Sequences and Series (Type 10)

Posted on March 27, 2020 by Lin McMullin

AP Questions Type 10: Sequences and Series (BC Only)

The last BC question on the exams usually concerns sequences and series. The question may ask students to write a Taylor or Maclaurin series and to answer questions about it and its interval of convergence, or about a related series found by differentiating or integrating. The topics may appear in other free-response questions and in multiple-choice questions. Questions about the convergence of sequences may appear as multiple-choice questions. With about 8 multiple-choice questions and a full free-response question this is one of the largest topics on the BC exams.

Convergence tests for series appear on both sections of the BC Calculus exam. In the multiple-choice section, students may be asked to say if a sequence or series converges or which of several series converge.

The Ratio test is used most often to determine the radius of convergence and the other tests to determine the exact interval of convergence by checking the convergence at the end points. Click here for a convergence test chart students should be familiar with; this list is also on the resource page.

Students should be familiar with and able to write several terms and the general term of a Taylor or Maclaurin series. They may do this by finding the derivatives and constructing the coefficients from them, or they may produce the series by manipulating a known or given series. They may do this by substituting into a series, differentiating it, or integrating it.

The general form of a Taylor series is $\displaystyle \sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$ ; if a = 0, the series is called a Maclaurin series.

What Students Should be Able to Do

Use the various convergence tests to determine if a series converges. The test to be used is rarely given so students need to know when to use each of the common tests. For a summary of the tests click: Convergence test chart. and the posts “What Convergence Test Should I use?” Part 1 and Part 2
Understand absolute and conditional convergence. If the series of the absolute values of the terms of a series converges, then the original series is said to absolutely convergent (or converges absolutely). If a series is absolutely convergent, then it is convergent. If the series of absolute values diverges, then the original series may or may not converge; if it converges it is said to be conditionally convergent.
Write the terms of a Taylor or Maclaurin series by calculating the derivatives and constructing the coefficients of each term.
Distinguish between the Taylor series for a function and the function. DO NOT say that the Taylor polynomial is equal to the function (this will lose a point); say it is approximately equal.
Determine a specific coefficient without writing all the previous coefficients.
Write a series by substituting into a known series, by differentiating or integrating a known series, or by some other algebraic manipulation of a series.
Know (from memory) the Maclaurin series for sin(x), cos(x), e^x and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into one of these.
Find the radius and interval of convergence. This is usually done by using the Ratio test to find the radius and then checking the endpoints.
Be familiar with geometric series, its radius of convergence, and be able to find the number to which it converges, $\displaystyle {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$ . Re-writing a rational expression as the sum of a geometric series and then writing the series has appeared on the exam.
Be familiar with the harmonic and alternating harmonic series. These are often useful series for comparison.
Use a few terms of a series to approximate the value of the function at a point in the interval of convergence.
Determine the error bound for a convergent series (Alternating Series Error Bound and Lagrange error bound). See my posts on Error Bounds and the Lagrange Highway
Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).

This list is quite long, but only a few of these items can be asked in any given year. The series question on the free-response section is usually quite straightforward. Topics and convergence test may appear on the multiple-choice section. As I have suggested before, look at and work as many past exam questions to get an idea of what is asked and the difficulty of the questions. Click on Power Series in the “Posts by Topic” list on the right side of the screen to see previous posts on Power Series or any other topic you are interested in.

Free-response questions:

2004 BC 6 (An alternate approach, not tried by anyone, is to start with $\displaystyle \sin \left( {5x+\tfrac{\pi }{4}} \right)=\sin (5x)\cos \left( {\tfrac{\pi }{4}} \right)+\cos (5x)\sin \left( {\tfrac{\pi }{4}} \right)$ )
2011 BC 6 (Lagrange error bound)
2016 BC 6
2017 BC 6
2019 BC 6

Multiple-choice questions from non-secure exams:

2008 BC 4, 12, 16, 20, 23, 79, 82, 84
2012 BC 5, 9, 13, 17, 22, 27, 79, 90,

These question come from Unit 10 of the 2019 CED.

