Category Archives: Sequences and series

Power Series 1

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This is a BC topic

POWER SERIES (Maclaurin series and Taylor series)

Introducing Power Series 1 (2-8-2013) Making better approximations

Introducing Power Series 2 (2-11-2013) Graphing and seeing the interval of convergence

Introducing Power Series 3 (2-13-2013) Questions pointing the way to power series

Graphing Taylor Polynomials (2-7-2017) Using a graphing calculator to graphs Taylor series

New Series from Old 1 (2-15-2013) Substituting

New Series from Old 2 (2-18-2013) Differentiating and Integrating

New Series from Old 3 (2-20-2013) Rational functions as geometric series

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

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam:  In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and $35/non-members. You can register for the event by following this link: http://eventreg.collegeboard.org/d/xbqbjz

 

 

 

 

 

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Sequences

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This is a BC topic.

SEQUENCES

Everyday series (1-17-2017) The most familiar series: Numbers

Amortization (2-9-2015) An important use of a (finite) series – Find you mortgage payment without calculus.

Convergence Test List A summary of the tests. Download and copy for your students (and yourself)

Which Convergence Test Should I Use? Part 1 (2-9-2018) You have a big choice

Which Convergence Test Should I Use? Part 2 (2-16-2018) Making the best choice.

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

The College Board is pleased to offer a new live online event for new and experienced AP Calculus teachers on March 5th at 7:00 PM Eastern.

I will be the presenter.

The topic will be AP Calculus: How to Review for the Exam:  In this two-hour online workshop, we will investigate techniques and hints for helping students to prepare for the AP Calculus exams. Additionally, we’ll discuss the 10 type questions that appear on the AP Calculus exams, and what students need know and to be able to do for each. Finally, we’ll examine resources for exam review.

Registration for this event is $30/members and $35/non-members. You can register for the event by following this link: http://eventreg.collegeboard.org/d/xbqbjz

 

 

 

 

 

Good Question 16

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I had an email last week from a teacher asking, how come I can use a substitution to find a power series for  \cos \left( {2x} \right), and for  {{e}^{{\left( {x-1} \right)}}}, but not for  \cos \left( {3x+\frac{\pi }{6}} \right)?

The answer is that you can. Substituting (2x) into the cosine’s series give you a Taylor series centered at x = 0, a Maclaurin Series. Substituting (x – 1) into the series for ex gives you a Taylor series centered at x = 1. And substituting \left( {3x+\frac{\pi }{6}} \right) into the cosine series gives you a Taylor series centered at  x=-\frac{\pi }{{18}}. I suspect that she was hoping for or was asked to find a Maclaurin series, not one with such a strange center.

The center of a Taylor series is the value of x that makes its argument zero.

AP Exam Question 2004 BC 6(a)

This brought to mind the AP Exam question 2004 BC 6(a) where students were asked to write the third-degree Taylor polynomial about x = 0 for the function f\left( x \right)=\sin \left( {5x+\frac{\pi }{4}} \right). The intended method was for students to find the first three derivative and substitute them into the general form for a Taylor series. That’s what students who got this correct did. This is the only time I can remember when students were expected to do that; usually they manipulate a given series or substitute into a known series.

A number of students tried to substitute \left( {5x+\frac{\pi }{4}} \right) into the series for the sine. This gets a very nice Taylor series centered at  x=-\frac{\pi }{{20}}. This earned no credit since a Maclaurin series was required.

But there is another way! (I originally wrote, “But there is an easier way!” but it’s only easier if you see how to do it.)

Trigonometry to the Rescue!

\sin \left( {5x+\frac{\pi }{4}} \right)=\sin (5x)\cos \left( {\frac{\pi }{4}} \right)+\cos \left( {5x} \right)\sin \left( {\frac{\pi }{4}} \right)=\frac{{\sqrt{2}}}{2}\left( {\sin \left( {5x} \right)+\cos \left( {5x} \right)} \right)

Then using the first two terms each from the series for sine and cosine you get the correct answer:

\displaystyle \frac{{\sqrt{2}}}{2}\left( {\left( {5x-{{{\frac{{\left( {5x} \right)}}{{3!}}}}^{3}}} \right)+\left( {1-\frac{{{{{\left( {5x} \right)}}^{2}}}}{{2!}}} \right)} \right)=\frac{{\sqrt{2}}}{2}+\frac{{5\sqrt{2}}}{2}x-\frac{{25\sqrt{2}}}{{2\left( {2!} \right)}}{{x}^{2}}-\frac{{125\sqrt{2}}}{{2\left( {3!} \right)}}{{x}^{3}}

This brings us to \cos \left( {3x+\frac{\pi }{6}} \right), which can be approached the same way. Here is the entire Maclaurin series.

