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Sequences and Series, Essays (university) of Mathematics

International Institute for Higher Education Mathématiques

Sequences and Series : Applications + Exer

Typology: Essays (university)

2022/2023

Uploaded on 09/26/2023

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Sequences and Series

Consider the following sum:

2+1

4+1

8+1

16 +···+1

2i+···

The dots at the end indicate that the sum goes on forever. Does this make sense? Can

we assign a numerical value to an infinite sum? While at first it may seem difficult or

impossible, we have certainly done something similar when we talked about one quantity

getting “closer and closer” to a fixed quantity. Here we could ask whether, as we add more

and more terms, the sum gets closer and closer to some fixed value. That is, look at

2=1

4=1

2+1

8=1

2+1

4+1

16 =1

2+1

4+1

8+1

and so on, and ask whether these values have a limit. It seems pretty clear that they do,

namely 1. In fact, as we will see, it’s not hard to show that

2+1

4+1

8+1

16 +···+1

2i=2i−1

2i= 1 −1

253

Discover Essays (university) of Mathématiques International Institute for Higher Education

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Sequences and Series

Consider the following sum:

i

The dots at the end indicate that the sum goes on forever. Does this make sense? Can

we assign a numerical value to an infinite sum? While at first it may seem difficult or

impossible, we have certainly done something similar when we talked about one quantity

getting “closer and closer” to a fixed quantity. Here we could ask whether, as we add more

and more terms, the sum gets closer and closer to some fixed value. That is, look at

and so on, and ask whether these values have a limit. It seems pretty clear that they do,

namely 1. In fact, as we will see, it’s not hard to show that

i

254 Chapter 11 Sequences and Series

and then

lim

i→∞

i

There is one place that you have long accepted this notion of infinite sum without really

thinking of it as a sum:

for example, or

Our first task, then, to investigate infinite sums, called series, is to investigate limits of

sequences of numbers. That is, we officially call

i=

i

a series, while

i

is a sequence, and

i=

i

= lim

i→∞

i

that is, the value of a series is the limit of a particular sequence.

11.1 Sequen es

While the idea of a sequence of numbers, a

, a

,... is straightforward, it is useful to

think of a sequence as a function. We have up until now dealt with functions whose domains

are the real numbers, or a subset of the real numbers, like f (x) = sin x. A sequence is a

function with domain the natural numbers N = { 1 , 2 , 3 ,.. .} or the non-negative integers,

Z

= { 0 , 1 , 2 , 3 ,.. .}. The range of the function is still allowed to be the real numbers; in

symbols, we say that a sequence is a function f : N → R. Sequences are written in a few

different ways, all equivalent; these all mean the same thing:

a

, a

{a

n

n=

{f (n)}

n=

As with functions on the real numbers, we will most often encounter sequences that

can be expressed by a formula. We have already seen the sequence a

i

= f (i) = 1 − 1 / 2

i

256 Chapter 11 Sequences and Series

closer to a single value, but take on all values between −1 and 1 over and over. In general,

whenever you want to know lim

n→∞

f (n) you should first attempt to compute lim

x→∞

f (x),

since if the latter exists it is also equal to the first limit. But if for some reason lim

x→∞

f (x)

does not exist, it may still be true that lim

n→∞

f (n) exists, but you’ll have to figure out

another way to compute it.

It is occasionally useful to think of the graph of a sequence. Since the function is

defined only for integer values, the graph is just a sequence of dots. In figure 11.1.1 we see

the graphs of two sequences and the graphs of the corresponding real functions.

0 5 10

f (x) = 1/x

0 5 10

f (n) = 1/n

− 1

f (x) = sin(xπ)

− 1

1 2 3 4 5 6 7 8

f (n) = sin(nπ)

Figure 11.1.1 Graphs of sequences and their corresponding real functions.