Revised March 12, 2021

A Curiosity

Posted on February 4, 2020 by Lin McMullin

Thoughts on the power series for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ , which I found curious.

Last week someone asked me a question about the Maclaurin series for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ . Finding the Maclaurin series is straightforward:

$\displaystyle \cos \left( x \right)=1-\frac{{{{x}^{2}}}}{{2!}}+\frac{{{{x}^{4}}}}{{4!}}-\frac{{{{x}^{6}}}}{{6!}}+\cdots +{{\left( {-1} \right)}^{n}}\frac{{{{x}^{{2n}}}}}{{\left( {2n} \right)!}}+\cdots$

Substituting $\displaystyle \sqrt{{x}}$ for x gives

$\displaystyle \cos \left( {\sqrt{x}} \right)=R\left( x \right)=1-\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{2!}}+\frac{{{{{\left( {\sqrt{x}} \right)}}^{4}}}}{{4!}}-\frac{{{{{\left( {\sqrt{x}} \right)}}^{6}}}}{{6!}}+\cdots {{\left( {-1} \right)}^{n}}\frac{{{{{\left( {\sqrt{x}} \right)}}^{{2n}}}}}{{\left( {2n} \right)!}}+\cdots$

$\displaystyle \cos \left( {\sqrt{x}} \right)=R\left( x \right)=1-\frac{x}{{2!}}+\frac{{{{x}^{2}}}}{{4!}}-\frac{{{{x}^{3}}}}{{6!}}\cdots +{{\left( {-1} \right)}^{n}}\frac{{{{x}^{n}}}}{{\left( {2n} \right)!}}+\cdots$

We can find the radius and interval of convergence by using the Ratio test:

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left| {\frac{{\frac{{{{x}^{{n+1}}}}}{{(2(n+1))!}}}}{{\frac{{{{x}^{n}}}}{{\left( {2n} \right)!}}}}} \right|=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left| {\frac{x}{{\left( {2x+2} \right)\left( {2n+1} \right)}}} \right|=0$

This indicates that Maclaurin series converges for all Real numbers. However, the original function $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ is not defined for negative numbers, but the series is. This can be accounted for by the fact that the series contains only even powers of x, and for all Real numbers x, $\displaystyle {{\left( {\sqrt{x}} \right)}^{2}}$ is a Real number. In addition, since the function ends at x = 0, how can the Maclaurin series be centered there? Since it is not defined to the left of zero, how can it have derivatives at zero?

This is in conformance with the graph. You can see the graph and experiment with here on Desmos. Use the slider and note the exaggerated scales. Also note that the power series extends steeply up to the left from the point (0, 1).

$\displaystyle \cos \left( {\sqrt{x}} \right)$ in red largely covered by its Maclaurin series (with n = 14) in blue.

The curiosity is that $\displaystyle \cos \left( {\sqrt{x}} \right)$ is not defined for negative numbers and is not differentiable at x = 0 (because the two-sided limit defining the derivative does not exist to the left of x = 0. But, but the Maclaurin series is continuous and differentiable for all Real numbers. The Maclaurin series is a good approximation for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ but approximates a larger function to the left of x = 0.

The explanation is that there is a larger function (that is, one defined for all Real numbers with the appropriate derivatives) that includes $latex \displaystyle \cos \left( {\sqrt{x}} x > 0 as part of it. The series is

$\displaystyle R\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\cos \left( {\sqrt{x}} \right)} & {x\ge 0} \\ {\cosh \left( {\sqrt{{-x}}} \right)} & {x<0} \end{array}} \right.$

(Note: $\displaystyle \cosh \left( {\sqrt{{-x}}} \right)=1+\frac{x}{{2!}}+\frac{{{{x}^{2}}}}{{4!}}+\frac{{{{x}^{3}}}}{{6!}}+\cdots =\frac{{{{e}^{{\sqrt{{-x}}}}}+{{e}^{{-\sqrt{{-x}}}}}}}{2}$ )

I wish to thank Louis A. Talman, Ph.D,,Emeritus Professor of Mathematics Metropolitan State University of Denver for helping me understand this function better and correcting some of my early ideas. He is the one who Developed the piecewise defined series above. An explanation of the reasoning and a longer discussion of this series can be found in this note “On $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ .” click here. In that note he shows that the Maclaurin series R(x) approximates this piecewise defined function. The two pieces form a function that is continuous and differentiable everywhere including at x = 0. (The pieces join smoothly the point (0, 1).

