AP Sf Calculus Infinte Series

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PROFESSIONAL DEVELOPMENT

Special Focus

AP CalculusInfnite Series

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Contents

1. Infnite Series in Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Jim Hartman

2. Setting the Stage with Geometric Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Dan Kennedy

3. Convergence o Taylor and Maclaurin Series. . . . . . . . . . . . . . . . . . . . . . . . . 17

Ellen Kamischke

4. Overview o Tests or Convergent o Infnite Series . . . . . . . . . . . . . . . . . . 49

Mark Howell

5. Instructional Unit: Manipulation o Power Series. . . . . . . . . . . . . . . . . . . . .63

Jim Hartman

6. Applications o Series to Probability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Ben Klein

7. Approximating the Sum o Convergent Series. . . . . . . . . . . . . . . . . . . . . . . .93

Larry Riddle

8. Positively Mister Gallagher. Absolutely Mister Shean.. . . . . . . . . . . . . . 103

Steve Greenfeld

9. About the Editor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

10. About the Authors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119



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Innite Series In Calculus

Jim Hartman

The College o Wooster

Wooster, Ohio

In the study o calculus, the topic o innite series generally occurs near the end o

the second semester in a typical two-semester sequence in single variable calculus.

This seems to be one o the most dicult topics or students to understand and or

teachers to explain clearly. It should not be surprising that these ideas are as dicult

to grasp as the use o proos or limits, since these conceptslimits and innite

seriesare related, and both took hundreds o years to ormulate. Thus, even though

now we have a better perspective rom which to start, it is still much to ask o our

students to gain a ull understanding o innite series in the two to our weeks given

to their study in a beginning calculus course.

Some History o Infnite Series

Concepts surrounding innite series were present in ancient Greek mathematics as

Zeno, Archimedes, and other mathematicians worked with nite sums. Zeno posed

his paradox in about 450 BCE, and Archimedes ound the area o a parabolic segment

in approximately 250 BCE by determining the sum o the innite geometric series

with constant ratio 14 (Stillwell 1989, 170). One cannot credit Archimedes (or the

Greeks) with discovering innite series, since Archimedes worked only with nite

sums and determined that certain nite sums underapproximated the area and others

overapproximated the area, leaving the common limit (not known to him as a limit)

as the area o the parabolic segment. However, this does point out one o the twomotivations or the development o innite series: (1) to approximate unknown areas,

and (2) to approximate the value o.A nongeometric series appeared inLiber calculationum by Richard Suiseth,

known as The Calculator (Stillwell 1989, 118), in approximately 1350. Suiseth



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SPEcIAL FOcuS: Calculus

2

indicated that1

2

2

2

3

2 22

2 3+ + + + + =L L

kk . At about the same time, Nicole Oresme,

a bishop in Normandy (Cajori 1919, 127), also ound this innite sum along with

similar ones, and proved the divergence o the harmonic series using

11

2

1

3

1

4

1

5

1

6

1

7

1

81

1

2

1

4

1

4

1

8

1

8

1+ + + + + + + + > + + + + + +L

88

1

8+ +L

However, in the same century, Madhava o Sangamagramma (c. 13401425)

and ellow scholars o the Kerala school in southern India were making even

more important discoveries about innite series. Madhava, an astronomer and

mathematician, used the now easily obtained series = +

41

1

3

1

5

1

7+L to estimate

, and, in addition, transormed this series into the more rapidly converging series

12 1 1

3 3

1

5 3

1

7 32 3

+

+ L or (Joseph, 290). More generally, he discovered what

we now call the Taylor series or arctangent, sine, and cosine (Joseph 2000, 289293;

Katz 2004, 152156). One theory is that these ideas may have been carried to Europe

by Jesuit missionaries to India (Katz 2004, 156).

Moving to Europe, Portuguese mathematician Alvarus Thomas considered

geometric series in 1509 (Cajori 1919, 172). Pietro Mengoli o Bologna treated particular

innite series inNov quadraturae arithmeticae in 1650, nding1

11 n nn +( )=

along

with proving the divergence o the harmonic series. In 1668, the theory o power series

began with the publication o the series or ln 1 +( )x by Nicolaus Mercator, who did

this by integrating1

1+x(Stillwell 1989, 120).

Newtons general binomial theorem in 1665 aided the nding o series or many

unctions. Newton, Gregory, and Leibniz all used interpolation ideas to lead to their

important results and went rom nite approximations to innite expansions. There

is strong evidence that James Gregory, who was the rst to publish a proo o the

Fundamental Theorem o Calculus in 1668, used Taylor series in 1671, 44 years prior to

Brook Taylors results in 1715.

In 1734, Leonhard Euler gave new lie to innite series by nding that 1

621

2

nn=

= ,

ater attempts by Mengoli and the Bernoulli brothers had ailed. In 1748, Euler used



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3

some ideas rom his work in 1734 to generate what we now call the Riemann zeta

unction znzn

( ) ==

11

. This unction, named ater Bernhard Riemann because he

was the rst to use complex numbers in the domain o this unction, has become

important because o its relationship to the distribution o prime numbers and holdsthe distinction today o pertaining to one o the most amous unsolved problems in

mathematics, the Riemann Hypothesis.

During this time, issues o convergence o series were barely considered, which

oten led to conusing and conficting statements concerning innite series. The

rst important and rigorous treatment o innite series was given by Karl Friedrich

Gauss in his study o hypergeometric series in 1812 (Cajori 1919, 373). In 1816, Bernard

Bolzano exhibited clear notions o convergence. Augustin-Louis Cauchy shared these

ideas with the public in 1821 in his textbook Cours danalyse de lcole Polytechnique.

The ratio and root tests and the idea o absolute convergence were included in this

text. Uniorm convergence was studied in the middle o the nineteenth century, and

divergent series were studied in the late nineteenth century.

Today, innite series are taught in beginning and advanced calculus courses.

They are heavily used in the study o dierential equations. They are still used to

approximate as illustrated by the BBP (Bailey, Borwein, and Ploue) ormulas. One o

these, discovered by Ploue in 1995, gives the base 16-digit extraction algorithm or

using =+

+

+

+

=

48 1

2

8 4

1

8 5

1

8 6

1

160 n n n n

n

n

.

This historical background would not be complete without mentioning

Fourier series, which attempt to give values or a unction using an innite sum

o trigonometric unctions. These are named ater Joseph Fourier, a scientist and

mathematician who used them inLa Thorie Analytique de la Chaleurin 1822 to

study the conduction o heat.

You can see that the development o the concepts surrounding innite series took

a long time and involved many dierent mathematicians with many dierent ideas.

Its not surprising that our students dont ully comprehend their nuances in just one

semester o study.

The Heart o Infnite Series

At the heart o innite series are three concepts:

(1) the denition o convergence o an innite series,



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4

(2) positive term series, and

(3) absolute convergence o series.

While absolute convergence does not appear specically in the AP syllabus,

power series cannot be ully considered without this idea, nor can one take ulladvantage o the ratio and root tests.

Given students diculty with understanding the concept o innite series,

I believe we requently rush to get to the tests or convergence and never really

require students to ully understand the notion o convergence. More time should be

spent with students computing partial sums and attempting to nd the limits o these

sequences o partial sums.

For positive term series, convergence o the sequence o partial sums is simple.

Since or a positive term series the sequence o partial sums is nondecreasing,

convergence o the sequence o partial sums occurs i and only i that sequence isbounded above. We probably should spend more time nding upper bounds or the

sequence o partial sums o a positive term series or showing that there is no such

upper bound. Nearly all o the convergence tests are ounded on this one idea. The

comparison test, limit comparison test, and integral test all lead directly to upper

bounds or the sequence o partial sums or show that there is no such upper bound.

Even the ratio and roots tests essentially are a limit comparison test with a geometric

series, and show convergence i the comparison is with a geometric series whose

common ratio has an absolute value o less than 1. Thus, the ratio and root tests are

just ormalized versions o a limit comparison test with a geometric series. Relating

these tests back to upper bounds or the sequence o partial sums might help our

students see the one common thread or all these tests.

