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# INFINITE SERIES

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C H A P T E R

## 7

In our daily lives, we are increasingly seeing the impact of digital technologies.

You needn’t go far to observe this phenomenon. For instance, the dominant media for the entertainment industry are now CDs and DVDs; we have digital video and still cameras and the Internet gives us easy access to a virtual world of digital information. An essential ingredient in this digital revolution is the use of Fourier analysis, a mathematical idea that is introduced in this chapter.

In this digital age, we have learned to represent information in a variety of ways. The ability to easily transform one representation into another gives us tremendous problem-solving powers. As an example, consider the music made

by a saxophone. The music is initially represented as a series of notes on sheet music, but the musician brings her own special interpretation to the music. Such an individual performance can then be recorded, to be copied and replayed later. While this is easily accomplished with conventional analog technology, the advent of digital technology has allowed us to record the performance with a previously unknown

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fidelity. The key to this is that the music is broken down into its component parts, which are individually recorded and then reassembled on demand to recreate the original sound. Think for a moment how spectacular this feat really is. The complex rhythms and intonations gener- ated by the saxophone reed and body are somehow converted into a relatively small number of digital bits (zeroes and ones). The bits are then turned back into music by a CD player.

The basic idea behind any digital technology is to break down a complex whole into a set of component pieces. To digitally capture a saxophone note, all of the significant features of the saxophone waveform must be captured.

Done properly, the components can then be recombined to reproduce each original note.

In this chapter, we learn how series of numbers combine and how functions can be broken down into a series of component functions. As part of this discussion, we will explore how music synthesizers work, but we will also see how calculators can quickly approximate a quantity like sin 1.23 and how equations can be solved using functions for which we don’t even have names. This chapter opens up a new world of important applications.

### 7.1 SEQUENCES OF REAL NUMBERS

The mathematical notion of sequence is not much different from the common English usage of the word. For instance, if you were asked to describe the sequence of events that led up to a traffic accident, you’d not only need to list the events, but you’d need to do so in a specific order (hopefully, the order in which they actually occurred). In mathematics, we use the term sequence to mean an infinite collection of real numbers, written in a specific order.

We have already seen sequences several times now (although we have not formally in- troduced the notion). For instance, you have found approximate solutions to nonlinear equa- tions like tan x− x = 0, by first making an initial guess, x0and then using Newton’s method to compute a sequence of successively improved approximations, x1, x2,. . . , xn,. . . .

By sequence, we mean any function whose domain is the set of integers starting with some integer n0(often 0 or 1). For instance, the function a(n)=n1, for n= 1, 2, 3,. . . , defines the sequence

Definition of sequence

1 1,1

2,1 3,1

4,. . . .

Here,11is called the first term,12is the second term and so on. We call a(n)= 1nthe general term, since it gives a (general) formula for computing all the terms of the sequence. Further, we usually use subscript notation instead of function notation and write aninstead of a(n).

EXAMPLE 1.1 The Terms of a Sequence

Write out the terms of the sequence whose general term is given by an = n+ 1 n , for n= 1, 2, 3,. . . .

Solution We have the sequence a1= 1+ 1

1 = 2

1, a2 =2+ 1

2 = 3

2, a3=4

3, a4= 5 4,. . . .



We often use set notation to denote a sequence. For instance, the sequence with general term an = 1

n2, for n= 1, 2, 3,. . . , is denoted by {an}n=1=

 1 n2



n=1,

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an

n 0.04

5 10 15 20

0.08 0.12

FIGURE 7.1 an= 1

n2.

or equivalently, by listing the terms of the sequence:

1 1, 1

22, 1 32,. . . ,1

n2,. . .

 .

To graph this sequence, we plot a number of discrete points, since a sequence is a function defined only on the integers (see Figure 7.1). You have likely already noticed something about the sequence

1 n2



n=1

. As n gets larger and larger, the terms of the sequence, an = 1 n2 get closer and closer to zero. In this case, we say that the sequence converges to 0 and write

nlim→∞an = lim

n→∞

1 n2 = 0.

In general, we say that the sequence{an}n=1converges to L

 i.e., limn

→∞an = L if we can make anas close to L as desired, simply by making n sufficiently large. We call L the limit of the sequence. You may notice that this language parallels that used in the definition of the limit

xlim→∞f (x)= L

for a function of a real variable x (given in Chapter 1). The only difference is that n can take on only integer values, while x can take on any real value (integer, rational or irrational).

Most of the usual rules for computing limits of functions of a real variable also apply to computing the limit of a sequence, as we see in the following result.

THEOREM 1.1

Suppose that{an}n=n0and{bn}n=n0both converge. Then (i) lim

n→∞(an+ bn)= lim

n→∞an+ lim

n→∞bn, (ii) lim

n→∞(an− bn)= lim

n→∞an− lim

n→∞bn, (iii) lim

n→∞(anbn)=

nlim→∞an

 

nlim→∞bn

 and

(iv) lim

n→∞

an

bn

= lim

n→∞an nlim→∞bn

(assuming lim

n→∞bn = 0).