Not surprisingly, the properties of limits of real functions translate into properties of

sequences quite easily. Theorem 2.3.6 about limits becomes

THEOREM 11.1.2 Suppose that lim

n→∞

a

n

= L and lim

n→∞

b

n

= M and k is some constant.

Then

lim

n→∞

ka

n

= k lim

n→∞

a

n

= kL

lim

n→∞

(a

n

+ b

n

) = lim

n→∞

a

n

+ lim

n→∞

b

n

= L + M

lim

n→∞

(a

n

− b

n

) = lim

n→∞

a

n

− lim

n→∞

b

n

= L − M

lim

n→∞

(a

n

b

n

) = lim

n→∞

a

n

· lim

n→∞

b

n

= LM

lim

n→∞

a

n

b

n

lim

n→∞

a

n

lim

n→∞

b

n

L

M

, if M is not 0

Likewise the Squeeze Theorem (4.3.1) becomes

11.1 Sequences 257

THEOREM 11.1.3 Suppose that a

n

≤ b

n

≤ c

n

for all n > N , for some N. If lim

n→∞

a

n

lim

n→∞

c

n

= L, then lim

n→∞

b

n

= L.

And a final useful fact:

THEOREM 11.1.4 lim

n→∞

|a

n

| = 0 if and only if lim

n→∞

a

n

This says simply that the size of a

n

gets close to zero if and only if a

n

gets close to

zero.

EXAMPLE 11.1.5 Determine whether

n

n + 1

n=

converges or diverges. If it con-

verges, compute the limit. Since this makes sense for real numbers we consider

lim

x→∞

x

x + 1

= lim

x→∞

x + 1

Thus the sequence converges to 1.

EXAMPLE 11.1.6 Determine whether

ln n

n

n=

converges or diverges. If it con-

verges, compute the limit. We compute

lim

x→∞

ln x

x

= lim

x→∞

1 /x

using L’Hˆopital’s Rule. Thus the sequence converges to 0.

EXAMPLE 11.1.7 Determine whether {(−1)

n

n=

converges or diverges. If it con-

verges, compute the limit. This does not make sense for all real exponents, but the sequence

is easy to understand: it is

and clearly diverges.

EXAMPLE 11.1.8 Determine whether {(− 1 /2)

n

n=

converges or diverges. If it con-

verges, compute the limit. We consider the sequence {|(− 1 /2)

n

n=

n

n=

. Then

lim

x→∞

x

= lim

x→∞

x

so by theorem 11.1.4 the sequence converges to 0.

11.1 Sequences 259

THEOREM 11.1.12 If a sequence is bounded and monotonic then it converges.

We will not prove this; the proof appears in many calculus books. It is not hard to

believe: suppose that a sequence is increasing and bounded, so each term is larger than the

one before, yet never larger than some fixed value N. The terms must then get closer and

closer to some value between a

and N. It need not be N , since N may be a “too-generous”

upper bound; the limit will be the smallest number that is above all of the terms a

i

EXAMPLE 11.1.13 All of the terms (

i

are less than 2, and the sequence is

increasing. As we have seen, the limit of the sequence is 1: 1 is the smallest number that

is bigger than all the terms in the sequence. Similarly, all of the terms (n + 1)/n are bigger

than 1/2, and the limit is 1: 1 is the largest number that is smaller than the terms of the

sequence.

We don’t actually need to know that a sequence is monotonic to apply this theorem—

it is enough to know that the sequence is “eventually” monotonic, that is, that at some

point it becomes increasing or decreasing. For example, the sequence 10, 9, 8, 15, 3, 21, 4,

3 /4, 7/8, 15/16, 31/ 32 ,... is not increasing, because among the first few terms it is not.

But starting with the term 3/4 it is increasing, so the theorem tells us that the sequence

3 / 4 , 7 / 8 , 15 / 16 , 31 / 32 ,... converges. Since convergence depends only on what happens as

n gets large, adding a few terms at the beginning can’t turn a convergent sequence into a

divergent one.