For series that have both positive and negative terms, the idea o absolute

convergence becomes helpul. I a series converges absolutely, then it must converge.

While this idea is the one needed most requently, our students sometimes xate on

the alternating series test, which is a very specialized test guaranteeing convergence

o a particular type o innite series. I believe we sometimes overemphasize the

importance o this test because we want to make clear the distinction between

absolute convergence and convergence. That is, we want to give examples o series

that converge but do not converge absolutely. It is relatively dicult to do that without

giving examples o series satisying the hypotheses o the alternating series test.

This is because it is dicult to show that a series not satisying the hypotheses is

convergent when it is not absolutely convergent. In act, showing convergence o

an arbitrary series can be quite dicult. No matter what rules we might develop to



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5

determine the convergence o a series, a series can be invented or which the rule

ails to give a decisive result. (Bromwich 1965, 46). Thus, it is not surprising that

innite series is a dicult topic or our students.

Once students understand the concept o convergence and divergence o innite

series, there are two basic questions one can ask about a specic series:

(1) Does that series converge?

(2) I it converges, to what does it converge; and i it diverges, why?

We have already talked about the rst question above. The second question

really leads to the study o power series. Power series dene unctions on their

intervals o convergence, and the challenge is to identiy these unctions. The other

role o power series is to allow us to express a given unction as a power series and

then use that expression as a means to approximate the unction with a polynomial

unction, which uses only simple arithmetic (addition and multiplication) toapproximate unctional values.

Infnite Series in AP Calculus

The May 2008 syllabus or AP Calculus BC lists the ollowing items:

Polynomial Approximations and Series

Concept o series

A series is dened as a sequence o partial sums, and convergence is dened interms o the limit o the sequence o partial sums. Technology can be used to explore

convergence and divergence.

Series o constants

Motivatingexamples,includingdecimalexpansion

Geometricserieswithapplications

Theharmonicseries

Alternatingserieswitherrorbound

Termsofseriesasareasofrectanglesandtheirrelationshiptoimproperintegrals, including the integral test and its use in testing the convergence

op-series

Theratiotestforconvergenceanddivergence

Comparingseriestotestforconvergenceordivergence



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6

Taylor series

Taylorpolynomialapproximationwithgraphicaldemonstrationof

convergence (or example, viewing graphs o various Taylor polynomials o

the sine unction approximating the sine curve) MaclaurinseriesandthegeneralTaylorseriescenteredatx=a

Maclaurinseriesforthefunctionsex, sin(x), cos(x), and1

1x FormalmanipulationofTaylorseriesandshortcutstocomputingTaylor

series, including substitution, dierentiation, antidierentiation, and the

ormation o new series rom known series

Functionsdenedbypowerseries

Radiusandintervalofconvergenceofpowerseries

LagrangeerrorboundforTaylorpolynomials

These indicate both what students should know about innite series and what

they should be able to do with innite series. Its possible to see these items being

tested by examining those ree-response questions that have appeared on the AP

Calculus BC Exam over the years. The ollowing table identies questions pertaining

to the particular ideas mentioned in the syllabus using the AP Calculus BC Exams

rom 1969 to the present. Questions can occur under multiple sections below.

Topi Exam Qestions

Geometric Series 1981-3a, 1981-3b, 1981-3c, 2001- 6d, 2002-6c

Harmonic Series 1972-4b, 1975-4b, 1982-5b, 2002-6a, 2005-6c

Alternating Series Test 1970-6b, 1972-4b, 1975-4b, 1982-5b, 2002-6a

Alternating Series Error1970-6c, 1971-4d, 1979-4c, 1982-5c, 1984-4c, 1990-5c, 1994-5c,

2000-3c, 2003-6b, 2006-6d, 2007-6d

Integral Comparisons 1969-7a

Integral Test 1969-7bc,1973-6c, 1992-6b

Ratio Test 1975-4a

nth Term Test (Divergence) 1973- 6a

Comparison Test 1972-4a, 1973 -6b, 1977-5a, 1980 -3a, 1980 -3c, 1992-6a, 1992-6c

Taylor Polynomials 1995 -4c, 1997-2a, 1998 -3a, 1999 -4a, 2000 -3a, 2004 -6a, 2005 -6a

Taylor Polynomial Approximation 1971-4c, 1984-4b, 1995-4a, 1995-4b, 1997-2d, 1998-3d

Lagrange Error Bound 1976-7c, 1999- 4b, 2004- 6c

Taylor or Maclaurin Series1971-4a, 1982-5a, 1986-5a, 1990-5a, 1996-2a, 2003-6a, 2004-6b,

2005-6b

Radius o Convergence 1984-4a, 2000-3b



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7

Interval o Convergence1971-4b, 1972-4b, 1975-4b, 1976-7b, 1979-4b, 1982-5b, 1983-5b,

1987-4b, 1988- 4, 1991-5c, 1994-5b, 2001-6a, 2002-6a, 2005-6c

Known Maclaurin Series 1976-7a, 1979- 4a

Manipulation o Known Series

1982-5d, 1986-5b, 1987-4a, 1987-4c, 1991-5a, 1993-5a, 1993-5b,

1994-5a, 1996-2b, 1996-2c, 1998-3b, 1999-4c, 2001-6b, 2003-6c,2006-6a, 2006-6b, 2007-6a, 2007-6b

Dierentiation o Series 1983-5c, 1986-5c, 1993-5c, 1997-2b, 2002-6b, 2003-6c, 2006-6a

Antidierentiation o Series1990-5b, 1991-5b, 1997-2c, 1998-3c, 2001-6c, 2004- 6d, 2006-6c,

2007-6c

The table above indicates that one o the most common types o questions involves

the use o known series that are then modied through dierentiation, integration,

substitution (e.g., nding a Maclaurin series or s in(x2) using the known series or

sin(x)), and/or algebraic manipulation. These are all questions that manipulate the

known series specied by the course description or manipulate a series given in thestem or another part o the problem.

A second item commonly tested is that o interval o convergence. This most

requently involves using the ratio test to nd the radius o convergence and then

checking individually each o the endpoints o the interval generated by that radius

o convergence to determine convergence or divergence. The endpoint checking has

oten involved the use o the alternating series test and knowledge o the convergence

or divergence o well-known series, such as the harmonic series.

The third most common question involves the use o the alternating series error-

bound theorem. The typical orm or this question is to use the rst several terms o a

Taylor polynomial to approximate the value o a unction at a point and then either ask

the question o how much error could have been made or have the student show that

the error made in the approximation is less than a specied amount.

In terms o the multiple-choice portion o the exam, the ollowing types o

questions were asked or the years when exams were released.

1997

Mltiple-choie Topis

1998

Mltiple-choie Topis

2003

Mltiple-choie Topis

Sum o a Geometric Series Taylor Polynomial Approximation Geometric Series

Sequence Limit Question Convergence TestsSeveral Manipulation o Series

Taylor Polynomial Integral Test Manipulation o Series or ex

Interval o Convergence Dierentiation o Series nth Term Test and pSeries

Dierentiation o Series Geometric and Alternating

Series Test

Dierentiation o Series and

Taylor Series



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CalculatorError in

Approximation QuestionTaylor Polynomials

Interval o Convergence

Exponential Series

In regard to teaching innite series, one thing is certain: They should be taught in the

context o giving a dierent way o expressing a unction, o their use as approximating

(Taylor) polynomials, and o the error made in that approximationwhether it be given

by the Lagrange error bound or by the alternating series error theorem.

Bibliography

Bailey, D. H., P. B. Borwein, and S. Ploue. On the Rapid Computation o Various

Polylogarithmic Constants.Mathematics o Computation 66 (199: 90313.)

Bromwich, T. J.An Introduction to the Theory o Infnite Series. New York: St. Martin's

Press, 1965.

Cajori, F.A History o Mathematics. New York: The Macmillan Company, 1919.

Joseph, G. G. The Crest o the Peacock: Non-European Roots o Mathematics.

Princeton, NJ: Princeton University Press, 2000.

Katz, V. J.A History o Mathematics: Brie Edition. Boston: Pearson Addison Wesley,

2004.

Stillwell, J.Mathematics and Its History. New York: Springer-Verlag, 1989.