The proof of Theorem 1.1 is virtually identical to the proof of the corresponding theorem about limits of a function of a real variable (see Theorem 3.1 in section 1.3 and Appendix G) and is omitted.

### REMARK 1.1

To find the limit of a sequence, you should work largely the same as when computing the limit of a function of a real variable, but keep in mind that sequences are defined only for integer values of the variable.

### NOTES

If you (incorrectly) apply l’Hˆopital’s Rule in example 1.2, you get the right answer. (Go ahead and try it; nobody’s looking.) Unfortunately, you will not always be so lucky. It’s a lot like trying to cross a busy highway: while there are times when you can successfully cross with your eyes closed, it’s not generally recommended.

Theorem 1.2 describes how you can safely use l’Hˆopital’s Rule.

EXAMPLE 1.2 Finding the Limit of a Sequence Evaluate lim

n→∞

5n+ 7 3n− 5.

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an

n

5 10 15 20

1 2 3 4 5 6

FIGURE 7.2 an = 5n+ 7

3n− 5.

Solution Of course, this has the indeterminate form ∞

. From the graph in Figure 7.2, it looks like the sequence tends to some limit around 2. Note that we cannot apply l’Hˆopital’s Rule here, since the functions in the numerator and the denominator are not continuous. (They are only defined for integer values of n, even though you could define these expressions for any real values of n.) You can, of course, use the simpler method of dividing numerator and denominator by the highest power of n in the denominator. We have

nlim→∞

5n+ 7 3n− 5 = lim

n→∞

(5n+ 7)1

n

 (3n− 5)1

n

 = limn→∞5+n7 3−n5 =5

3.



In example 1.3, we see a sequence that diverges by virtue of its terms tending to+∞.

EXAMPLE 1.3 A Divergent Sequence Evaluate lim

n→∞

n2+ 1 2n− 3. an

n

5 10 15 20

2 4 6 8 10 12

FIGURE 7.3 an = n2+ 1

2n− 3.

Solution Again, this has the indeterminate form ∞

∞, but from the graph in Figure 7.3, the sequence appears to be increasing without bound. Dividing top and bottom by n (the highest power of n in the denominator) we have

nlim→∞

n2+ 1 2n− 3 = lim

n→∞

(n2+ 1)1

n

 (2n− 3)1

n

 = limn→∞n+1n 2−n3 = ∞ and so, the sequence

n2+ 1 2n− 3



n=1

diverges.



10 15

5

1 1 an

n

FIGURE 7.4 an = (−1)n.

In example 1.4, we see that a sequence doesn’t need to tend to ±∞ in order to diverge.

EXAMPLE 1.4 A Divergent Sequence Whose Terms Do Not Tend to

Determine the convergence or divergence of the sequence{(−1)n}n=1. Solution If we write out the terms of the sequence, we have

{−1, 1, −1, 1, −1, 1,. . .}.

That is, the terms of the sequence alternate back and forth between−1 and 1 and so, the sequence diverges. To see this graphically, we plot the first few terms of the sequence in Figure 7.4. Notice that the points do not approach any limit (a horizontal line). 

You can use an advanced tool like l’Hˆopital’s Rule to find the limit of a sequence, but you must be careful. Theorem 1.2 says that if f (x)→ L as x → ∞ through all real values, then f (n) must approach L, too, as n→ ∞ through integer values. (See Figure 7.5 for a graphical representation of this.)

THEOREM 1.2 Suppose that lim

x→∞ f (x)= L. Then, limn

→∞ f (n)= L, also.

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5 10 1

2 an

n

FIGURE 7.5

an= f (n), where f (x) → 2, as x→ ∞.

### REMARK 1.2

The converse of Theorem 1.2 is false. That is, if lim

n→∞ f (n)= L, it need not be true that lim

x→∞ f (x)= L. This is clear from the following observation. Note that

nlim→∞cos (2πn) = 1, since cos (2πn) = 1 for every integer n (see Figure 7.6).

However,

xlim→∞cos (2πx) does not exist,

since as x→ ∞, cos (2πx) oscillates between −1 and 1 (see Figure 7.7).

5 10 15 20

0.2 0.4 0.6 0.8 1.0 1.2 an

n

10

0.5

1 0.5 1

x y

FIGURE 7.6 an= cos (2πn).

FIGURE 7.7 y= cos (2πx).

EXAMPLE 1.5 Applying L’Hˆopital’s Rule to a Related Function Evaluate lim

n→∞

n+ 1 en .

5 10 15 20

0.2 0.4 0.6 0.8 an

n

FIGURE 7.8 an= n+ 1 en .

