n an.The Ratio Test can show that ∞
n=1an converges, even though the individual terms are hard to compute explicitly. Perhaps use the Ratio Test to check if
∞
n=1bn converges, where{bn} is the recursive sequence defined by b1= 1, bn+1=
1+1
n
n bn. Answer: lim
n→∞
bn+1
bn = e > 1, so the series diverges.
• The topic of rearrangements of conditionally convergent series can be both counterintuitive and beautiful.
After going over the example in the text, perhaps have the students try to approximate numbers such as 2, √
2 and e only using distinct terms from the alternating harmonic series. Working on these problems
“hands on” will help the students to understand this result.
WORKSHOP/DISCUSSION
• When stating the Ratio Test, be sure to indicate that limn→∞
an+1 an
= 1 gives no information. Present examples of series for which the Ratio Test is not helpful, such as the p-series ∞
n=1
1
np for various p.
• Apply the Ratio Test to ∞
n=1
(1.1)n n2 , ∞
n=1
n+ 2
n· n!, and ∞
n=2
1
n ln n. (The Ratio Test is inconclusive in the last case. The Integral Test would have been a better choice.)
Answer: Diverges, converges, diverges (by the Integral Test)
• Test the convergence of ∞
n=1
(2n)!
nn , ∞
n=1
1 nn, ∞
n=1
en
n!, and ∞
n=1
2n
n2, using either the Ratio Test or the Root Test, as appropriate.
Answer: Divergent, convergent, convergent, divergent GROUP WORK 1: Exploring Infinite Series (Part 2)
Several of the questions in this exercise have no answer. If the students are struggling, tell them that “No such series exists” is a possible answer.
Answers: (Answers to Problems 1, 4, 5, and 6 will vary.) 1.
1,12,13,14,15, . . .
2. This is not possible, due to the Test for Divergence.
3. This is not possible, due to the Test for Divergence.
4. {−1, 2, −3, 4, −5, 6, . . .}. The Ratio Test requires that
an+1 an
< 1.
5. an =
10 if n≤ 100 0 if n> 100 6. f (x) = sin (πx)
SECTION 12.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
GROUP WORK 2: What’s Your Ratio?
This group work foreshadows power series and the notion of the radius and interval of convergence.
Answers:
1. (−1, 1), by the Ratio Test
2. (−∞, 1) ∪ (1, ∞), by the Ratio Test
3. The Ratio Test is inconclusive for x = ±1. The Alternating Series Test and the Integral Test give us that the series is convergent for both of these values of x.
HOMEWORK PROBLEMS
Core Exercises: 2, 7, 11, 12, 17, 25, 29, 38
Sample Assignment: 2, 5, 7, 8, 11, 12, 17, 21, 24, 25, 29, 31, 34, 36, 38
Exercise D A N G
2 ×
5 ×
7 ×
8 ×
11 ×
12 ×
17 ×
21 ×
24 ×
25 ×
29 ×
31 ×
34 ×
36 ×
38 × ×