\cos \left( {3x+\frac{\pi }{6}} \right)=\cos \left( {3x} \right)\cos \left( {\frac{\pi }{6}} \right)-\sin \left( {3x} \right)\sin \left( {\frac{\pi }{6}} \right)

\displaystyle =\frac{{\sqrt{3}}}{2}\cos \left( {3x} \right)-\frac{1}{2}\sin \left( {3x} \right)

\displaystyle =\frac{{\sqrt{3}}}{2}\sum\limits_{{n=0}}^{\infty }{{\frac{{{{{\left( {3x} \right)}}^{{2n}}}}}{{\left( {2n} \right)!}}}}-\frac{1}{2}\sum\limits_{{n=0}}^{\infty }{{\frac{{{{{\left( {3x} \right)}}^{{2n+1}}}}}{{\left( {2n+1} \right)!}}}}

\displaystyle =\sum\limits_{{n=0}}^{\infty }{{\left( {\frac{{\sqrt{3}\left( {{{3}^{{2n}}}} \right)}}{{2\left( {2n} \right)!}}{{x}^{{2n}}}-\frac{{1\left( {{{3}^{{2n+1}}}} \right)}}{{2\left( {2n+1} \right)!}}{{x}^{{2n+1}}}} \right)}}

Moral: Trig can be very useful.

Here is a previous post, Geometric Series – Far Out, that shows a “mistake” you may find interesting.

Good Question 14

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Good Question 14 – The Integral Test

I have no criteria for what constitutes a “Good Question” for this series of occasional posts. They are just questions that I found interesting, or that seem more than usually instructive, or that I learn something from. I cannot quote this question (2016 BC 92) since it is on a secure exam. What made it interesting is that to answer it students pretty much needed to know the proof of the Integral Test and the figures that go with it.

I recall only one AP question from many years ago that asked students to “prove” something – usually students are asked to show that a result was true by citing the theorem that applied and showing the hypotheses were met. The directions are often “justify your answer.”

Doing an original proof is not, in my opinion, a fair question and proving some known theorem is just a matter of memorization. For these reasons, students are not asked to prove things on the exams. So, should you prove things in class? Probably, yes.

Here is the usual proof of the integral test. Afterwards I’ll discuss the question from the exam.

The Integral Test

Hypotheses: Let f\left( x \right) be a function that is positive, decreasing, and continuous for x\ge 1 ; and let {{a}_{n}}=f\left( n \right) for x\ge 1

In the first drawing the rectangles have a height of an and a width of 1. The area is of each is an, and the sum of their areas the series is \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}.

Part 1: Notice that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}} Assume that the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

  • Conclusion 1: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}}  diverges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges.
  • Conclusion 2: (The contrapositive of conclusion 1) If the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges.

Part 2: In the second drawing below, assume that the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges. The sum of the areas of the rectangles is \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}}. (NB: this series starts at n = 2.) Since \sum\limits_{{n=2}}^{\infty }{{{{a}_{n}}}} is less than the convergent improper integral it will also converge. Adding {{a}_{1}} to this gives the original series, \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}; this series also converges.

  • Conclusion 3: If the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} converges, then the series \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} converges.
  • Conclusion 4: (The contrapositive of conclusion 3) If the series   \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}} diverges, then the improper integral \int_{1}^{\infty }{{f\left( x \right)dx}} diverges.

Putting the four conclusions together is the Integral Test: If the hypotheses above are met, then the series and the improper integral will both converge, or both diverge.

To answer the multiple-choice question (2106 BC 92) on the exams students were told that the improper integral converges. Therefore, the associated series converges. They then had to determine whether the series or the improper integral has the greater value. Stop here and see if you can figure that out.

Return to the first figure above, only this time assume that the improper integral and the series converge. It is pretty obvious that \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}>\int_{1}^{\infty }{{f\left( x \right)dx}}.

So, even though students were not asked to prove anything, a familiarity with the proof and its figures is necessary to answer the question. That’s why I liked it,

On the other hand, it is kind of an obscure point and I’m not sure it has any practical value.

What Convergence Test Should I Use? Part 2

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In last Friday’s post I really didn’t answer this question. Rather, I tried to show that there is not only one convergence test that must be used on a given series. Nevertheless, the form of a series suggests a test that is likely to work. In this post, I’ll try to give some suggestions as to what test to try first based on the form of the series.