Solution This has the indeterminate form ∞

∞, but from the graph in Figure 7.8, it appears that the sequence converges to 0. However, there is no obvious way to resolve this, except by l’Hˆopital’s Rule (which does not apply to limits of sequences). So, we instead consider the limit of the corresponding function of a real variable to which we may apply l’Hˆopital’s Rule. (Be sure you check the hypotheses.) We have

xlim→∞

x+ 1 ex = lim

x→∞

d

d x(x+ 1) d d x(ex)

= lim

x→∞

1 ex = 0.

From Theorem 1.2, we now have

nlim→∞

n+ 1

en = 0, also.



Although we now have a few tools for computing the limit of a sequence, most inter- esting sequences resist our attempts to find their limit. In many cases (including infinite series, which we study throughout the remainder of this chapter), we don’t even have an explicit formula for the general term. In such circumstances, we must test the sequence

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for convergence in some indirect way. The first indirect tool we present corresponds to the result (of the same name) for limits of functions of a real variable presented in section 1.3.

THEOREM 1.3 (Squeeze Theorem)

Suppose{an}n=n0and{bn}n=n0are convergent sequences, both converging to the limit, L. If there is an integer n1≥ n0such that for all n≥ n1, an ≤ cn≤ bn, then {cn}n=n0

converges to L, too.

In example 1.6, we demonstrate how to apply the Squeeze Theorem to a sequence.

Observe that the trick here is to find two sequences, one on either side of the given sequence (i.e., one larger and one smaller) that have the same limit.

EXAMPLE 1.6 Applying the Squeeze Theorem to a Sequence Determine the convergence or divergence of

sin n n2



n=1.

5 10 15 20

0.05 0.25 0.20 0.15 0.10 0.05 an

n

FIGURE 7.9 an = sin n

n2 .

Solution From the graph in Figure 7.9, the sequence appears to converge to 0, despite the oscillation. Further, note that you cannot compute this limit using the rules we have established so far. (Try it!) However, recall that

−1 ≤ sin n ≤ 1, for all n.

Dividing through by n2gives us

−1 n2sin n

n2 ≤ 1

n2, for all n ≥ 1.

Finally, observe that

nlim→∞

−1

n2 = 0 = lim

n→∞

1 n2. From the Squeeze Theorem, we now have that

nlim→∞

sin n n2 = 0, also. 

The following useful result follows immediately from Theorem 1.3.

COROLLARY 1.1 If lim

n→∞|an| = 0, then lim

n→∞an= 0, also.

PROOF

Notice that for all n,

−|an| ≤ an ≤ |an| and

nlim→∞|an| = 0 and lim

n→∞(−|an|) = − limn

→∞|an| = 0.

So, from the Squeeze Theorem, lim

n→∞an = 0, too.

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Corollary 1.1 is particularly useful for sequences with both positive and negative terms, as in example 1.7.

EXAMPLE 1.7 A Sequence with Terms of Alternating Signs Determine the convergence or divergence of

(−1)n n



n=1. 10

5 15 20

1

0.5 0.5 an

n

FIGURE 7.10 an= (−1)n

n .

Solution From the graph of the sequence in Figure 7.10, it seems that the sequence oscillates but still may be converging to 0. Since (−1)noscillates back and forth between

−1 and 1, we cannot compute limn

→∞

(−1)n

n directly. However, notice that

(−1)n n

 = 1 n and

nlim→∞

1 n = 0.

From Corollary 1.1, we get that lim

n→∞

(−1)n

n = 0, too.



We remind you of Definition 1.1, which we use throughout the chapter.

DEFINITION 1.1

For any integer n≥ 1, the factorial, n! is defined as the product of the first n positive integers,

n!= 1 · 2 · 3 · · · n.

We define 0!= 1.

Example 1.8 shows a sequence whose limit would be extremely difficult to find without the Squeeze Theorem.

EXAMPLE 1.8 An Indirect Proof of Convergence Investigate the convergence of

n!

nn



n=1

.

5 10 15 20

0.2 0.4 0.6 0.8 1.0 an

n

FIGURE 7.11 an= n!

nn.

Solution First, notice that we have no means of computing lim

n→∞

n!

nn directly. (Try this!) From the graph of the sequence in Figure 7.11, it appears that the sequence is converging to 0. Notice that the general term of the sequence satisfies

0< n!

nn = 1· 2 · 3 · · · n n· n · n · · · n

n factors

= 1

n

2· 3 · · · n n· n · · · n

n− 1 factors

≤ 1

n

 (1)= 1

n. (1.1)

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From the Squeeze Theorem and (1.1), we have that since

nlim→∞

1

n = 0 and lim

n→∞0= 0, then

nlim→∞

n!

nn = 0, also.



Just as we did with functions of a real variable, we need to distinguish between se- quences that are increasing and decreasing. The definitions are quite straightforward.

