infinite sequences and seriesrfrith.uaa.alaska.edu/m201/chapter11/chap11_sec3.pdf · 2013. 4....

11INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES

In general, it is difficult to find the exact sum of a series.

We were able to accomplish this for geometric series and the series ∑ 1/[n(n+1)].

This is because, in each of these cases, we can find a simple formula for the nth partial sum sn.

Nevertheless, usually, it isn’t easy to compute lim nns

→∞


So, in the next few sections, we develop several tests that help us determine whether a series is convergent or divergent without explicitly finding its sum.

In some cases, however, our methods will enable us to find good estimates of the sum.


Our first test involves improper integrals.

11.3The Integral Test and Estimates of Sums

In this section, we will learn how to:

Find the convergence or divergence of

a series and estimate its sum.


We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers:

There’s no simple formula for the sum snof the first n terms.

INTEGRAL TEST

2 2 2 2 2 21

1 1 1 1 1 1 ...1 2 3 4 5n n

∞

=

= + + + + +∑

INTEGRAL TEST

However, the computer-generated values given here suggest that the partial sums are approaching near 1.64 as n → ∞.

So, it looks as if the series is convergent.

We can confirm this impression with a geometric argument.

INTEGRAL TEST

This figure shows the curve y = 1/x2

and rectangles that lie below the curve.

INTEGRAL TEST

The base of each rectangle is an interval of length 1.

The height is equal to the value of the function y = 1/x2 at the right endpoint of the interval.

INTEGRAL TEST

Thus, the sum of the areas of the rectangles is:

2 2 2 2 2 21

1 1 1 1 1 1...1 2 3 4 5 n n

∞

=

+ + + + + =∑

If we exclude the first rectangle, the total area of the remaining rectangles is smaller than the area under the curve y = 1/x2 for x ≥ 1, which is the value of the integral

In Section 7.8, we discovered that this improper integral is convergent and has value 1.

INTEGRAL TEST

2

1(1/ )x dx

∞

∫

So, the image shows that all the partial sums are less than

Therefore, the partial sums are bounded.

INTEGRAL TEST

2 21

1 1 21

dxx

∞+ =∫

We also know that the partial sums are increasing (as all the terms are positive).

Thus, the partial sums converge (by the Monotonic Sequence Theorem).

So, the series is convergent.

INTEGRAL TEST

The sum of the series (the limit of the partial sums) is also less than 2:

INTEGRAL TEST

2 2 2 2 21

1 1 1 1 1 ... 21 2 3 4n n

∞

=

= + + + + <∑

The exact sum of this series was found by the mathematician Leonhard Euler (1707–1783) to be π2/6.

However, the proof of this fact is quite difficult.

See Problem 6 in the Problems Plus, following Chapter 15.

INTEGRAL TEST

Now, let’s look at this series: INTEGRAL TEST

1

1 1 1 1 1 1 ...1 2 3 4 5n n

∞

=

= + + + + +∑

INTEGRAL TESTThe table of values of sn suggests that the partial sums aren’t approaching a finite number.

So, we suspect that the given series may be divergent.

Again, we use a picture for confirmation.

INTEGRAL TEST

The figure shows the curve y = 1/ .

However, this time, we use rectangles whose tops lie above the curve.

x

INTEGRAL TEST

The base of each rectangle is an interval of length 1.

The height is equal to the value of the function y = 1/ at the left endpoint of the interval. x

INTEGRAL TEST

So, the sum of the areas of all the rectangles is:

1

1 1 1 1 1 1...1 2 3 4 5 n n

∞

=

+ + + + + =∑

This total area is greater than the area under the curve y = 1/ for x ≥ 1, which is equal to the integral

However, we know from Section 7.8 that this improper integral is divergent.

In other words, the area under the curve is infinite.

INTEGRAL TEST

1(1/ )x dx

∞

∫x

Thus, the sum of the series must be infinite.

That is, the series is divergent.

INTEGRAL TEST

The same sort of geometric reasoning that we used for these two series can be used to prove the following test.

The proof is given at the end of the section.

INTEGRAL TEST

Suppose f is a continuous, positive, decreasing function on [1, ∞) and let an = f(n).

Then, the series is convergent if and only if the improper integral is convergent.

THE INTEGRAL TEST

1n

na

∞

=∑

1( )f x dx

∞

∫

In other words,

i. If is convergent, then is convergent.

ii. If is divergent, then is divergent.

THE INTEGRAL TEST

1( )f x dx

∞

∫1

nn

a∞

=∑

1( )f x dx

∞

∫1

nn

a∞

=∑

When we use the Integral Test, it is not necessary to start the series or the integral at n = 1.

For instance, in testing the series

we use

NOTE

24

1( 3)n n

∞

= −∑24

1( 3)

dxx

∞

−∫

Also, it is not necessary that f be always decreasing.

What is important is that f be ultimately decreasing, that is, decreasing for x larger than some number N.

Then, is convergent.