EXAMPLE 11.1.14 Show that {n

1 /n

} converges.

We first show that this sequence is decreasing, that is, that n

1 /n

> (n+1)

1 /(n+1)

. Consider

the real function f (x) = x

1 /x

when x ≥ 1. We can compute the derivative, f

(x) =

x

1 /x

(1−ln x)/x

, and note that when x ≥ 3 this is negative. Since the function has negative

slope, n

1 /n

> (n + 1)

1 /(n+1)

when n ≥ 3. Since all terms of the sequence are positive, the

sequence is decreasing and bounded when n ≥ 3, and so the sequence converges. (As it

happens, we can compute the limit in this case, but we know it converges even without

knowing the limit; see exercise 1.)

EXAMPLE 11.1.15 Show that {n!/n

n

} converges.

Again we show that the sequence is decreasing, and since each term is positive the sequence

converges. We can’t take the derivative this time, as x! doesn’t make sense for x real. But

we note that if a

n+

/a

n

< 1 then a

n+

< a

n

, which is what we want to know. So we look

at a

n+

/a

n

a

n+

a

n

(n + 1)!

(n + 1)

n+

n

n!

(n + 1)!

n!

n

(n + 1)

n+

n + 1

n

n + 1

n

n + 1

n

260 Chapter 11 Sequences and Series

(Again it is possible to compute the limit; see exercise 2.)

Exercises 11.1.

Compute lim

x→∞

1 /x

. ⇒

Use the squeeze theorem to show that lim

n→∞

n

= 0.

Determine whether {

√

n + 47 −

√

n=

converges or diverges. If it converges, compute the

limit. ⇒

Determine whether

{

(n + 1)

}

n=

converges or diverges. If it converges, compute the limit.

⇒

Determine whether

{

n + 47

√

}

n=

converges or diverges. If it converges, compute the

limit. ⇒

Determine whether

{

n

}

n=

converges or diverges. ⇒

11.2 Series

While much more can be said about sequences, we now turn to our principal interest,

series. Recall that a series, roughly speaking, is the sum of a sequence: if {a

n

n=

is a

sequence then the associated series is

i=

a

n

= a

+ a

Associated with a series is a second sequence, called the sequence of partial sums

{s

n

n=

s

n

i=

a

i

So

s

= a

, s

= a

+ a

, s

= a

+ a

A series converges if the sequence of partial sums converges, and otherwise the series

diverges.

EXAMPLE 11.2.1 If a

n

= kx

n

n=

a

n

is called a geometric series. A typical partial

sum is

s

n

= k + kx + kx

+ kx

+ · · · + kx

n

= k(1 + x + x

+ x

+ · · · + x

n

262 Chapter 11 Sequences and Series

(a

n

+ b

n

) is convergent and

(a

n

+ b

n

a

n

b

n

The converses of the two parts of this theorem are subtly different. Suppose that

a

n

diverges; does

ca

n

also diverge if c is non-zero? Yes: suppose instead that

ca

n

converges; then by the theorem,

(1/c)ca

n

converges, but this is the same as

a

n

, which

by assumption diverges. Hence

ca

n

also diverges. Note that we are applying the theorem

with a

n

replaced by ca

n

and c replaced by (1/c).

Now suppose that

a

n

and

b

n

diverge; does

(a

n

+ b

n

) also diverge? Now the

answer is no: Let a

n

= 1 and b

n

= −1, so certainly

a

n

and

b

n

diverge. But

(a

n

b

n

0 = 0. Of course, sometimes

(a

n

+ b

n

) will also diverge, for

example, if a

n

= b

n

= 1, then

(a

n

+ b

n

2 diverges.

In general, the sequence of partial sums s

n

is harder to understand and analyze than

the sequence of terms a

n

, and it is difficult to determine whether series converge and if so

to what. Sometimes things are relatively simple, starting with the following.