The sequence{an}n=1is increasing if

a1≤ a2≤ · · · ≤ an ≤ an+1≤ · · · . The sequence{an}n=1is decreasing if

a1≥ a2≥ · · · ≥ an ≥ an+1≥ · · · .

If a sequence is either increasing or decreasing, it is called monotonic.

There are any number of ways to show that a sequence is monotonic. Regardless of which method you use, you will need to show that either an≤ an+1for all n (increasing) or an+1≤ anfor all n (decreasing). One very useful method is to look at the ratio of the two successive terms anand an+1. We illustrate this in example 1.9.

5 10 15 20

0.2 0.4 0.6 0.8 1.0 an

n

FIGURE 7.12 an = n

n+ 1.

EXAMPLE 1.9 An Increasing Sequence Investigate whether the sequence

 n

n+ 1



n=1

is increasing, decreasing or neither.

Solution From the graph in Figure 7.12, it appears that the sequence is increasing.

However, you should not be deceived by looking at the first few terms of a sequence.

Instead, we look at the ratio of two successive terms. So, if we define an = n n+ 1, we have an+1 =n+ 1

n+ 2 and

an+1

an

=

n+ 1 n+ 2



n

n+ 1

 = n+ 1

n+ 2

 n+ 1 n



= n2+ 2n + 1

n2+ 2n = 1 + 1

n2+ 2n > 1. (1.2)

Since an > 0, notice that we can multiply both sides of (1.2) by an, to obtain an+1> an,

for all n and so, the sequence is increasing. As an alternative, notice that you can always consider the function f (x)= x

x+ 1 (of the real variable x) corresponding to the se- quence. Observe that

f(x)=(x+ 1) − x

(x+ 1)2 = 1

(x+ 1)2 > 0,

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which says that the function f (x) is increasing. From this, it follows that the corresponding sequence an = n

n+ 1 is also increasing.



EXAMPLE 1.10 A Sequence That Is Increasing for n≥ 2 Investigate whether the sequence

n!

en



n=1

is increasing, decreasing or neither.

10 5

50 100 150 200 an

n

FIGURE 7.13 an= n!

en.

Solution From the graph of the sequence in Figure 7.13, it appears that the sequence is increasing (and rather rapidly, at that). Here, for an = n!

en, we have an+1 =(n+ 1)!

en+1 , so that

an+1

an =

(n+ 1)!

en+1



n!

en

 = (n+ 1)!

en+1 en n!

= (n+ 1)n!en

e(en)n! =n+ 1

e > 1, for n ≥ 2. Since (n+ 1)! = (n + 1) · n!

and en+1= e · en. (1.3) Multiplying both sides of (1.3) by an > 0, we get

an+1 > an, for n ≥ 2.

Notice that in this case, although the sequence is not increasing for all n, it is increasing for n ≥ 2. Keep in mind that it doesn’t really matter what the first few terms do, anyway.

We are only concerned with the behavior of a sequence as n → ∞. 

We need to define one additional property of sequences. We say that the sequence {an}n=n0is bounded if there is a number M> 0 (called a bound) for which |an| ≤ M, for all n.

There is often some slight confusion here. A bound is not the same as a limit, although the two may coincide. The limit of a convergent sequence is the value that the terms of the se- quence are approaching as n→ ∞. On the other hand, a bound is any number that is greater than or equal to the absolute value of every term. This says that a given sequence may have any number of bounds (e.g., if|an| ≤ 10 for all n, then |an| ≤ 20, for all n, too). However, a sequence may have only one limit (or no limit, in the case of a divergent sequence).

EXAMPLE 1.11 A Bounded Sequence Show that the sequence

3− 4n2 n2+ 1



n=1

is bounded.

Solution We use the fact that 4n2− 3 > 0, for all n ≥ 1, to get

|an| =

3− 4n2 n2+ 1

 =4n2− 3

n2+ 1 < 4n2

n2+ 1 < 4n2 n2 = 4.

So, this sequence is bounded by 4. (We might also say in this case that the sequence is bounded between−4 and 4.) Further, note that we could also use any number greater than 4 as a bound. 

We should emphasize that the reason we are considering whether a sequence is mono- tonic or bounded is that very often we cannot compute the limit of a given sequence directly

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and must rely on indirect means to determine whether or not the sequence is convergent.

Theorem 1.4 provides a powerful tool for the investigation of sequences.

THEOREM 1.4

Every bounded, monotonic sequence converges.

A typical bounded and increasing sequence is illustrated in Figure 7.14a, while a bounded and decreasing sequence is illustrated in Figure 7.14b. In both figures, notice that a bounded and monotonic sequence has nowhere to go and consequently, must converge.

The proof of Theorem 1.4 is rather involved and can be found in a more advanced text.