A similar curious situation, where a series, but not a Taylor/Maclaurin series, approximates a function is discussed in Geometric Series – Far Out.

What’s the “Best” Error Bound?

Posted on January 28, 2020 by Lin McMullin

A know a lot of people like mathematics because there is only one answer, everything is exact. Alas, that’s not really the case. Numbers written as non-terminating decimals are not “exact;” they must be rounded or truncated somewhere. Even things like $\sqrt{7},\pi ,$ and 5/17 may look “exact,” but if you ever had to measure something to those values, you’re back to using decimal approximations.

There are many situations in mathematics where it is necessary to find and use approximations. Two if these that are usually considered in introductory calculus courses are approximating the value of a definite integral using the Trapezoidal Rule and Simpson’s Rule and approximating the value of a function using a Taylor or Maclaurin polynomial.

If you are using an approximation, you need and want to know how good it is; how much it differs from the actual (exact) value. Any good approximation technique comes with a way to do that. The Trapezoidal Rule and Simpson’s Rule both come with expressions for determining how close to the actual value they are. (Trapezoidal approximations, as opposed to the Trapezoidal Rule and Simpson’s Rule per se, are tested on the AP Calculus Exams. The error is not tested.) The error approximation using a Taylor or Maclaurin polynomial is tested on the exams.

The error is defined as the absolute value of the difference between the approximated value and the exact value. Since, if you know the exact value, there is no reason to approximate, finding the exact error is not practical. (And if you could find the exact error, you could use it to find the exact value.) What you can determine is a bound on the error; a way to say that the approximation is at most this far from the actual value. The BC Calculus exams test two ways of doing this, the Alternating Series Error Bound (ASEB) and the Lagrange Error Bound (LEB). These two techniques are discussed in my previous post, Error Bounds. The expressions used below are discussed there.

Examining Some Error Bounds

We will look at an example and the various ways of computing an error bound. The example, which seems to come up this time every year, is to use the third-degree Maclaurin polynomial for sin(x) to approximate sin(0.1).

Using technology to twelve decimal places sin(0.1) = 0.099833416647

The Maclaurin (2n – 1)^th-degree polynomial for sin(x) is

$\displaystyle x-\frac{1}{{3!}}{{x}^{3}}+\frac{1}{{5!}}{{x}^{5}}-+\cdots \frac{1}{{\left( {2n-1} \right)!}}{{x}^{{2n-1}}}$

So, using the third degree polynomial the approximation is

$\sin \left( {0.1} \right)\approx 0.1-\frac{1}{6}{{\left( {0.1} \right)}^{3}}=0.099833333333...$

The error to 12 decimal places is the difference between the approximation and the 12 place value. The error is:

$\displaystyle 0.00000008331349=8.331349\times {{10}^{{-8}}}=Error$

Using the Alternating Series Error Bound:

Since the series meets the hypotheses for the ASEB (alternating, decreasing in absolute value, and the limit of the n^th term is zero), the error is less than the first omitted term. Here that is

$\displaystyle \frac{1}{{5!}}{{\left( {0.1} \right)}^{5}}\approx 0.0000000833333\approx 8.33333\times {{10}^{-8}}={{B}_{1}}$

The actual error is less than B₁ as promised.

Using the Legrange Error Bound:

For the Lagrange Error Bound we must make a few choices. Nevertheless, each choice gives an error bound larger than the actual error, as it should.