So, is convergent by Note 4 of Section 11.2

NOTE

nn Na∞

=∑

1 nna∞

=∑

Test the series for convergence or divergence.

The function f(x) = 1/(x2 + 1) is continuous, positive, and decreasing on [1, ∞).

INTEGRAL TEST Example 1

21

11n n

∞

= +∑

So, we use the Integral Test:INTEGRAL TEST Example 1

]

2 21 1

11

1

1 1lim1 1

lim tan

lim tan4

2 4 4

t

t

t

t

t

dx dxx x

x

t π

π π π

∞

→∞

−

→∞

−

→∞

=+ +

=

= −

= − =

∫ ∫

So, is a convergent integral.

Thus, by the Integral Test, the series ∑ 1/(n2 + 1) is convergent.


11/( 1)x dx

∞+∫

For what values of p is the series convergent?


1

1p

n n

∞

=∑

If p < 0, then

If p = 0, then

In either case,

So, the given series diverges by the Test for Divergence (Section 11.2).


lim(1/ )p

nn

→∞= ∞

lim(1/ ) 1p

nn

→∞=

lim(1/ ) 0p

nn

→∞≠

If p > 0, then the function f(x) = 1/xp

is clearly continuous, positive, and decreasing on [1, ∞).


In Section 7.8 (Definition 2), we found that :

Converges if p > 1

Diverges if p ≤ 1

INTEGRAL TEST

1

1p dx

x∞

∫

Example 2

It follows from the Integral Test that the series ∑ 1/np converges if p > 1 and diverges if 0 < p ≤ 1.

For p = 1, this series is the harmonic series discussed in Example 7 in Section 11.2


To use the Integral Test, we need to be able to evaluate

Therefore, we have to be able to find an antiderivative of f.

Frequently, this is difficult or impossible.

So, we need other tests for convergence too.

INTEGRAL TEST

1( )f x dx

∞

∫

The series in Example 2 is called the p-series.

It is important in the rest of this chapter.

So, we summarize the results of Example 2 for future reference—as follows.

p-SERIES

The p-series

is convergent if p > 1 and divergent if p ≤ 1

p-SERIES Result 1

1

1p

n n

∞

=∑

The series

is convergent because it is a p-series with p = 3 > 1

p-SERIES Example 3 a

3 3 3 3 31

1 1 1 1 1 ...1 2 3 4n n

∞

=

= + + + +∑

The series

is divergent because it is a p-series with p = ⅓ < 1.

p-SERIES Example 3 b

1/3 3 33 31 1

1 1 1 1 11 ...2 3 4n nn n

∞ ∞

= =

= = + + + +∑ ∑

We should not infer from the Integral Test that the sum of the series is equal to the value of the integral.

In fact, whereas

Thus, in general,

NOTE

2

21

16n nπ∞

=

=∑

11

( )nn

a f x dx∞ ∞

=

≠∑ ∫

21

1 1dxx

∞=∫

Determine whether the series converges or diverges.

The function f(x) = (ln x)/x is positive and continuous for x > 1 because the logarithm function is continuous.

However, it is not obvious that f is decreasing.


1

lnn

nn

∞

=∑

So, we compute its derivative:

Thus, f’(x) < 0 when ln x > 1, that is, x > e.

It follows that f is decreasing when x > e.


2 2

(1/ ) ln 1 ln'( ) x x x xf xx x− −

= =

So, we can apply the Integral Test:

Since this improper integral is divergent, the series Σ (ln n)/n is also divergent by the Integral Test.

INTEGRAL TEST

2

1 11

2

ln ln (ln )lim lim2

(ln )lim2

tt

t t

t

x x xdx dxx x

t

∞

→∞ →∞

→∞

= =

= = ∞

∫ ∫

Example 4

Suppose we have been able to use the Integral Test to show that a series ∑ an

is convergent.

Now, we want to find an approximation to the sum s of the series.s

ESTIMATING THE SUM OF A SERIES

Of course, any partial sum sn is an approximation to s because

How good is such an approximation?


lim nns

→∞=

To find out, we need to estimate the size of the remainder

Rn = s – sn = an+1 + an+2 + an+3+ …

The remainder Rn is the error made when sn, the sum of the first n terms, is used as an approximation to the total sum.


We use the same notation and ideas as in the Integral Test, assuming that fis decreasing on [n, ∞).


ESTIMATING THE SUM OF A SERIESComparing the areas of the rectangles with the area under y = f(x) for x > n in the figure, we see that:

1 2 ... ( )n n n nR a a f x dx

∞

+ += + + ≤ ∫

ESTIMATING THE SUM OF A SERIESSimilarly, from this figure, we see that:

1 2 1... ( )n n n n

R a a f x dx∞

+ + += + + ≥ ∫

Thus, we have proved the following error estimate.


Suppose f(k) = ak, where f is a continuous, positive, decreasing function for x ≥ n and ∑ an is convergent.