THEOREM 11.2.3 If

a

n

converges then lim

n→∞

a

n

Proof. Since

a

n

converges, lim

n→∞

s

n

= L and lim

n→∞

s

n− 1

= L, because this really says

the same thing but “renumbers” the terms. By theorem 11.1.2,

lim

n→∞

(s

n

− s

n− 1

) = lim

n→∞

s

n

− lim

n→∞

s

n− 1

= L − L = 0.

But

s

n

− s

n− 1

= (a

+ a

+ · · · + a

n

) − (a

+ a

+ · · · + a

n− 1

) = a

n

so as desired lim

n→∞

a

n

This theorem presents an easy divergence test: if given a series

a

n

the limit lim

n→∞

a

n

does not exist or has a value other than zero, the series diverges. Note well that the

converse is not true: If lim

n→∞

a

n

= 0 then the series does not necessarily converge.

EXAMPLE 11.2.4 Show that

n=

n

n + 1

diverges.

We compute the limit:

lim

n→∞

n

n + 1

Looking at the first few terms perhaps makes it clear that the series has no chance of

converging:

11.2 Series 263

will just get larger and larger; indeed, after a bit longer the series starts to look very much

like · · · + 1 + 1 + 1 + 1 + · · ·, and of course if we add up enough 1’s we can make the sum

as large as we desire.

EXAMPLE 11.2.5 Show that

n=

n

diverges.

Here the theorem does not apply: lim

n→∞

1 /n = 0, so it looks like perhaps the series con-

verges. Indeed, if you have the fortitude (or the software) to add up the first 1000 terms

you will find that

n=

n

so it might be reasonable to speculate that the series converges to something in the neigh-

borhood of 10. But in fact the partial sums do go to infinity; they just get big very, very

slowly. Consider the following:

and so on. By swallowing up more and more terms we can always manage to add at least

another 1/2 to the sum, and by adding enough of these we can make the partial sums as

big as we like. In fact, it’s not hard to see from this pattern that

n

so to make sure the sum is over 100, for example, we’d add up terms until we get to

around 1/ 2

, that is, about 4 · 10

terms. This series,

(1/n), is called the harmonic

series.

Exercises 11.2.

Explain why

∑

n=

2 n

diverges. ⇒

Explain why

∑

n=

1 /n

diverges. ⇒

Explain why

∑

n=

diverges. ⇒

11.3 The Integral Test 265

0 1 2 3 4 5

A = 1

A = 1/ 4

Figure 11.3.1 Graph of y = 1/x

with rectangles.

area under the curve, and so of course any sum of rectangle areas is less than the area

under the entire curve, that is, all the way to infinity. There is a bit of trouble at the left

end, where there is an asymptote, but we can work around that easily. Here it is:

s

n

x

dx < 1 +

x

dx = 1 + 1 = 2,

recalling that we computed this improper integral in section 9.7. Since the sequence of

partial sums s

n

is increasing and bounded above by 2, we know that lim

n→∞

s

n

= L ≤ 2, and

so the series converges to some number at most 2. In fact, it is possible, though difficult,

to show that L = π

We already know that

1 /n diverges. What goes wrong if we try to apply this

technique to it? Here’s the calculation:

s

n

x

dx < 1 +

x

dx = 1 + ∞.

The problem is that the improper integral doesn’t converge. Note well that this does

not prove that

1 /n diverges, just that this particular calculation fails to prove that it

converges. A slight modification, however, allows us to prove in a second way that

1 /n

diverges.

EXAMPLE 11.3.2 Consider a slightly altered version of figure 11.3.1, shown in fig-

ure 11.3.2.

The rectangles this time are above the curve, that is, each rectangle completely contains

the corresponding area under the curve. This means that

s

n

n+

x

dx = ln x

n+

= ln(n + 1).

As n gets bigger, ln(n + 1) goes to infinity, so the sequence of partial sums s

n

must also

go to infinity, so the harmonic series diverges.