University o Manchester. Indians Predated Newton Discovery By 250 Years,

Scholars Say. ScienceDaily, August 15, 2007. http://www.sciencedaily.com/

releases/2007/08/070813091457.htm.



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Setting the Stage with Geometric Series

Dan Kennedy

Baylor School

Chattanooga, Tennessee

One o the dicult things about teaching innite series at the end o an AP Calculus

BC course is trying to make the students see the topic as something other than a

our-week detour down a side track just when the train ought to be coming into the

station. It is particularly dicult to disabuse students o this notion i the teacher

secretly shares their concern. That is why it is easier or both teacher and students i

the emphasis is on unctions rom the beginning, and i calculus becomes part o the

picture shortly thereater.

There are good reasons to talk about series in an introductory calculus course,

and it is helpul to keep them in mind when thinking about how to teach the topic.

Innite series are another important application o limits. Moreover, as limits they

are easier to understand than either the derivative or the integral because they do not

involve those mysterious dierentials.

The construction o Taylor series is not only a nice application o the derivative

but also a nice review o such topics as linear approximation, slope, and concavity.

I a unction can be represented by a power series, it can be dierentiated or

integrated as easily as a polynomial. This provides another approach to evaluating

expressions like e dxx2 that would be dicult or impossible otherwise. Series

gured prominently in the early history o calculus, so they ought to play some role in

an introductory course.

Series gure prominently in higher-level analysis courses, so it is useul to lay thegroundwork or our better students as soon as we can.

Notice that most o these good reasons or teaching series in a calculus course

involve series asunctions, not series asinfnite sums o numbers. In act, series o

numbers are only important when considering the various tests or convergence. One



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10

school o thought is that students need to see all those tests beore seeing their rst

power series, necessitating a detour away rom unctions and calculus that is dicult

to motivate. A more intuitive (and historically aithul) approach is to play with power

series rom the beginning and see what can be done with them. The question o

convergence eventually must be conronted, but in the meantime it is sucient or

students to know that convergence is an issue.

Happily, there is a way to introduce students to power series right away and

simultaneously make them aware o the question o convergence, all while building on

their existing knowledge o a previous topic: geometric series.

Geometric Series Basics

Even i students have not studied geometric series by name, they have encountered

them in various convergent orms.

For example, a 1"-by-1" square can be cut into

two halves. One hal can then be cut into two

quarters, one quarter can be cut into two eighths,

and so on ad innitum. This process o innite

subdivision, the basis o some o Zenos ancient

paradoxes, leads to the inevitable conclusion that

1 1

4

1

8

1

21+ + + +

+ =

n

.

For another example, some well-known rational numbers have amiliar decimal

expansions that are actually convergent geometric series:

0 33

10

3

100

3

10003

1

10

1

3. = + + + +

+ =L L

n

.

So innite sums can converge to nite numbers, but obviously not all o them

do. For example, 1 + 1 + 1 + is innite, and 1 1 1 1 11 + + + ++L L( )n is at least

ambiguous. The latter series shows the necessity o dening thesum o an innite

series as the limit o its sequence o partial sums.

There is a ormula or thenth partial sum o a geometric series in whichr 1:

a ar ar ar ar a ar

rn

n

+ + + + + =

2 3 1

1L .

L L



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11

(There are several ways to prove this, and most precalculus texts do.) The limit

o thenth partial sum as n depends entirely on the ate orn, which goes to zero iand only i r < 1 . I r 1 , the limit diverges.

Thus, whether or not students have ormally studied geometric series,they can be led very quickly to the realization that a series o the orm

a ar ar ar ar n+ + + + + +2 3 1L L converges i and only i r < 1, in which case the sum

isa

r1 .

At this point, students are ready orx.

A Geometric Series or xx

( ) =1

1-

I x < 1, then1

1 xis the sum o the geometric series 12 3 1

+ + + + + +

x x x xn

L L.

The latter expression looks like a polynomial o innite degree, but since there is

no such thing we must give it a new name. We call it apower series (a series o

powers ox). It is an interesting example o a unction with domain (1, 1), since we

can technically plug in values oxoutside the domain, but we get expressions like

1 2 2 22 3+ + + +L that are simply meaningless. I, on the other hand, we plug in avalue oxinside the domain, we get

1

1 x. The interval (1, 1) is called theinterval o

convergence.

The partial sums o a power series are polynomials. I we graph them, we get adramatic visualization o why the interval o convergence matters, as in the graphs

below o the ourth and th partial sums o 1 2 3 1+ + + + + +x x x xnL L compared to

the graph o1

1 x:



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12

The quartic and quintic polynomial partial sums do a very good job o approximating

1

1 xin the interval o convergence (1, 1), but outside that interval they are not even

close. The partial sums o higher degree approximate the curve progressively more

closely on the interval o convergence, but with no better success outside that interval.

Exploring the Implications

Findapowerseriestorepresent1

1+xand give its interval o convergence.

Solution: We use a geometric series with rst term 1 and ratio x.

1 12 3 1 1 + + + + x x x xn nL L( ) .

It converges or


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13

The interval o convergence o this series is (1, 1).

Forxin the interval o convergence, we can write the equation:

1

1 1 122 4 6 1 2 2

+ = + + + +

x x x x xn n

L L( )

The antiderivatives should then dier by a constant:

tan ( ) ( )

+ = + + +

13 5 7

12 1

3 5 71

2 1x C x

x x x x

nn

n

L ++L

Now we can nd the constant by lettingx=0:

tan ( ) ( )

+ = + + +

13 5 7

12 1

0 00

3

0

5

0

71

0

2 1C

nn

n

L ++

+ =

=

L

0 0

0

C

C

So tan ( ) ( )

= + + +

+13 5 7

12 1

3 5 71

2 1x x

x x x x

nn

n

L L.

We expect this equation at least to be valid on the interval (1, 1), but in act we

have also picked up convergence at the endpoints. For example, pluggingx=1

into the series yields

11

3

1

5

1

71

1

2 1

1 + + +

+L L( )nn

.

This is an example o an alternatingseries o terms odiminishing magnitude that

tend to a limit o 0. Such series always converge. (This is the alternating series

test.) Students can easily be convinced o this act by tracking the partial sums o

this series on a number line:

L



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14

The partial sums bounce back and orth on the number line, but because each added

or subtracted term is smaller than the previous one, they get closer to a limit in the

process. Since the terms tend to zero, there is a limit L that becomes the sum.

It seems logical (and it can be proved using a deeper understanding o limits) that

the number to which 11

3

1

5

1

71

1

2 1

1 + + +

+( )nn

converges ought to be

tan ( )1 1 , which we know to be

4. This series can, in act, be used to compute , but

it is so close to the threshold o divergence that it converges too slowly to be o much

practical value. (Students can convince themselves o this act by trying the partial

sums on their calculators. You know that some o them will.)

Beginning with geometric series, you will have thus exposed students to the idea

that unctions can be represented by power series, that such series have intervalso convergence or which such representations are valid, and that series can be

manipulated using calculus to yield new unctions. You will have shown them power

series that they will eventually recognize as the Maclaurin series or three dierent

unctions (along with their intervals o convergence), and you will even have exposed

them to a valuable convergence test or series o constants. In all likelihood you will

have been able to accomplish this during a single class period.

Reaping the Benefts

Once students have seen that a power series can represent a unction on some

interval o convergence, they can discover some signicant results on their own. Here

are just two examples.

Challenge: Consider the unctiondened by the innite series

f x xx x x

n

n

( )! ! !

.= + + + + + +12 3

2 3

(a) Find(0).

(b) Find'(x). What is interesting about it?

(c) What can you conclude about the unction?

Students will easily guess that the unction is(x) - ex, but only the best o them will

recognize that they have the inormation required to conclude that(x)must be ex. In

part (b) they discover the dierential equation(x) = (x), and in part (a) they discover

the initial condition(0)=1.Theuniquesolutiontothisinitialvalueproblemis(x)=ex.