For the third-degree Maclaurin polynomial, the LEB is given by

$\displaystyle \left| {\frac{{\max {{f}^{{(4)}}}\left( z \right)}}{{4!}}{{{(0.1)}}^{4}}} \right|$ for some number z between 0 and 0.1.

The fourth derivative of sin(x) is sin(x) and its maximum absolute value between 0 and 0.1 is |sin(0.1)|. So, the error bound is

$\displaystyle \left| {\frac{{\sin (0.1)}}{{4!}}{{{(0.1)}}^{4}}} \right|\approx 4.15973...\text{ }\!\!\times\!\!\text{ }{{10}^{-7}}={{B}_{2}}$

However, since we’re approximating sin(0.1) we really shouldn’t use it. In a different example, we probably won’t know it.

What to do?

The answer is to replace it with something larger. One choice is to use 0.1 since 0.1 > sin(0.1). This gives

$\displaystyle \left| {\frac{{0.1}}{{4!}}{{{(0.1)}}^{4}}} \right|\approx 4.166666666\times {{10}^{{-7}}}={{B}_{3}}$

The usual choice for sine and cosine situations is to replace the maximum of the derivative factor with 1 which is the largest value of the sine or cosine.

$\displaystyle \left| {\frac{1}{{4!}}{{{(0.1)}}^{4}}} \right|\approx 4.166666666\times {{10}^{{-6}}}={{B}_{4}}$

Since the 4^th degree term is zero, the third-degree Maclaurin Polynomial is equal to the fourth-degree Maclaurin Polynomial. Therefore, we may use the fifth derivative in the error bound expression, $\displaystyle \left| {\frac{{\max {{f}^{{(5)}}}\left( z \right)}}{{5!}}{{{(0.1)}}^{5}}} \right|$ to calculate the error bound. The 5^th derivative of the sin(x) is cos(x) and its maximum value in the range is cos(0) =1.

$\displaystyle \left| {\frac{{\cos (0)}}{{5!}}{{{(0.1)}}^{5}}} \right|\approx 8.33333333\times {{10}^{{-8}}}={{B}_{5}}$

I could go on ….

Since B₁, B₂, B₃, B₄, and B₅are all greater than the error, which should we use? Or should we use something else? Which is the “best”?

The error is what the error is. Fooling around with the error bound won’t change that. The error bound only assures you your approximation is, or is not, good enough for what you need it for. If you need more accuracy, you must use more terms, not fiddle with the error bound.

2019 CED Unit 10: Infinite Sequences and Series

Posted on January 14, 2020 by Lin McMullin

Unit 10 covers sequences and series. These are BC only topics (CED – 2019 p. 177 – 197). These topics account for about 17 – 18% of questions on the BC exam.

Topics 10.1 – 10.2

Timing

The suggested time for Unit 9 is about 17 – 18 BC classes of 40 – 50-minutes, this includes time for testing etc.

Previous posts on these topics :

Introducing Power Series 1

A Lesson on Sequences

Posted on July 9, 2019 by Lin McMullin

This blog post describes a lesson that investigates some ideas about sequences that are not stressed in the AP Calculus curriculum. The lesson could be an introduction to sequences. I think the lesson would work in an Algebra I course and is certainly suitable for a pre-calculus course. The investigation is of irrational numbers and their decimal representation. The successive decimal approximations to the square root of 2 is an example of a non-decreasing sequence that is bounded above and therefore converges.

Students do not need to know any of that as it will be developed in the lesson. Specifically, don’t even mention square roots, the square root of 2, or even irrational numbers until a student mention something of the sort.

This is not an efficient algorithm for finding square roots. There are far more efficient ways.

We begin with some preliminaries.

Preliminaries

There is a blank table that you can copy for students to use here.
There is a summary of the new terms used and completed table Sequence Notes. Do not give this out until after the lesson is completed.
We will be working with some rather long decimal numbers that will need to be squared. Scientific and graphing calculators usually compute with 14 digits and give their results rounded to 12 digits. Since ours will quickly get longer than that, I suggest you use WolframAlpha. This can be used with a computer online (at wolframalpha.com) or with an app available for smart phones and tablets. It is best if students have this website or app for their individual use.