266 Chapter 11 Sequences and Series

0 1 2 3 4 5

A = 1

A = 1/ 2

A = 1/ 3

Figure 11.3.2 Graph of y = 1/x with rectangles.

The important fact that clinches this example is that

lim

n→∞

n+

x

dx = ∞,

which we can rewrite as

x

dx = ∞.

So these two examples taken together indicate that we can prove that a series converges

or prove that it diverges with a single calculation of an improper integral. This is known

as the integral test, which we state as a theorem.

THEOREM 11.3.3 Suppose that f (x) > 0 and is decreasing on the infinite interval

[k, ∞) (for some k ≥ 1) and that a

n

= f (n). Then the series

n=

a

n

converges if and only

if the improper integral

f (x) dx converges.

The two examples we have seen are called p-series; a p-series is any series of the form

1 /n

p

. If p ≤ 0, lim

n→∞

1 /n

p

= 0, so the series diverges. For positive values of p we can

determine precisely which series converge.

THEOREM 11.3.4 A p-series with p > 0 converges if and only if p > 1.

Proof. We use the integral test; we have already done p = 1, so assume that p 6 = 1.

x

p

dx = lim

D→∞

x

1 −p

1 − p

D

= lim

D→∞

D

1 −p

1 − p

If p > 1 then 1 − p < 0 and lim

D→∞

D

1 −p

= 0, so the integral converges. If 0 < p < 1 then

1 − p > 0 and lim

D→∞

D

1 −p

= ∞, so the integral diverges.

268 Chapter 11 Sequences and Series

We can compute the integral:

N

x

dx =

N

so N = 100 is a good starting point. Adding up the first 100 terms gives approximately

1 .634983900, and that plus 1/100 is 1.644983900, so approximating the series by the value

halfway between these will be at most 1/200 = 0.005 in error. The midpoint is 1.639983900,

but while this is correct to ± 0 .005, we can’t tell if the correct two-decimal approximation

is 1.63 or 1.64. We need to make N big enough to reduce the guaranteed error, perhaps to

around 0.004 to be safe, so we would need 1/N ≈ 0 .008, or N = 125. Now the sum of the

first 125 terms is approximately 1.636965982, and that plus 0.008 is 1.644965982 and the

point halfway between them is 1.640965982. The true value is then 1. 640965982 ± 0 .004, and

all numbers in this range round to 1.64, so 1.64 is correct to two decimal places. We have

mentioned that the true value of this series can be shown to be π

/ 6 ≈ 1 .644934068 which

rounds down to 1.64 (just barely) and is indeed below the upper bound of 1.644965982,

again just barely. Frequently approximations will be even better than the “guaranteed”

accuracy, but not always, as this example demonstrates.

Exercises 11.3.

Determine whether each series converges or diverges.

∑

n=

⇒ 2.

∑

n=

⇒

∑

n=

ln n

⇒ 4.

∑

n(ln n)

⇒

Find an N so that

∑

n=

is between

N

∑

n=

and

N

∑

n=

0.005. ⇒

Find an N so that

∑

n=

ln n

0.005. ⇒

Find an N so that

11.4 Alternating Series

Next we consider series with both positive and negative terms, but in a regular pattern:

they alternate, as in the alternating harmonic series for example:

n=

n− 1

n

In this series the sizes of the terms decrease, that is, |a

n

| forms a decreasing sequence,

but this is not required in an alternating series. As with positive term series, however,

when the terms do have decreasing sizes it is easier to analyze the series, much easier, in

fact, than positive term series. Consider pictorially what is going on in the alternating

harmonic series, shown in figure 11.4.1. Because the sizes of the terms a

n

are decreasing,

the partial sums s

, s

, and so on, form a decreasing sequence that is bounded below

by s

, so this sequence must converge. Likewise, the partial sums s

, s

, and so on,

form an increasing sequence that is bounded above by s

, so this sequence also converges.