5 10 15 20

0.2 0.4 0.6 0.8 1.0

n an

5 10 15 20

2 4 6

n an

FIGURE 7.14a

A bounded and increasing sequence.

FIGURE 7.14b

A bounded and decreasing sequence.

In the very common case where we do not know how to compute the limit of a sequence, this theorem says that if we can show that a sequence is bounded and monotonic, then it must also be convergent, although we may have little idea of what its limit might be. Once we establish that a sequence converges, we can approximate its limit by computing a sufficient number of terms, as in example 1.12.

EXAMPLE 1.12 An Indirect Proof of Convergence Investigate the convergence of the sequence

2n n!



n=1

.

5 10 15 20

0.5 1.0 1.5 2.0

n an

FIGURE 7.15 an = 2n

n!.

n an2n n!

2 2

4 0.666667 6 0.088889 8 0.006349 10 0.000282 12 0.0000086 14 1.88 × 10−7 16 3.13 × 10−9 18 4.09 × 10−11 20 4.31 × 10−13

Solution First, note that we do not know how to compute lim

n→∞

2n

n!. This has the inde- terminate form∞

∞, but we cannot apply l’Hˆopital’s Rule to it. (Why not?) The graph in Figure 7.15 suggests that the sequence converges to some number close to 0. To confirm this suspicion, we first show that the sequence is monotonic. We have

an+1

an

=

 2n+1 (n+ 1)!



2n n!

 = 2n+1 (n+ 1)!

n!

2n

= 2(2n) n!

(n+ 1) n!2n = 2

n+ 1 ≤ 1, for n ≥ 1. Since 2n+1= 2 · 2nand

(n+ 1)! = (n + 1) · n! (1.4)

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Multiplying both sides of (1.4) by angives us an+1 ≤ an, for all n and so, the sequence is decreasing. Next, since we have already shown that the sequence is decreasing, observe that

0< 2n n! ≤ 21

1! = 2,

for n≥ 1 (i.e., the sequence is bounded by 2). Since the sequence is both bounded and monotonic, it must be convergent, by Theorem 1.4. To get an approximation of the limit of the sequence, we display a number of terms of the sequence in the accompanying table.

From the table, it appears that the sequence is converging to approximately 0. We can make a slightly stronger statement, though. Since we have established that the sequence is decreasing and convergent, we have from our computations that

0≤ an ≤ a20≈ 4.31 × 10−13, for n ≥ 20.

Further, the limit L must also satisfy the inequality 0≤ L ≤ 4.31 × 10−13.

So, even if we can’t conclude that the sequence converges to 0, we can conclude that it converges to some number extremely close to zero. For most purposes, such an estimate is entirely adequate. Of course, if you need greater precision, you can always compute a few more terms. 

### REMARK 1.3

Do not underestimate the importance of Theorem 1.4. This indirect way of testing a sequence for convergence takes on additional significance when we study infinite series (a special type of sequence that is the topic of the remainder of this chapter).

### EXERCISES 7.1

WRITING EXERCISES

1. Compare and contrast lim

x→∞sinπx and lim

n→∞sinπn. Indicate the domains of the two functions and how this affects the limits.

2. Explain why Theorem 1.2 should be true, taking into account the respective domains and their effect on the limits.

3. In words, explain why a decreasing bounded sequence must converge.

4. A sequence is said to diverge if it does not converge. The word

“diverge” is well chosen for sequences that diverge to∞, but is less descriptive of sequences such as{1, 2, 1, 2, 1, 2,. . .} and {1, 2, 3, 1, 2, 3,. . .}. Briefly describe the limiting behavior of these sequences and discuss other possible limiting behaviors of divergent sequences.

In exercises 1–4, write out the terms a1, a2, . . . , a6of the given sequence.

1. an= 2n− 1

n2 2. an = 3

n+ 4 3. an= 4

n! 4. an = (−1)n n

n+ 1

In exercises 5–10, (a) find the limit of each sequence and (b) plot the sequence on a calculator or CAS.

5. an = 1

n3 6. an= 2

n2 7. an = n

n+ 1 8. an= 2n+ 1

n 9. an = 2

n 10. an= 4

n+ 1

In exercises 11–30, determine whether the sequence converges or diverges. If it converges, determine the limit.

11. an = 3n2+ 1

2n2− 1 12. an= 5n3− 1

2n3+ 1 13. an = n2+ 1

n+ 1 14. an= n2+ 1

n3+ 1 15. an = n+ 2

3n− 1 16. an= n+ 4

n+ 1 17. an = (−1)n n+ 2

3n− 1 18. an= (−1)nn+ 4 n+ 1 19. an = (−1)n n+ 2

n2+ 4 20. an= (−1)n 4 n+ 1

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21. an= cos πn 22. an= sin πn

23. an= ne−n 24. an= cos n

en 25. an= en+ 2

e2n− 1 26. an= e2n

en+ 1 27. an= 3n

en+ 1 28. an= n2n

3n 29. an= cos n

n! 30. an= n!