L L

L L



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15

Challenge: Construct a th-degree polynomialPsuch that:

P

P

P

P

P

P

( )

( )

( )

( )

( )( )

(

0 2

0 3

0 5

0 7

0 114

5

=

=

= =

=))( )0 13=

Students will usually succeed at this, and in the process they will see that the

coecient oxn must beP

n

n( )( )

!

0. You are then ready to ask them to build a polynomial

whose rstn derivatives match the rstn derivatives o a unctionat 0 (the sine

unction, or example). The coecient o eachxn will then be n

n( )

( )!0 . Your students

will have discovered Maclaurin series!

Challenge: BC-3 rom the 1981 AP Examination.

Let Sbe the series St

t

n

n

=+

=

10

where t 0.

(a) Find the value to which Sconverges when t=1.

(b) Determine the values otor which Sconverges. Justiy your answer.

(c) Find all values ot

that make the sum o the seriesS

greater than 10.Students are ready or this as soon as they understand geometric series. Curiously,

as they learn more about series they become overqualied to solve something this

simple.

Challenge BC-3 Solution

(a) When t=1, Sn

n

=

=

120

. This is a geometric series with rst term a=1and

common ratio r=1

2. Hence S

a

r=

=

=1

1

1 12

2 .

(b) The series will converge i and only it

t11

+< where t 0. This will be true or t

> 0 and or < 10 when t> 9.

Once your students have grasped the general idea o what series are and how they

behave, they will be ready to tackle the rest o the topics or innite serieseventually

even those other tests or convergence!



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Convergence o Taylor and Maclaurin Series

Ellen Kamishke

Interlochen Arts Academy

Interlochen, Michigan

Instructional Unit Overview

Focus: How to determine the radius and interval o convergence o a Taylor or

Maclaurin series

Audience: AP Calculus BC students

In this three-day unit students are introduced to the idea o the interval and radius o

convergence o a Maclaurin or Taylor series. The rst lesson has students determine

this interval visually and by checking suspected endpoints o the interval. Students

start by considering the series or sin(x) and also the series or1

1 xsince this

expression can be viewed as the sum o a geometric series. Next they investigate a

series that is not geometric but appears to have a limited interval o convergence. In

the process o exploring this series, the harmonic series is introduced along with the

integral test. In a homework problem, students will encounter an alternating harmonic

series and need to reason out its convergence. An investigation is also provided that

leads students through much o this material independently or in a small-group setting

rather than in a large class ormat.

The second lesson introduces the ratio test and ormalizes the alternating series

test. Time is spent practicing these tests on portions o previous AP ree-response

problems. An investigation is provided that asks students to explore the ratio o the

terms o a series in the limit and generalize the ratio test rom their results.

The third lesson addresses some interesting questions about series, such as

Does a Taylor series always converge to the unction rom which it was constructed?

Can a series converge to more than one unction?, and Is every series that converges



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18

to a unction the Taylor series or that unction? Refecting on questions like these

and extending beyond the normal range o material helps students gain perspective

on the larger picture.

Assumed skills and knowledge: Students should be able to use a calculator

or computer to graph a unction and evaluate it at a point. They should understand the

meaning o convergence and divergence or an innite series. Students should know

the ormula or the sum o a geometric series. Students should be able to construct a

Taylor or Maclaurin series or a given unction. They should know the Lagrange orm

o the error and be able to use it to estimate the accuracy o an approximation made

with a Taylor polynomial.

Background inormation on Taylor series: When nding the Taylor series

or a unction, it is important to determine or what values oxthe series approximates

the generating unction. Finding this interval o convergence is sometimes a very

straightorward procedure. Oten the greatest challenges come in determining

convergence at the endpoints o the interval.

The Taylor series or a unction aboutx=a is constructed according to the ormula

x a a x a a

x a a

( ) = ( ) + ( ) ( ) +( )

( ) +( )

22

! 333

! !x a

a

nx a

nn( ) + +

( )( ) +L L.

Students should recognize the rst two terms o this series as the linear

approximation or a unction, or the equation o the line tangent to(x)atx=a. I like

to check the students understanding o the inormation presented in the rst ew

terms o the series by using exercises like the ollowing:

1. I the Taylor series is constructed

or the unction shown at right,

centered at the point (1, 3), what

can be determined about the

coecients o the rst three

terms o the series?

Solution: The rst term is 2, the value o(1). The coecient o the second term is

positive because the rst derivative or slope o the tangent is positive. The coecient o

the third term is also positive because the unction is concave up near the point (1, 2).



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19

2. The third-degree Taylor polynomial Taboutx=1forafunction,(x), is

T x x x ( ) = + ( ) ( )3 2 1 4 12 3 . What inormation do you know about(x)?

Solution: From the rst term you know that(1)=3.Thereisnolineartermsothe

rst derivative at the point (1, 3) is 0, and this is a critical point. The coecient o thesecond-degree term is positive, so the unction is concave up near the point (1, 3),

making this critical point a minimum. The second derivative has a value o 4 and the

third derivative has a value o -24.

Other problems that test students understanding o the construction o Taylor

series give an expression or thenth derivative o the unction. In student work one

common error is to neglect to include the actorials in the denominators. A nice

example o this sort o problem comes rom the 2005 AP Calculus BC Exam, question 6.

2005 BC#6

Letbe a unction with derivatives o all orders and or which(2)=7.Whenn is odd,

thenth derivative oatx=2is0.Whenn is even andn 2, thenth derivative oat

x=2isgivenbynn

n

( ) ( ) =( )

21

3

!.

a. Write the sixth-degree Taylor polynomial oraboutx=2.

Solution: P x x x6 22

4

4

67

1

3

1

22

3

3

1

42

5

3( ) = + ( ) + ( ) +

!

!

!

!

!11

62 6

!x( )

Students must pay close attention to the act that the odd derivatives are 0. They

must also realize that the sixth-degree Taylor polynomial includes terms up to and

including (x 2)6, and not necessarily six terms.

b. For the Taylor series oraboutx=2,whatisthecoefcientof(x 2)2n or

n 1?

Solution:2 1

3

1

2

1

3 22 2n

n nn n( )

( )=

( )!

!

Students can either substitute 2n orn in the general term o the series, or

observe the pattern in the terms o the series with powers o 2 0, 2 1, 2 2, and 2 3,

then create a general term based on that pattern. Once again they must be careul toremember to include the actorial in the denominator and not just state the value o

the derivative.

c. Find the interval o convergence o the Taylor series oraboutx=2.

Show the work that leads to your answer.



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20

The rst two parts o this question deal with constructing the Taylor polynomial and

series, and can be done beore studying the interval o convergence. The problem o

determining the interval o convergence is the ocus o the next three lessons.

Day 1: Introducing the Interval o Convergence

To introduce the idea o the interval o convergence I have students look at the Taylor

series or(x)=sin(x) and graph partial sums on their calculators. We then discuss

how it appears that as one adds more terms o the series, the partial sums t the

unction better and better. I ask them i they continued this orever, would the series

t the entire unction? Or is there somexvalue beyond which the series just wont t,

no matter how many terms are added on? How do they know? This usually prompts

a good discussion and provides a motivation or looking or thesexvalues where the

series ts or converges to the unction.

Next I like to introduce a second example that has a limited interval o convergence,

or instance, the unction g xx

( ) =1

1, and look at several o its partial sums. In

this case it becomes readily apparent that this series is a good t only in a small

interval. Students easily see that it is not good or any approximations whenxis

greater than 1, and they may also guess the lower bound o the interval to be 1. I

ask them why this happens, and since we have discussed innite geometric seriespreviously, we can usually come to the realization that this unction can be thought

o as the sum o a geometric series with rst term 1 and ratiox. They then make the

connection that it converges only when x < 1 .



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21

The next issue we consider is what happens i the series doesnt appear to

converge or all values oxbut is not geometric. How can we determine the interval oconvergence?

We are now ready to return to part (c) o 2005 BC, #6. I we take our answer

rom part (b) and write the general term o the series we get1

3 22

2

2

n

n

nx

( )( ) .