If Rn = s – sn, then

REMAINDER ESTIMATE (INT. TEST)

1( ) ( )nn n

f x dx R f x dx∞ ∞

+≤ ≤∫ ∫

Estimate 2

a. Approximate the sum of the series ∑ 1/n3

by using the sum of the first 10 terms. Estimate the error involved.

b. How many terms are required to ensure the sum is accurate to within 0.0005?

REMAINDER ESTIMATE Example 5

In both parts, we need to know

With f(x) = 1/x3, which satisfies the conditions of the Integral Test, we have:


( )n

f x dx∞

∫

3 2

2 2 2

1 1lim21 1 1lim

2 2 2

t

n tn

t

dxx x

t n n

∞

→∞

→∞

= −

= − + =

∫

As per the remainder estimate 2, we have:

So, the size of the error is at most 0.005

REMAINDER ESTIMATE Example 5 a

103 3 3 3 31

1 1 1 1 1 1.19751 2 3 10n

sn

∞

=

≈ = + + + ⋅⋅⋅+ ≈∑

10 3 210

1 1 12(10) 200

R dxx

∞≤ = =∫

Accuracy to within 0.0005 means that we have to find a value of n such that Rn ≤ 0.0005

Since

we want

REMAINDER ESTIMATE Example 5 b

3 2

1 12n n

R dxx n

∞≤ =∫

2

1 0.00052n

<

Solving this inequality, we get:

We need 32 terms to ensure accuracy to within 0.0005

REMAINDER ESTIMATE

2 1 1000 or 1000 31.60.001

n n> = > ≈

Example 5 b

If we add sn to each side of the inequalitiesin Estimate 2, we get

because sn + Rn = s

REMAINDER ESTIMATE

1( ) ( )n nn n

s f x dx s s f x dx∞ ∞

++ ≤ ≤ +∫ ∫

Estimate 3

The inequalities in Estimate 3 give a lower bound and an upper bound for s.

They provide a more accurate approximation to the sum of the series than the partial sum sn does.

REMAINDER ESTIMATE

Use Estimate 3 with n = 10 to estimate the sum of the series


31

1n n

∞

=∑

The inequalities in Estimate 3 become:


10 103 311 10

1 1s dx s s dxx x

∞ ∞+ ≤ ≤ +∫ ∫

From Example 5, we know that:

Thus,

REMAINDER ESTIMATE

3 2

1 12n

dxx n

∞=∫

10 102 2

1 12(11) 2(10)

s s s+ ≤ ≤ +

Example 6

Using s10 ≈ 1.197532, we get:

1.201664 ≤ s ≤ 1.202532

If we approximate s by the midpoint of this interval, then the error is at most half the length of the interval.

Thus,


31

1 1.2021 with error 0.0005n n

∞

=

≈ <∑

If we compare Example 6 with Example 5, we see that the improved estimate 3 can be much better than the estimate s ≈ sn.

To make the error smaller than 0.0005, we had to use 32 terms in Example 5, but only 10 terms in Example 6.

REMAINDER ESTIMATE

PROOF OF THE INTEGRAL TEST

We have already seen the basic idea behind the proof of the Integral Test for the series ∑ 1/n2 and ∑ 1/ . n

For the general series ∑ an, consider these figures.


The area of the first shaded rectangle in this figure is the value of f at the right endpoint of [1, 2], that is, f(2) = a2.


So, comparing the areas of the rectangles with the area under y = f(x) from 1 to n, we see that:

Notice that this inequality depends on the fact that fis decreasing.


2 3 1( )

n

na a a f x dx+ + ⋅⋅⋅+ ≤ ∫

Estimate 4

Likewise, the figure shows that:PROOF OF THE INTEGRAL TEST

1 2 11( )

n

nf x dx a a a −≤ + + ⋅⋅⋅ +∫

Estimate 5

If is convergent, then Estimate 4 gives

since f(x) ≥ 0.


1( )f x dx

∞

∫

1 12

( ) ( )n n

ii

a f x dx f x dx∞

=

≤ ≤∑ ∫ ∫

Case i

Therefore,

Since sn ≤ M for all n, the sequence {sn} is bounded above.


1 1 12

( )

,

n

n ii

s a a a f x dx

M say

∞

=

= + ≤ +

=

∑ ∫

Case i

Also,

since an+1 = f(n + 1) ≥ 0.

Thus, {sn} is an increasing bounded sequence.


1 1n n n ns s a s+ += + ≥

Case i

Thus, it is convergent by the Monotonic Sequence Theorem (Section 11.1).

This means that ∑ an is convergent.

PROOF OF THE INTEGRAL TEST Case i

If is divergent,

then

as n → ∞ because f(x) ≥ 0.


1( )f x dx

∞

∫

1( )

nf x dx →∞∫

Case ii

However, Estimate 5 gives:

Hence, sn–1→∞.

This implies that sn → ∞, and so ∑ an diverges.


1

111

( )nn

i ni

f x dx a s−

−=

≤ =∑∫

Case ii

infinite sequences and seriesrfrith.uaa.alaska.edu/m201/chapter11/chap11_sec3.pdf · 2013. 4....

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