Students will enter their numbers as shown below. Specifying “30 digits” will produces answers long enough for our purpose. To speed things up, students can edit the current number by changing the last digit in the entry line. When you get started you may have to show students how to do this. Students will need internet access.

Computer Smart phone

The Lesson

The style of the lesson is Socratic. You, the teacher, will present the problem, explain how they are to go about it, and ask leading questions as appropriate. Some questions are suggested; be ready to ask others. Later, you will have to explain (define) some new words, but as much as possible let the class suggest what to do. Drag things out of them, rather than telling them.

To begin – Produce some data

Explain to the class that they are going to generate and investigate two lists of numbers (technically called sequences). Each new member of the lists will be a number with one more decimal place than the preceding number.

The first list, whose members are called L_n, will be the largest number with the given number of decimal places, n, whose square is less than two. The subscript, n, stands for the number of decimal places in the number.

Ask: “What is the largest integer whose square is less than 2?” Answer 1, so, L₀ = 1. Ask: What is the largest one place decimal whose square is less than 2?” Answer L₁ =1.4.

The second list, G_n, will be the smallest number whose square is greater than 2. So, G₀ = 2 and G₁ = 1.5. Notice that 1.4² = 1.96 < 2 and 1.5² = 2.25 > 2

Divide the class into 10 groups named Group 0, Group 1, Group 2, …, Group 9. In each round the groups will append their “name” to the preceding decimal and square the resulting number. Group 0 squares 1.40, group 1, squares 1.41, group 2 squares 1.42, etc. using WolframAlpha.

Ask which groups have squares less than two and enter the largest in L_n, the next number will be the smallest number whose square is greater than 2; enter it in G_n.

Complete the table by entering the largest number whose square is less than 2 in the L_n column and the smallest number whose square is greater than 2 in the G_n column. At each stage, each group appends their digit to the most recent L_n . Project or write the table on the board. Students may fill in their own copy. A completed table is here: Sequence Notes and definitions

Next – Lead a discussion

When the table is complete, prompt the students to examine the lists and come up with anything and everything they observe whether it seems important or not. Accept and discuss each observation and let the others say what they think about each observation. (Obviously, don’t deprecate or laugh at any answer – after all at this point, we don’t know what is and is not significant.)

There are (at least) three observations that are significant to what we will consider next. Hopefully, someone will mention them; keep questioning them until they do. They are these, although students may use other terms:

L_n is non-decreasing. Students may first say L_n is increasing. Pause if they do and look at L₁₂ and L₁₃, and L₁₅ and L₁₆. Ask how they know L_n is non-decreasing (because each time we add a digit on the end, you get a bigger number).
Likewise, G_n is non-increasing.
For all n, L_n < G_n, and the numbers differ only in the last digit, and with the last digits differing by 1.

Direct instruction: Explain these ideas and terms (definition)

A sequence is a list or set of numbers in a given order.
A sequence is bounded above if there exists a number greater than or equal to all the terms of the sequence. The smallest upper bound of a sequence is called its least upper bound (l.u.b.)
A sequence is bounded below if there exists a number less than or equal to all the terms of the sequence. The largest lower bound is called the greatest lower bound (g.l.b.)

More questions: Apply these terms to the sequence L_n with questions like these:

Is L_n bounded above, below, or not bounded? (Bounded above)
Give an example of a number greater than all the terms of L_n. (Many answers: 1,000,000, 4, 2, 1.415, etc. and, in fact, any and every number in G_n)
What is the l.u.b. of L_n? Can you think of the smallest number that is an upper bound of this sequence? (Yes, $\sqrt{2}$ . Don’t tell them this – drag it out of them if necessary.) Why? How do you know this?
Make the class convince you that for all n, $\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\left\{ {{{L}_{n}}} \right\}=\sqrt{2}$

Ask similar questions about G_n.