2n

In exercises 31–34, use the Squeeze Theorem and Corollary 1.1 to prove that the sequence converges to 0 (given that lim

n→∞

1 n

nlim→∞

1 n2 0).

31. an= cos n

n2 32. an= cos nπ

n2 33. an= (−1)ne−n

n 34. an= (−1)nln n

n2

In exercises 35–42, determine whether the sequence is increas- ing, decreasing or neither.

35. an= n+ 3

n+ 2 36. an= n− 1

n+ 1 37. an= en

n 38. an= n

2n 39. an= 2n

(n+ 1)! 40. an= 3n

(n+ 2)!

41. an= 10n

n! 42. an= n!

5n In exercises 43–46, show that the sequence is bounded.

43. an= 3n2− 2

n2+ 1 44. an = 6n− 1

n+ 3 45. an= sin (n2)

n+ 1 46. an = e1/n

47. Numerically estimate the limits of the sequences a n = 1+1nn

and bn = 1−1nn

. Compare the answers to e and e−1.

48. Numerically estimate the limits of the sequences a n = 1+2nn

and bn= 1−2nn

. Compare the answers to e2 and e−2.

49. Given that lim

n→∞

1+n1n

= e, show that lim

n→∞

1+rnn

= er for any constant r . (Hint: Make the substitution n= rm.) 50. Evaluate lim

n→∞

(n+ 1)n

nn and lim

n→∞

(n+ 1)n+1 nn .

51. Numerically estimate the limit of the sequence defined by a1=√

2, a2= 2√

2, a3=

 2

2√

2, and so on.

52. To verify your conjecture from exercise 51, note that the terms form the pattern a2=√

2a1, a3=√

2a2, and so on. If the

sequence converges, then an+1≈ anor√

2an≈ an. Solve the equation√

2a= a to determine the values at which this can happen.

53. Find all values of p such that the sequence an= 1

pn converges.

54. Find all values of p such that the sequence an = 1

np converges.

55. A packing company works with 12square boxes. Show that for n= 1, 2, 3,. . . , a total of n2disks of diameter12nfit into a box. Let anbe the wasted area in a box with n2disks. Com- pute an.

56. The pattern of a sequence can’t always be determined from the first few terms. Start with a circle, pick two points on the circle and connect them with a line segment. The circle is divided into a1= 2 regions. Add a third point, connect all points and show that there are now a2= 4 regions. Add a fourth point, connect all points and show that there are a3= 8 regions. Is the pattern clear? Show that a4= 16 and then compute a5for a surprise!

57. You have heard about the “population explosion.” The follow- ing dramatic warning is adapted from the article “Doomsday:

Friday 13 November 2026” by Foerster, Mora and Amiot in Science (Nov. 1960). Start with a0= 3.049 to indicate that the world population in 1960 was approximately 3.049 billion.

Then compute a1= a0+ 0.005a02.01 to estimate the popula- tion in 1961. Compute a2= a1+ 0.005a12.01 to estimate the population in 1962, then a3= a2+ 0.005a22.01 for 1963, and so on. Continue iterating and compare your calculations to the actual populations in 1970 (3.721 billion), 1980 (4.473 billion) and 1990 (5.333 billion). Then project ahead to the year 2035.

Frightening, isn’t it?

58. The so-called hailstone sequence is defined by xk=

3xk−1+ 1 if xk−1is odd

1

2xk−1 if xk−1is even.

If you start with x1= 2n for some positive integer n, show that xn+1= 1. The question (an unsolved research problem) is whether you eventually reach 1 from any starting value. Try several odd values for x1and show that you always reach 1.

59. A different population model was studied by Fibonacci, an Italian mathematician of the thirteenth century. He imagined a population of rabbits starting with a pair of newborns. For one month, they grow and mature. The second month, they have a pair of newborn baby rabbits. We count the number of pairs of rabbits. Thus far, a0= 1, a1= 1 and a2= 2. The rules are: adult rabbit pairs give birth to a pair of newborns every month, newborns take one month to mature and no rabbits die.

Show that a3= 3, a4= 5 and in general an= an−1+ an−2. This sequence of numbers, known as the Fibonacci sequence, occurs in an amazing number of applications.

60. In this exercise, we visualize the Fibonacci sequence. Start with two squares of side 1 placed next to each other (see Figure A).

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Place a square on the long side of the resulting rectangle (see Figure B); this square has side 2. Continue placing squares on the long sides of the rectangles: a square of side 3 is added in Figure C, then a square of side 5 is added to the bottom of Figure C, and so on.

FIGURE A FIGURE B FIGURE C

Argue that the sides of the squares are determined by the Fibonacci sequence of exercise 59.