Or by rearranging a bit we can write it as1

2

2

3

2

n

x n

. In this orm it looks like

some relative o a geometric series. We look at the graphs o some o the partial

sums and try to make an estimate as to where the series will converge. Based on

the graphs students usually agree that the series converges orx=3andx=4,but

might wonder aboutx=5andanythinggreater.Sowetryoutoneormoreofthesexvalues to see what the series looks like i we substitute them in place ox.



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22

For instance, ix=3thegeneraltermbecomes1

2

1

3

2

n

n

. The second actor here

is thenth term o a convergent geometric series. Students can then reason that the

rst actor is less than 1 and thus makes the terms o the actual series smaller than

those o the convergent geometric series. So the series converges whenx=3.Ihave

students investigate other values ox. They quickly realize that orx> 5 the geometric

part o this series diverges and the rst actor probably doesnt compensate or this

divergence. A good discussion usually surrounds the case ox=5.Inthiscasethe

rst actor essentially becomes one-hal thenth term o the harmonic series. I like to

encourage this discussion as students wrestle with this important series. Eventually

we are led to looking at the series graphically and thinking about the integral

associated with the unction xx

( ) =1

.

The areas o the rectangles represent the terms o the harmonic series1

1nn=

. Thesum o their areas is clearly larger than the area under the curve. We have studied

this unction and the improper integral1

1x

dx

and know that it is divergent. So it islogical that the harmonic series is also divergent. We oten have to go back and redo

this integral, but since this is such an important result, I think the time spent is worth

it. Students are now ready to state the upper bound o the interval o convergence as 5

(with divergence atx=5).Bysimilarreasoningtheyrealizethattheserieswilldivergeix< 1. The next question is whether it converges at x=1.Becauseofthe2n in the

exponent, the series is identical to that whenx=5soitdivergeshereaswell.



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23

At this point I like students to try these ideas on their own, and I ask them to

investigate the Taylor series or cos(x) along with another series such asx

n

n

nn

=

1

1 3 or

their homework. They should be able to use the graphs o the partial sums and their

reasoning about geometric series, along with the harmonic series to nd the intervals

o convergence or these series. Inx

n

n

nn

=

1

1 3, one endpoint o the interval creates an

alternating series. I ask students to think about this and try to reason out on their own

whether it will converge. They usually decide that i the partial sums are oscillating

and those oscillations are getting smaller, the series will converge. This inormal

understanding o the ideas behind the alternating series test helps them when we

study it ormally later. They are also motivated to learn the ratio test the next day in

order to make this search or the interval o convergence a bit easier.

Day 2: A Procedure or Determining the Intervalo Convergence

During the next class we look at the ratio test and practice its use. I do not ormally

prove the test, but rather present it with the sort o inormal reasoning outlined below.

The Ratio Test

Suppose the limit limn

n

n

a

a+ =1 L either exists or is innite.

Then

a. IL < 1, the series ann=

1

converges.

b. IL > 1, the series ann=

1

diverges.

c. IL=1,thetestisinconclusive.

IL < 1, this means that successive terms are getting larger and it is logical that

the series diverges.

IL < 1, this means that successive terms are getting smaller and it is possible

that the series converges. Look a bit more closely at the terms o a power series

c xnn

n=

0

, where an=c

nxn. The power onxincreases regularly. This is like a geometric



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24

series. O course, the coecients dont always change by the same ratio, but we can

still think o this as a sort o cousin to a geometric series. The ratio ormed in the limit

is less than some numberror a geometric series. Since the limit is less than 1, you

can pick a number orrthat is also less than 1 and above the ratioa

an

n

+1

. So our seriesis less than this convergent geometric series and it makes sense that iL < 1, the

series will converge.

WhenL=1theresultisinconclusive.Intheharmonicseries,L=1andtheseries

diverges. However, the series1

21nn=

hasL=1 again, but this series converges. So iL=1, we cant say whether the series converges.

We go back and apply the ratio test to part (c) rom 2005 AP Calculus BC, #6, and

get the ollowing:

lim

n

n

n

n

nx

nx

+( )+( )

+( )( )

12 1

13

2

1

2

1

3

2 1

2 1

222

2

2 1

3

3 32

2

2

2 2

2

( )=

+( )( ) =

n n

n

n n

n

nxlim lim

+( )

=( )n

n

x x

1

2

9

2

9

2 2

.

The series converges when this limit is less than 1:

xx x

( )< ( ) < 1, the series an

n=1

diverges.c. IL =1,thetestisinconclusive.

Investigation 3: A Series That Converges to TwoDierent Functions (in Form)

1. Write the rst our nonzero terms and general term o the Maclaurin series or

(x)=cos(x).

2. You know that you can create a series or cos(g(x))by replacing eachxin the

series with g(x). You have done this with such unctions as g(x)=x2. Now use

this idea to write a series or

f x( )= cos x( ) .

3. Find the interval o convergence or your series.



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38

4. Veriy by graphing that the series seems to converge to the original unction or

x 0. Sketch a graph o f x( )= cos x( )and two dierent series approximations toshow this convergence.

5. What happens whenx< 0? The interval o convergence indicates that the series

will still converge, but f x( )= cos x( ) is not dened orx< 0. What will the seriesconverge to then? We can nd an answer to this question by examining the

unction (called the hyperbolic cosine unction) cosh xe ex x

( ) =+

2. Write the

series or ex here.

6. Write the series or e-x here.

7. Use the two series rom steps 5 and 6 to get a series or cosh(x).

8. Use the series rom step 7 to get a series or cos x( ) .

9. Forx< 0, use the series rom step 8 to get a series or cosh ( )x .

10. Leth (x) be the unction represented by the series in step 2. Expressh (x) in

terms o the cos x( ) and cosh ( )x . Veriy by graphing cosh ( )x and twodierent series approximations to illustrate convergence orx< 0.



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39

Teacher Notes on the Investigations

Investigation 1: Convergence o a Taylor Series

1. Write the Taylor series or(x)=sin(x).

x x

x x x x

nn

n

( ) = + + + ( )+( )

+3 5 7 2 1

3 5 71

2 1! ! !...

!++ ...

2. Graph the ollowing Taylor polynomials or(x)=sin(x) along with(x)=sin(x):

T5(x), T

11(x), T

17(x). Record a sketch o your graphs here.

3. Based on your graph above, or what values oxdo you think you could use the

series to approximate the values o(x)=sin(x)? Explain.

It appears that as you add more terms to the polynomial, it ts a larger portion o

the curve. Since the series is innitely long, it is reasonable that the series would

t the entire unction. So you could use it to approximate values o(x)=sin(x) or

all values ox. However, i thexvalue is ar rom zero, it would take many terms to

have a good approximation.

4. Write the Maclaurin series or f x( )= 11 x

. (Hint: Think o this unction as the sum

o an innite geometric series.)

x x x x xn( ) = + + + + + +1 2 3 ... ...



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40

5. Graph the ollowing Taylor polynomials or f x( )= 11 x

along with f x( )= 11 x

:

T4(x), T

7(x), T

12(x). Record a sketch o your graphs here.

6. Based on your graph above, or what values oxdo you think you could use the

series to approximate the values o f x( )= 11 x

? Justiy your answer using ideas

o convergent geometric series.

The series appears to match the unction well in the interval 1


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41

Based on the graphs o partial sums students may estimate the interval o

convergence to be 4


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42

Ix=3,thisseriesbecomestheharmonicseriesandwilldiverge.Forany

positive value less than 3, however, the series will converge.

10. I you substitute a negative value orx, the terms o the series will

alternate in sign. You will learn an ocial test or alternating seriesconvergence later, but or now think about what happens when you

add up a string o numbers that alternate in sign. I the terms you are

summing are getting closer and closer to zero, then the partial sums

behave something like the picture below, where the upward arrows

represent positive terms and the downward arrows represent negative

terms. I more terms were added you can imagine that the series would

converge to some point between the high and low, something like a

spring bouncing and nally coming to rest at an equilibrium position.

For the series

xn1

n3nn=1

, what negative values oxwould result in the

convergence o the series? In other words, what is the lower bound o theinterval o convergence?

When trying to nd the lower bound by substituting values ox, students

will nd that i they use a number less than 3, the geometric actor in

the series is divergent. Usingx=3,theycreateanalternatingharmonic

series. Students should be able to reason that this will converge because the

terms being added alternate in sign and get smaller in absolute value. When

discussing this question, you may wish to ormally state the alternating

series test and show how it applies here.