Is G_nbounded above, below, or not bounded? (Bounded below)
Give an example of a number less than all the terms of L_n. (Many answers: any negative number, zero, 1, 1.414, etc. Any and every number in L_n)
What is the g.l.b. of G_n? Can you think of the greatest number that is a lower bound of this sequence? (Yes, $\sqrt{2}$ ) Why? How do you know this?
Make the class convince you that for all n, $\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\left\{ {{{G}_{n}}} \right\}=\sqrt{2}$

Summing Up

Ask, “What’s happening with the numbers in the L_n sequence?” and “What’s happening to the numbers in the G_nsequence?”

The answer you want is that they are getting closer to $\sqrt{2}$ , one from below, the other from above. (As always, wait for a student to suggest this and then let the others discuss it.)

Once everyone is convinced, explain how mathematicians say and write, “gets closer to”:

Mathematicians say that $\sqrt{2}$ is the limiting value (or limit) of both sequences. They write $\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\left\{ {{{L}_{n}}} \right\}=\sqrt{2}$ and $\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\left\{ {{{G}_{n}}} \right\}=\sqrt{2}$ .

Explain very carefully that while $n\to \infty$ is read, “n approaches infinity,” that infinity, $\infty$ , is not a number. The symbol $n\to \infty$ means that n gets larger without bound or that n gets larger than all (any, every) positive numbers.

In a more technical sense there is an infinite series $\displaystyle \sum\limits_{{n=0}}^{\infty }{{{{a}_{n}}\cdot {{{10}}^{{-n}}}}}$ where $\displaystyle {{a}_{n}}$ is one of the digits 0, 1, 2, 3, …, 9, but there is no formula for listing the values of $\displaystyle {{a}_{n}}$ . However, the sequence of partial sum of this series is the sequence $\displaystyle \left\{ {{{L}_{n}}} \right\}$ which converges to $\displaystyle \sqrt{2}$ . Therefore, $\displaystyle \sum\limits_{{n=0}}^{\infty }{{{{a}_{n}}\cdot {{{10}}^{{-n}}}=\sqrt{2}}}$

$\displaystyle \sqrt{2}$ is an Irrational number, but this same procedure may be used to find decimal approximation of roots of rational numbers as well. However, for Rational numbers, there are easier ways.

Finally, Irrational numbers are exactly those that cannot be written as repeating (or terminating) decimals. They “go on forever” with no pattern. The decimals you can calculate eventually stop and are rounded to the last digit. Even WolframAlpha and similar computers must eventually do this. Irrational numbers are the limits of sequences like the one we looked at today.

Exercises

Follow the procedure above to find the sequence whose limit is $\sqrt{{\frac{{16}}{{121}}}}$ . Find this number the usual way (simplify and use long division) and compare the results.
Follow the procedure above to find the sequence whose limit is $\sqrt{{0.390625}}$ . Find this number the usual way and compare the results.
Using WolframAlpha determine if the computer is using L_n, G_n. both, or neither when it gives a value for $\sqrt{2}$ . (Hint: enter “square root 2 to 5 digits” and change to 6, 7, and 8 digits; compare the answer with the sequences, you found.)

Answers:

0.363636…
0.625
For n = 5 and 6 the numbers are from L_n, for n = 7 and 8 they are from G_n. WolframAlpha is using a different algorithm to compute the square root of 2; the numbers appear from both sequences due to the rounding of the answers. To see WolframAlpha’s algorithm type “square root algorithm” on the entry line. This method also produces a sequence of approximations a/b.

Revised July 28, 2021

Type 10: Sequences and Series Questions

Posted on April 5, 2019 by Lin McMullin

The last BC question on the exams usually concerns sequences and series. The question usually asks students to write a Taylor or Maclaurin series and to answer questions about it and its interval of convergence, or about a related series found by differentiating or integrating. The topics may appear in other free-response questions and in multiple-choice questions. Questions about the convergence of sequences may appear as multiple-choice questions. With about 8 multiple-choice questions and a full free-response question this is one of the largest topics on the BC exams.