Since all the even numbered partial sums are less than all the odd numbered ones, and

since the “jumps” (that is, the a

i

terms) are getting smaller and smaller, the two sequences

must converge to the same value, meaning the entire sequence of partial sums s

, s

converges as well.

1 = s

= a

= −

Figure 11.4.1 The alternating harmonic series.

There’s nothing special about the alternating harmonic series—the same argument

works for any alternating sequence with decreasing size terms. The alternating series test

is worth calling a theorem.

THEOREM 11.4.1 Suppose that {a

n

n=

is a non-increasing sequence of positive

numbers and lim

n→∞

a

n

= 0. Then the alternating series

n=

n− 1

a

n

converges.

Proof. The odd numbered partial sums, s

, s

, and so on, form a non-increasing

sequence, because s

2 k+

= s

2 k+

− a

2 k+

+ a

2 k+

≤ s

2 k+

, since a

2 k+

≥ a

2 k+

. This

11.5 Comparison Tests 271

Exercises 11.4.

Determine whether the following series converge or diverge.

∑

n=

(−1)

n− 1

2 n + 5

⇒ 2.

∑

n=

(−1)

n− 1

√

n − 3

⇒

∑

n=

(−1)

n− 1

3 n − 2

⇒ 4.

∑

n=

(−1)

n− 1

ln n

⇒

Approximate

∑

n=

(−1)

n− 1

to two decimal places. ⇒

Approximate

∑

n=

(−1)

n− 1

to two decimal places. ⇒

11.5 Comparison Tests

Sequences and Series, Essays (university) of Mathematics

Related documents

Partial preview of the text

Download Sequences and Series and more Essays (university) Mathematics in PDF only on Docsity!

Sequences and Series

Consider the following sum:

i

The dots at the end indicate that the sum goes on forever. Does this make sense? Can

we assign a numerical value to an infinite sum? While at first it may seem difficult or

impossible, we have certainly done something similar when we talked about one quantity

getting “closer and closer” to a fixed quantity. Here we could ask whether, as we add more

and more terms, the sum gets closer and closer to some fixed value. That is, look at

and so on, and ask whether these values have a limit. It seems pretty clear that they do,

namely 1. In fact, as we will see, it’s not hard to show that

i

i

i

i

254 Chapter 11 Sequences and Series

and then

lim

i→∞

i

There is one place that you have long accepted this notion of infinite sum without really

thinking of it as a sum:

for example, or

Our first task, then, to investigate infinite sums, called series, is to investigate limits of

sequences of numbers. That is, we officially call

i=

i

i

a series, while

i

i

is a sequence, and

i=

i

= lim

i→∞

i

i

that is, the value of a series is the limit of a particular sequence.

While the idea of a sequence of numbers, a

, a

, a

,... is straightforward, it is useful to

think of a sequence as a function. We have up until now dealt with functions whose domains

are the real numbers, or a subset of the real numbers, like f (x) = sin x. A sequence is a

function with domain the natural numbers N = { 1 , 2 , 3 ,.. .} or the non-negative integers,

Z

= { 0 , 1 , 2 , 3 ,.. .}. The range of the function is still allowed to be the real numbers; in

symbols, we say that a sequence is a function f : N → R. Sequences are written in a few

different ways, all equivalent; these all mean the same thing:

a

, a

, a

{a

n

n=

{f (n)}

n=

As with functions on the real numbers, we will most often encounter sequences that

can be expressed by a formula. We have already seen the sequence a

i

= f (i) = 1 − 1 / 2

i

256 Chapter 11 Sequences and Series

closer to a single value, but take on all values between −1 and 1 over and over. In general,

whenever you want to know lim

n→∞

f (n) you should first attempt to compute lim

x→∞

f (x),

since if the latter exists it is also equal to the first limit. But if for some reason lim

x→∞

f (x)

does not exist, it may still be true that lim

n→∞

f (n) exists, but you’ll have to figure out

another way to compute it.