61. Start with two circles C1 and C2 of radii r1 and r2, respec- tively, that are tangent to each other and each tangent to the x-axis. Construct the circle C3that is tangent to C1, C2and the x-axis. (See the figure.) If the centers of C1and C2are (c1, r1) and (c2, r2), respectively, show that (c2− c1)2+ (r2− r1)2= (r1+ r2)2and then|c2− c1| = 2√

r1r2. Find similar relation- ships for circles C1 and C3 and for circles C2 and C3. Show that the radius r3of C3is given by√

r3=

r1r2

r1+√ r2

. y

x

62. In exercise 61, construct a sequence of circles where C4 is tangent to C2, C3and the x-axis. Then, C5is tangent to C3, C4

and the x-axis. If you start with unit circles r1= r2= 1, find a formula for the radius rnin terms of Fn, the nth term in the Fibonacci sequence of exercises 59 and 60.

63. Let C be the circle of radius r inscribed in the parabola y= x2. (See the figure.) Show that for r>1/2the y-coordinate c of the center of the circle equals c= 14+ r2.

y

x 1

1 2 4

3

2 1 2

64. In exercise 63, let C1be the circle of radius r1= 1 inscribed in y= x2. Construct a sequence of circles C2, C3and so on, where each circle Cnrests on top of the previous circle Cn−1 (that is, Cnis tangent to Cn−1) and is inscribed in the parabola.

If rnis the radius of circle Cn, find a (simple) formula for rn. 65. Determine whether the sequence an =√n

n converges or di- verges. (Hint: Use n1/n= e(1/n) ln n.)

66. Suppose that a1= 1 and an+1= 12 an+a4n

. Show numer- ically that the sequence converges to 2. To find this limit analytically, let L= lim

n→∞an+1= lim

n→∞an and solve the equa- tion L= 12

L+L4 .

67. As in exercise 66, determine the limit of the sequence defined by a1= 1, an+1= 12

an+acn

for c> 0 and an> 0.

EXPLORATORY EXERCISES

1. Suppose that a ball is launched from the ground with initial ve- locityv. Ignoring air resistance, it will rise to a height of v2/(2g) and fall back to the ground at time t= 2v/g. Depending on how

“lively” the ball is, the next bounce will only rise to a fraction of the previous height. The coefficient of restitution r, defined as the ratio of landing velocity to rebound velocity, measures the liveliness of the ball. The second bounce has launch velocity rv, the third bounce has launch velocity r2v and so on. It might seem that the ball will bounce forever. To see that it does not, ar- gue that the time to complete two bounces is a2= 2vg(1+ r), the time to complete three bounces is a3= 2gv(1+ r + r2), and so on. Take r= 0.5 and numerically determine the limit of this sequence. (We study this type of sequence in detail in section 7.2.) In particular, show that (1+ 0.5) = 32, (1 + 0.5 + 0.52)= 74and (1+ 0.5 + 0.52+ 0.53)= 158, find a general ex- pression for anand determine the limit of the sequence. Argue that at the end of this amount of time, the ball has stopped bouncing.

2. A surprising follow-up to the bouncing ball problem of ex- ercise 1 is found in An Experimental Approach to Nonlinear Dynamics and Chaos by Tufillaro, Abbott and Reilly. Suppose the ball is bouncing on a moving table that oscillates up and down according to the equation A cosωt for some amplitude A and frequencyω. Without the motion of the table, the ball will quickly reach a height of 0 as in exercise 1. For different values of A andω, however, the ball can settle into an amazing variety of patterns. To understand this, explain why the colli- sion between table and ball could subtract or add velocity to the ball (What happens if the table is going up? down?). A sim- plified model of the velocity of the ball at successive collisions with the table isvn+1= 0.8vn− 10 cos (v0+ v1+ · · · + vn).

Starting withv0= 5, compute v1, v2,. . . ,v15. In this case, the ball never settles into a pattern; its motion is chaotic.

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### 7.2 INFINITE SERIES

Recall that we write the decimal expansion of13 as the repeating decimal 1

3 = 0.33333333,

where we understand that the 3s in this expansion go on for ever and ever. An alternative way to think of this is as

1

3 = 0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 + · · ·

= 3(0.1) + 3(0.1)2+ 3(0.1)3+ 3(0.1)4+ · · · + 3(0.1)k+ · · · . (2.1) For convenience, we write (2.1) using summation notation as

1 3 =

k=1

[3(0.1)k]. (2.2)

But, what exactly could we mean by the infinite sum indicated in (2.2)? Of course, you can’t add infinitely many things together. (You can only add two things at a time.) By this expression, we mean that as you add together more and more terms, the sum gets closer and closer to 13.