Investigation 2: The Ratio Test

When a series is geometric, it is easy to tell when it converges. However, i the series

is not geometric and not closely related to a geometric series, it can be more dicult

to nd this interval o convergence. In this investigation you will look or relationships



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43

based on the ideas o geometric series that will be useul in determining the

convergence o other series.

1. The terms o a geometric series have a constant ratio. That is,an+1

an

is always the

same, no matter which two consecutive terms you use. The series converges i

this ratio has an absolute value less than 1. In nongeometric series this ratio is

not constant; it can change with each pair o terms. However, since the series is

innite it is reasonable to look at the limit o this ratio to discover the behavior o

the terms in the long run.

Below are two lists o series. For each one determine limn

n

n

a

a+1 .

Converging Series Diverging Series

2

n2 + 1n=0

limn an+1

an

= 1 1n

32

n

n=1

limn an+1

an

= 32

6

7

n

n=0

limn

an+1

an

=6

7

n

n + 1n=0

limn

an+1

an

= 1

2n

n!n=1

limn

an+1

an

= 0 n!

n2n=1

limn

an+1

an

=

n!( )22n( )!n=0

limn

an+1

an

=1

4

2n

n2n=1

limn

an+1

an

= 2

2. I the limit o this ratio is less than 1, what seems to be true about the series? Why

is this reasonable?The series converges. When the limit o the ratio is less than 1, it means the

terms o the series are approaching 0 and are doing so in a way such that each

pair o successive terms has a smaller ratio than the pair beore. This is just like a

geometric series with a ratio less than 1, and thus the series converges.

3. I the limit o the ratio is more than 1, what seems to be true about the series? Why

is this reasonable?

The series diverges. When the limit o the ratio is greater than 1, it means the

terms o the series are growing in size. With each additional term the total

changes by more than it did when the previous term was added. Thus the sum

cannot be approaching a single value and the series diverges.

4. I the limit o the ratio is 1, what seems to be true about the series? Why is this

reasonable?



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44

The series might converge or it might diverge. Both outcomes are possible i the

limit o the ratio is one. When the limit o the ratio is one this means that the terms

are becoming more and more alike. I they are also becoming smaller at a ast

enough rate, the series will converge. However, i they are either not approaching a

limit o 0 or not doing so quickly enough, as in the harmonic series, the series will

diverge.

5. Using what you have discovered, or what interval o values do you think the series

x 2( )n10nn=0

will converge?

This is a geometric series with ratiox 210

so the series will converge when

x 210

< 1 or x 2 < 10 . This describes an interval o radius 10 centered at 2.

So the interval o convergence is 8 1, the series ann=1

diverges.c. IL=1,thetestisinconclusive.

Investigation 3: A Series That Converges to Two Dierent Functions

(in Form)

1. Write the rst our nonzero terms and general term o the Maclaurin series or(x)

=cos(x).

12 4 6

12

2 4 6 2

+ + + ( )( )

+x x x x

nn

n

! ! !...

!...



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45

2. You know that you can create a series or cos(g(x)) by replacing eachxin the

series with g(x). You have done this with such unctions as g(x) =x2. Now use

this idea to write the series or f x( )= cos x( ).

12 4 6

12

2 3

+ + + ( ) ( ) +x x x xnn

n

! ! !...

!...

3. Find the interval o convergence or your series.

Using the ratio test, the series converges or all values ox.

4. Veriy by graphing that the series seems to converge to the original unction orx

> 0. Sketch a graph o f x( )= cos x( )and two dierent series approximations toshow this convergence.

5. What happens whenx=0?Theintervalofconvergenceindicatesthattheseries

will still converge, but f x( )= cos x( ) is not dened or x < 0. What will the seriesconverge to then? We can nd an answer to this question by examining the

unction (called the hyperbolic cosine unction) cosh x( )= ex + ex

2. Write the series

or exhere.

ex =1+ x +x

2

2!

+x

3

3!

+x

4

4!

+

6. Write the series or exhere.

e

x =1 x +x

2

2!

x3

3!+

x4

4!

x5

5!+

L

L



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46

7. Use the two series rom steps 5 and 6 to get a series or cosh(x).

Adding the two series rom steps 5 and 6 we get

ex + ex =1+ x +x 2

2!+

x 3

3!+

x 4

4!+L+1 x +

x2

2!

x 3

3!+

x4

4!+L= 2 + 2

x2

2!+ 2

x 4

4!+L

.

Thus

cosh x( )= 12

ex + ex( )= 1+ x2

2!+

x4

4!+L =

x2n

2n( )!n= 0

.

8. Use the series rom step 7 to get a series or cosh x( ).

cosh x( )= 1+x( )22!

+x( )44!

+L=x( )2n

2n( )!n= 0

= xn

2n( )!n= 0

9. For x < 0 , use the series rom step 8 to get a series or cosh ( )x .

cosh x( )=x( )2n2n( )!

n= 0

= x( )n

2n( )!n= 0

= 1( )nx

n

2n( )!n= 0

10. Let h x( ) be the unction represented by the series in step 2. Expressh(x) in

terms o the cos x( ) andcosh ( )x . Veriy by graphing cosh ( )x and

two dierent series approximations to illustrate convergence orx< 0.

h x( )=cos x( ) ifx 0

cosh x

( )ifx < 0

=cos x( ) ifx 0

cosh x

( )ifx < 0

A graph is given by:

L

L

L

L L



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47

Students are likely to be unamiliar with the hyperbolic unctions. The connections

between the hyperbolic unctions and the trigonometric unctions are interesting

and somewhat unexpected. One can also use complex numbers to explore the

relationships between these two types o unctions.

For example, sincei2n equals 1 whenn is even and 1 whenn is odd, it is easy to

use the Maclaurin series or cos(x) to check that cos(ix)=cosh(x). Alternatively, this

identity ollows rom the identity ei=cos() +isin (), which plays a very important

role in complex analysis.

Many students nd complex numbers ascinating, and when they realize that

there is a meaning to the sine or cosine o an imaginary number they are intrigued.

Students interested in this topic could investigate what happens with the series or

sin(x)whenxis replaced by x . This will lead to the identity sin(ix)=isinh(x).



52/12808-1455.AP.SF.Calculus 0910.indd 48 9/10/08 10:22:0


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Overview o Tests or Convergence oInnite Series

Mark Howell

Gonzaga College High School

Washington, D.C.

To the Teacher

By the time students have seen all o the tests or convergence o innite series they

are usually overwhelmed by the sheer number o tests. Moreover, the number o ways

o asking a question that requires the use o a test or convergence can be daunting

as well. Beore applying a test or convergence or divergence o a series, students

need rst to recognize that they need to know whether a given series converges

(or, in the case o series where thenth term is a unction ox, the radius or interval

o convergence). Sometimes, this recognition is trivial: A question may simply ask

whether a series converges. Other times, a student may need to think a bit to see that

they need to know whether a series converges. Once thats done, he or she then needs

to select an appropriate test and apply it. This is not a simple task by any means.

This unit attempts to summarize the process or students. First, well look at

the contexts where the convergence tests are applied. Then well examine each

test, giving a rationale or why it works, and consider some questions that address

the conceptual sides o the tests. There are ample examples or straightorward

testing o convergence o series in textbooks; we wont provide many more o these.

Consequently, this unit should be seen as a supplement rather than a replacement or

textbook coverage o convergence tests.

By examining the AP questions identied in the Introduction to Innite Series

at the beginning o these materials, you can see some o the ways students have been

asked questions that require a test or convergence on past AP Exams.



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50

Note that there are two broad classes o series: those with constant terms, and

those whose terms depend on a variable. Tests or convergence can be applied to

both.

I. When do I need to test or convergence o an infnite series?

Whichofthefollowingseriesconverge?

Whichofthefollowingseriesdiverge?

Forwhatvaluesofkdoes a series whosenth term is a unction ok

converge?

Isitpossibletoevaluate(a) with arbitrary accuracy using its Taylor

Series expansion atx=b?