What Students Should be Able to Do

Use the various convergence tests to determine if a series converges. The test to be used is rarely given so students need to know when to use each of the common tests. For a summary of the tests click: Convergence test chart. and the posts “What Convergence Test Should I use?” Part 1 and Part 2
Understand absolute and conditional convergence. If the series of the absolute values of the terms of a series converges, then the original series is said to absolutely convergent (or converges absolutely). If the series of absolute values diverges, then the original series may or may not converge; if it converges it is said to be conditionally convergent.
Write the terms of a Taylor or Maclaurin series by calculating the derivatives and constructing the coefficients of each term.
Distinguish between the Taylor series for a function and the function. DO NOT say that the Taylor polynomial is equal to the function (this will lose a point); say it is approximately equal.
Determine a specific coefficient without writing all the previous coefficients.
Write a series by substituting into a known series, by differentiating or integrating a known series, or by some other algebraic manipulation of a series.
Know (from memory) the Maclaurin series for sin(x), cos(x), e^x and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into them.
Find the radius and interval of convergence. This is usually done by using the Ratio test and checking the endpoints.
Be familiar with geometric series, its radius of convergence, and be able to find the number to which it converges, $\displaystyle {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$ . Re-writing a rational expression as the sum of a geometric series and then writing the series has appeared on the exam.
Be familiar with the harmonic and alternating harmonic series. These are often useful series for comparison.
Use a few terms of a series to approximate the value of the function at a point in the interval of convergence.
Determine the error bound for a convergent series (Alternating Series Error Bound and Lagrange error bound). See my posts on Error Bounds and the Lagrange Highway
Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).

Free-response questions:

2004 BC 6 (An alternate approach, not tried by anyone, is to start with $\displaystyle \sin \left( {5x+\tfrac{\pi }{4}} \right)=\sin (5x)\cos \left( {\tfrac{\pi }{4}} \right)+\cos (5x)\sin \left( {\tfrac{\pi }{4}} \right)$ )
2016 BC 6
2017 BC 6

Multiple-choice questions from non-secure exams:

2008 BC 4, 12, 16, 20, 23, 79, 82, 84
2012 BC 5, 9, 13, 17, 22, 27, 79, 90,

Tuesday February 27 – AP Exam Review
Friday, March 2 – Resources for reviewing
Tuesday March 6 – Type 1 questions – Rate and accumulation questions
Friday March 9 – Type 2 questions – Linear motion problems
Tuesday March 13 – Type 3 questions – Graph analysis problems
Friday March 16 – Type 4 questions – Area and volume problems
Tuesday Match 20 Type 5 questions – Table and Riemann sum questions
Friday March 23 Type 6 questions – Differential equation questions
Tuesday March 27 – Type 7 questions – miscellaneous
Friday March 30 Type 8 questions – Parametric and vector questions (BC topic)
Tuesday April 2 – Type 9 questions – Polar equations (BC topic)
Friday April 5 – Type 10 questions – Sequences and Series (BC topic) (this post)

The concludes the series of posts on the type questions in review for the AP Calculus exams.

Power Series 2

Posted on February 12, 2019 by Lin McMullin

This is a BC topic

Good Question 16 (11-30-2018) What you get when you substitute.

Geometric Series – Far Out (2-14-2017) A very interesting and instructive mistake

Synthetic Summer Fun (7-10-2017) Finding the Taylor series coefficients without differentiating

Error Bounds (2-22-2013) The alternating series error bound, and the Lagrange error bound

The Lagrange Highway (5-20-15) a metaphor for the error bound

REVIEW NOTES Type 10: Sequence and Series Questions (4-6-2018) A summary for reviewing sequences and series.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Category Archives: Sequences and series

Sequences and Series (Type 10)

A Curiosity

What’s the “Best” Error Bound?