In general, for any sequence{ak}k=1, suppose we start adding the terms together. We define the individual sums by

S1= a1,

S2= a1+ a2= S1+ a2, S3= a 1+ a2

S2

+ a3= S2+ a3,

S4= a 1+ a 2+ a 3 S3

+ a4= S3+ a4, (2.3)

...

Sn= a 1+ a2+ · · · + a n−1

Sn−1

+ an = Sn−1+ an (2.4)

and so on. We refer to Sn as the nth partial sum. Note that we can compute any one of these as the sum of two numbers: the nth term, an and the previous partial sum, Sn−1, as indicated in (2.4).

For instance, for the sequence

1 2k



k=1, consider the partial sums S1 = 1

2, S2 = 1

2+ 1 22 =3

4, S3 = 3

4+ 1 23 =7

8, S4 = 7

8+ 1 24 =15

16,

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and so on. Look at these carefully and you might notice that S2 =3

4 = 1 − 1

22, S3=7 8 = 1− 1

23, S4=15

16 = 1 − 1

24 and so on, so that Sn= 1 − 1

2n, for each n = 1, 2,. . . . If we were to consider the convergence or divergence of the sequence{Sn}n=1 of partial sums, observe that we now have

nlim→∞Sn= lim

n→∞

1− 1

2n



= 1.

Think about what this says: as we add together more and more terms of the sequence

 1 2k



k=1, the partial sums are drawing closer and closer to 1. In this instance, we write

 k=1

1

2k = 1. (2.5)

It’s very important to understand what’s going on here. This new mathematical object,

 k=1

1

2k is called a series (or infinite series). It is not a sum in the usual sense of the word, but rather, the limit of the sequence of partial sums. Equation (2.5) says that as we add together more and more terms, the sums are approaching the limit of 1.

In general, for any sequence,{ak}k=1, we can write down the series a1+ a2+ · · · + ak+ · · · =

k=1

ak.

If the sequence of partial sums Sn =n

k=1akconverges (to some number S), then we say that the series 

k=1akconverges (to S). We write

 k=1

ak = lim

n→∞

n k=1

ak = lim

n→∞Sn= S.

Definition of infinite series

In this case, we call S the sum of the series. On the other hand, if the sequence of partial sums, {Sn}n=1 diverges (i.e., lim

n→∞Sn does not exist), then we say that the series diverges.

EXAMPLE 2.1 A Convergent Series Determine if the series

k=1

1

2k converges or diverges.

Solution From our work on the introductory example, observe that

 k=1

1 2k = lim

n→∞

n k=1

ak= lim

n→∞

1− 1

2n



= 1.

In this case, we say that the series converges to 1.  In example 2.2, we examine a simple divergent series.

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EXAMPLE 2.2 A Divergent Series

Investigate the convergence or divergence of the series

k=1k2. Solution Here, we have the nth partial sum

Sn =

n k=1

k2= 12+ 22+ · · · + n2 and

nlim→∞Sn = limn

→∞(12+ 22+ · · · + n2)= ∞.

Since the sequence of partial sums diverges, the series diverges also. 

Determining the convergence or divergence of a series is only rarely as simple as it was in examples 2.1 and 2.2.

EXAMPLE 2.3 A Series with a Simple Expression for the Partial Sums

Investigate the convergence or divergence of the series

k=1

1 k(k+ 1).

5 10 15 20

0.2 0.4 0.6 0.8 1.0

n Sn

FIGURE 7.16 Sn=n

k=1

1 k(k+ 1).

Solution In Figure 7.16, we have plotted the first 20 partial sums. In the accompanying table, we list a number of partial sums of the series.

From both the graph and the table, it appears that the partial sums are approaching 1, as n→ ∞. However, we must urge caution. It is extremely difficult to look at a graph or a table of any finite number of the partial sums and decide whether a given series is converging or diverging. In the present case, we are fortunate that we can find a simple expression for the partial sums. We leave it as an exercise to find the partial fractions decomposition of the general term of the series

1

k(k+ 1) =1

k − 1

k+ 1. (2.6)

n Sn n

k1

1 k(k 1)

10 0.90909091

100 0.99009901

1000 0.999001 10,000 0.99990001 100,000 0.99999 1× 106 0.999999 1× 107 0.9999999

Now, consider the nth partial sum. From (2.6), we have Sn =

n k=1

1 k(k+ 1) =

n k=1

1

k − 1

k+ 1



= 1

1−1 2

 +

1 2−1

3

 +

1 3 −1

4

 + · · · +

1

n− 1−1 n

 +

1

n − 1

n+ 1

 .

Notice how nearly every term in the partial sum is canceled by another term in the sum (the next term). For this reason, such a sum is referred to as a telescoping (or collapsing) sum. We now have

Sn= 1 − 1 n+ 1 and so,

nlim→∞Sn = lim

n→∞

1− 1

n+ 1



= 1.

This says that the series converges to 1, as conjectured from the graph and the table. 

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