Whatistheradiusorintervalofconvergenceforaparticularseries?

All o these are contexts that require you to choose and apply tests or

convergence. In some cases, the test or convergence isnt really much o a test atall. You may be expected to simplyknowthat a certain series converges or diverges.

Typical examples that rarely i ever require justication on the ree-response section

o the AP Exam include geometric series, harmonic and alternating harmonic series,

andp-series. Students could simply assert: This is the harmonic series. It diverges,

or This is the alternating harmonic series. It converges. The geometric series and

p-series tests are almost as easy to apply. To show that1

n

n

e

=

converges, a studentcan simply state, This is a geometric series with a common ratio whose absolute

value is less than 1. It converges. To show that3 21

2

n n

=diverges, its sucient to

say This is ap-series withp < 1. It diverges. It would be clearer i the student said,

1

n

n

e

=

converges to e e because it is a geometric series with common ratio

0 .

comparison Test

Suppose 0 an bn or all n N.

I1

n

n

b

= converges, then 1 nn a

= converges.I

1

n

n

a

= diverges, then

1

n

n

b

= diverges.

Ratio Test

I1lim 1n

nn

a

a

+

< , then

1

n

n

a

= converges absolutely.

I1lim 1n

nn

a

a

+

> , then

1

n

n

a

= diverges.

I1

lim 1n

nn

a

a

+

= , then no conclusion can be made about 1n

na

= .



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52

Alternating Series

Test

I 0na > , decreasing, and lim 0nn

a

= , then ( ) 11

1n

n

n

a

=

and

1

( )

n

an

n=1

converge.

Absolte conver-

gene TestI

1

n

n

a

= converges, then

1

n

n

a

= converges.

Additional tests or convergence include the limit comparison test and the root

test, but these are not tested on the AP Exam. O course, students can use these tests

on the exam, but no questions will require their application.

What Each Test Really Says

nth Term Test or Divergence Test

Indirect reasoning may help you understand why thenth term test works. Suppose

the limit o thenth term were a numberL 0. I that were the case, then eventually

the series would behave like a series where each term was a nonzero constant,L.

Essentially, the tail end o the series acts likeL + L + L + , and so the sum could be

made arbitrarily large.Note that having annth term that approaches 0 is a necessarybut not a sufcient

condition or convergence. A requent mistake is assuming that the converse o

the statement o thenth term test is also true. For example, a student might look at

the harmonic series1 1 1

1 ... ...2 3 n

+ + + + + and reason incorrectly that since1

0n

as

n , the series1 1 1

1 ... ...2 3 n

+ + + + + must converge. You should simply know thatthe harmonic series diverges, although this act is easily veried (see the discussion

in the section on the integral test). Having annth term that approaches 0 tells us

nothing about the convergence or divergence o a series. Having annth term thatdoes not approach 0 tells us that a series denitely diverges. Thats thenth term test.



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Overview o Tests or Convergence o Innite Series

53

Geometric Series

I a geometric series ari1

i=1

has rst term a and common ratiorthen the sum o

the rstn terms is given by ( )1

1

na r

r

. Looking at the behavior o this sum asn

inorms us about the convergence o that geometric series. The only way( )1

lim1

n

n

a r

r

can exist is i |r| < 1. When|r| < 1, the termrngoes to 0 asn, and the innite

geometric series converges to1

a

r. Note that this is the only test that tells us not only

that a series converges but whatthe series converges to.

Applying this test requires two steps:

Recognizingthatyouhaveageometricseries.

Findingthecommonratio.

Once youve done that, its a simple matter to compare the absolute value o

the common ratio to 1, and then make a conclusion about the series based on that

comparison.

Integral Test

The integral test is based on let and right Riemann sums. The pictures below give

ample justication or the test.

FIGuRE 1 FIGuRE 2



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54

First, you have to imagine extending the domain o the sequence(k) that generates

the series f k( )k=1

to include the positive real numbers. The graph o the resultingunction is shown in Figures 1 and 2. Notice in Figure 1 above that each let endpoint

rectangle has base 1 and height(k) ork=1,2,3....Thesumoftheareasofthese

rectangles is the same as the series and greater than the area under the graph o

y=(x) orxrom 1 to. So i the integral x dx( )

1

diverges, the series, which is

greater, must diverge as well.

Similarly in Figure 2, a right Riemann sum is illustrated. Again, the area o

each rectangle is a term in the series. The series sum is less than the area under the

curve, again rom 1 to , plus(1), so i the integral x dx( )

1

converges, the series

converges. Adding the rst term,(1), doesnt aect the convergence.

This is the essence o the integral test. I is eventuallycontinuous, decreasing

and positive (orxM), then ( )M

f x dx

and1

( )n

f n

= either both converge or both

diverge. You can begin the summation o the series atn=1insteadofatMbecause

you can remove any nite number o beginning terms o a series without aecting its

convergence or divergence.

The integral test is commonly used to show that the harmonic series

11

2

1

3

1+ + + + +... ...

ndiverges. Notice that the unction ( ) 1f x

x= is indeed

continuous, positive, and decreasing orx> 0. The integral test tells us that since1 1

1 1x

dxx

dx NN

N

N

=

= ( )( ) = lim lim ln , the harmonic series also diverges. Youll

see the integral test applied in the section covering p-series. As with all tests or

convergence, its important to veriy the hypotheses o the integral test beore blindly

applying the test.

Interesting Observation On the Side

You might guess that since the divergence o the harmonic series was determined

by the divergence o the natural logarithm unction as the inputs get innitely large,

there might be a connection between the sum o the rst n terms o the harmonic

series and ln(n). In act, it has been shown that 11

2

1

3

1+ + + +

( )... lnn

n approaches

a constant asn. That is, lim ... lnn n

n

+ + + + ( )

11

2

1

3

1exists. The value o that

08-1455.AP.SF.Calculus 0910.indd 54 9/10/08 10:22:


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55

limit is called Eulers constant, denoted by (the Greek letter gamma), and has the

approximate value o 0.5772. Its an open question whether is rational or not.

p-Series

A series o the orm1

1p

n n

= is, by denition, ap-series. Thep-series test says that

such a series converges as long as the exponent,p, is greater than 1. This test results

directly rom applying the integral test to the series. That is, since1

1p

dxx

convergesi and only ip > 1,

1

1p

n n

= converges i and only ip > 1. Note that orp > 0 andx> 0,

( ) 1pf x x= is indeed a continuous, positive, and decreasing unction oxso thehypotheses o the integral test are satised. Ip 0, the series

1

1p

n n

= diverges by the

nthterm test. For example, ip=2,wehave 22

1 1

1

n n

n

n

= =

=

. Theres no way that one

converges! In act, it ails thenth term test.

Applying thep-series test involves the same sort o knowledge as applying

the test or a geometric series:

Recognizingthatyouhaveap-series.

Findingthevalueofp.

Once youve done that, its a simple matter to compare the value op to 1, and

then make a conclusion about the series based on that comparison.

One remarkable act that becomes apparent when you think about thep-series

test is how special the value op=1is.Itestablishesamagicalboundarybetween

two wholly dierent classes o series. That is, sincep=0.99intheseries

1

n0.99

n=1

, itdiverges. So you can make the partial sums o

1

10.99+

1

20.99+

1

30.99+

1

40.99+ as large as

you want. Increasep just a little, and the behavior is antastically dierent. You cant

make partial sums o1

1

1

2

1

3

1

41 01 1 01 1 01 1 01. . . ....+ + + + as large as you want, or even any

larger than 101, as the ollowing discussion demonstrates. Amazing!


I you look just a little closer, the ideas behind the integral test let you establish an

upper bound or1

1

1

2

1

3

1

41 01 1 01 1 01 1 01. . . ....+ + + + . (The ollowing is not a required topic

or the AP Exam.) Adding the rst 10 terms o the series gives a sum o 2.902261. I



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57

approaches a number whose absolute value is less than 1, then the series converges.

A geometric series has a constant common ratio. Applying the ratio test checks

whether a nonconstant ratio has a limit asn. The conditions o the two tests arealmost the same.

Ageometricseriesconvergeswhen|r| < 1.

Aseriesconvergesabsolutelybytheratiotestwhen 1lim 1nn

n

a

a

+

< .

Ageometricseriesdivergeswhen|r| > 1.

Aseriesdivergesbytheratio test when 1lim 1nn

n

a

a

+

> .

The only dierence between the two is the case wherer=1.Ageometric series

withr=1 diverges. But the ratio test is inconclusive when 1lim 1nn

n

a

a

+

= . Notice that i

thenth term has an expression raised to thenth power (as it will in any power series),

when you calculate 1n

n

aa

+ the powers vanish. Moreover, when you divide a term with

( )1 !n + by a term withn!, the actorials vanish. So, it should come as no surprisethat the ratio test is invariably used to nd the radius o convergence o power series,

including Taylor series.

One other important act about the ratio test is that when 1lim 1nn

n

a

a

+

< , we know

more than just that

1

n

n

a

=

converges. In particular, we know that1

n

n

a

=

converges.

Alternating Series Test

The alternating series test has three parts to its hypothesis: The terms in the series

must alternate in sign, they must decrease in absolute value, and thenthterm must

approach 0 asn approaches innity. I all three conditions are met, the series converges.

The justication or the alternating series test typically involves a look at how the

sequence o partial sums,s1=a

1,s

2=a

1 a

2,s

3=a

1 a

2+ a

3, ..., behaves. (Here, were

assuming each o the terms ak

is positive.) Most texts have a graph like the ollowing:



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58

Adding a3

to the second partial sums2

results in a third partial sum that is less

than the rst,s1. This must be true because we subtracted more (a

2) roms

1than was

added back (a3). In this context, i eventually the amount a partial sum changes by

goes to 0, the partial sums must converge. This analysis shows why the alternating

series test requires that terms decrease to 0 in absolute value. I the terms didnt

decrease, then one term could bounce us past an earlier partial sum. And o course i

the terms dont approach 0, then the sequence o partial sums cant converge by the

nth term test.

Similar to the way the integral test gives rise to an upper bound or the error

in stopping an innite summation at thenth term (sometimes called the truncation

error), so too does the alternating series test allow us to determine an error bound.

This time, though, the error bound is a required topic or AP Calculus BC. I an

alternating series bnn=1

converges with bn

n=1

= S [note that here were usingb

n=(1)n+1a

nwhere the sequence (a

n) is positive and decreasing] and thenth

partial sum, bn

n=1

N

= SN , then the alternating series error bound says simply thatthe error |S S

N| satises |S S

N| < |b

N+1|. That is, the truncation error is no greater

in magnitude than the magnitude o the rst omitted term. IbN+1

> 0, then SN

< S

and ibN+1

< 0, then SN

> S. Heres a simple application o the test and error bound.

The alternating harmonic series,

( )= +

+

=

1

1

1

2

1

3

1

1

n

n n ..., has terms that alternate

and decrease to 0 in absolute value. The alternating series test guarantees that the

series converges. In act, it converges to ln(2) 0.693. Adding the rst ve terms gives

5

1 1 1 1 471 0.783

2 3 4 5 60S = + + = = . The error bound says that the truncation error is no

greater in magnitude than the rst omitted term, 1

6= 0.16 . Indeed, S

5is within 0.16

o ln(2).

Absolute Convergence Test

Though rarely a topic on recent AP Exams, the absolute convergence test is perhaps

the simplest test o all. It says that i1

n

n

a

= converges, then

1

n

n

a

= converges. In other

words, a series that converges absolutely must converge. You can also be sure that

1 1 1

n n n

n n n

a a a

= = =

.



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59

This test can be more useul than rst meets the eye. Both the integral test

and the comparison test require terms that are nonnegative. I youre presented with

a series1

n

n

a

= that doesnotsatisy this requirement, you could try testing whether

1

n

n

a

=

converges. I it does, this test guarantees that1

n

n

a

=

converges as well. Youmight also try the alternating series test.


When a series converges absolutely, the order in which the terms are added

makes no dierence. The sum will be the same. The same isnottrue or series

that converge but not absolutely. Once again, consider the alternating harmonic

series. Earlier, we noted that the sum( )

= + = ( )+

=

1 1 12

1

32

1

1

n

n n... ln . But i we

rearrange the terms (two positive terms and then a negative termrepeated)

1+ 13 12 + 15 + 17 14 + 19 + 111 16 +L well get a convergent series. Curiously, though,

this series doesnotconverge to ln(2). You can show that this series converges to

3ln 2( )2

by using the series orln(2) in ln 2( )+1

2ln 2( ). For a series like 1 1

4

1

9

1

16 + + ...

that does converge absolutely (the series o absolute values is ap-series withp=2),it

doesnt matter how you rearrange the terms; the series will still converge to the same

number.

Questions

1. I1

n

n

a

= diverges, which o the ollowing must be true?

I. lim 0nn

a

II. limn

an +1

an

1

2III. I(n)=a

n, then x dx( )

1

diverges.

2. I1

n

n

a

= converges, which o the ollowing is a valid conclusion?

I.1

lim2

nn

a

II. 1lim 2nn

n

a

a

+

III.

1

0nn

a

=

=

3. I lim 0nn

a

= , which o these must be true?

I.1

n

n

a

= converges II.

1

n

n

a

= diverges III. 1lim 1n

nn

a

a

+

0 or alln, then which o these must be true?

I. a1

+ a2

a3

a4

+ a5

+ a6

a7

a8

+ ... also converges.



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60

II. lim 0nn

a

=

III. 1lim 0nn

n

a

a

+

=

5. Ia1

+ a2

+ a3

+ ... converges, which o these must be true?

I. a1

+ a2

a3

a4

+ a5

+ a6

a7

a8

+ ... also converges.

II. lim 0nn

a

=

III. 1lim 0nn

n

a

a

+

=

6. How many terms o the series( )

= + +

=

1 1 12

1

3

1

1

n

n n... must be added beore the

alternating series error bound would guarantee that the sum is within 0.001 o

ln(2)?

Answers

1. Option I need not be true because a series can diverge even when itsnth term

goes to 0. For example,1

nn=1

diverges but limn

1

n= 0.

Option II must be true, because i limn

an +1

an

=1

2, then

1

n

n

a

=

would converge by theratio test.

Option III need not be true because we dont know i(x) is continuous, positive,

and decreasing. I(x)=an

orn x


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61

Notice the dierence between Options I and III.

3. None must be true. Having annth term that goes to zero tells us nothing about

the convergence o a series. Counterexamples to I, II, and III, respectively, are

1

nn=1

, 1n

2

n=1

, and 1n

n=1

.

4. Option I must be true. Since the series in I converges absolutely, it converges.

Option II must be true. When a series converges, itsnth term must go to 0 by the

nth term test.

Option 3 need not be true. All we know is that i 1lim nn

n

a

a

+

exists, then

10 lim 1nnn

a

a+ .

5. Notice the dierence between this question and the previous one. Here we are not

told that an

> 0 or alln, so we dont know i the given series converges absolutely.

The terms could really alternate in sign. The only option that must be true is

Option II.

6. By the alternating series error bound, adding 999 terms guarantees the

error is less than the absolute value o the thousandth term, or 0.001. In act,

1( )n +1

nn=1

999

= 0.6936 is within 0.001 o ln(2) 0.6931.

Java Applets or Series

http://www.slu.edu/classes/maymk/SeriesGraphs/SeriesGraphs.html

http://www.slu.edu/classes/maymk/SeqSeries/SeqSeries.html

http://www.scottsarra.org/applets/calculus/SeriesGrapherApplet.html

(This one oers a nice way to increase the degree and graph Taylor polynomial

approximations.)

Bibliography

Dick, Thomas P., and Charles M. Patton. Calculus o a Single Variable. Pacic Grove,

CA: Brooks/Cole Publishing, 1994.



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62

Hughes-Hallett, Deborah, and Andrew M. Gleason, et al. Calculus Single Variable,

Fourth Edition. New York: John Wiley and Sons, 2004.

Kline, Morr

Documents

AP Sf Calculus Infinte Series