For reference, click here for a table summarizing the common convergence tests.

The goal is for students to be able to decide which test to start with at a glance.

Start with the nth-term test for divergence. If the limit of the general term as n goes to infinity is not zero, the sequence will diverge. The \underset{{n\to \infty }}{\mathop{{\lim }}}\,{{a}_{n}}=0 is a necessary condition for convergence. It is not sufficient; if the limit is zero then the series may converge. Look for a convergence test.

If the series alternates plus and minus signs, it is an alternating series and if it satisfies the other hypotheses use the Alternating Series Test. If the series contains positive and negative signs that do not alternate, or one of the other hypotheses is not met, then a different test must be used.

If the series is geometric then the Geometric Series Test may be used. If the common ratio (the number multiplied by each term to get the next term) is between –1 and 1 the series converges. If the common ratio is greater than or equal to 1, or less than or equal to –1, the series diverges.

The remaining tests are for series with all positive terms. They are tests for absolute convergence. If you series has negative terms then you may ignore the signs and try one of the following tests. If your series is absolutely convergent, then it is convergent. (If not, it may still be convergent.)

If the general term (written with x’s) looks like something that you can integrate, use the Integral Test.

The Direct Comparison Test and the Limit Comparison Test are used if you can find a test to compare them with.

A p-series, \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{p}}}}}} converges if p>1  and diverges if p\le 1. A p-series is often a good test to use for comparison in the next two tests. However, any series whose convergence you are sure of may be used.

The Direct Comparison Test is used with fraction expressions. “Extra” factors in the denominator can often be ignored. Some examples

  •  \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{5}^{n}}\sqrt{n}}}}} would be a geometric series except for the radical. Compare it with the geometric series \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{5}^{n}}}}}}
  • \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}} can be compared with the p-series \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}. The hint here is that ignoring the lower power terms in the denominator and reducing we see that the original series looks like \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}. Both series converge. But be careful \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}-2n-1}}}} while similar, has terms greater than the terms of \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}.)
  • The terms of the series \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{{\left( {{{n}^{2}}+2} \right)}}^{{1/3}}}}}}} are larger than the harmonic series \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{n}}} a divergent p-series, so this series diverges.

The Limit Comparison Test may be used with the same kinds of series that are messy to use with direct comparison.

  • Returning to \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{2}}}}{{{{n}^{4}}+2n+1}}}}, try the limit comparison test with \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{{{n}^{2}}}}}}. The limit is 1, so both series converge.
  • \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{{\sqrt{{{{n}^{2}}+3}}}}}} Series with radicals also are candidates for the limit comparison test. Since the general terms is approximately \displaystyle {\frac{1}{n}} Compare this with \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{1}{n}}}. Both series diverge.

More complicated series, perhaps with exponential factors and/or factorials can be examined with the Ratio Test.

  • \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{3}^{n}}}}{{n!}}}} or \displaystyle \sum\limits_{{n=1}}^{\infty }{{\frac{{{{n}^{3}}}}{{{{5}^{n}}}}}} are candidates for the Ratio Test. Both Converge.
  • \displaystyle \sum\limits_{{n=1}}^{\infty }{{{{{\left( {-1} \right)}}^{n}}\frac{{n!}}{{{{{500}}^{n}}}}}} appears to be a candidate for the alternating series test. However, for large values of n > 530 the terms increase in absolute vale, so the alternating series test cannot be applied. The ratio test works here, but since the terms do not approach 0 as n increases, the nth-term test for divergence also works. This series diverges.

Practice, Practice, Practice

The AP Calculus BC exams rarely, if ever, specify which test to use. Often these are multiple-choice questions. If students can see whether the series converges or diverges, that is enough. But here again the key is practice, practice, practice. 

As you teach the various tests, pause to look at the form of the series in the exercises for each test that your book provides. Most books also have mixed sets of exercises where tests other than the one in that section are needed. One of the things you can do is assign these entire sets with the directions that students should determine what test they would try, and, for their comparison tests, to which series they would compare it. Discuss their opinions especially if there is more than one suggested or suggest others. Work only those those students are confused about or those for which they have divergent opinions; try to converge on a good test for each.

Revised July 18, 2021, January 29, 2023

Which Convergence Test Should I Use? Part 1

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One common question from students first learning about series is how to know which convergence test to use with a given series.  The first answer is: practice, practice, practice. The second answer is that there is often more than one convergence test that can be used with a given series.

I will illustrate this point with a look at one series and the several tests that may be used to show it converges. This will serve as a review of some of the tests and how to use them. For a list of convergence tests that are required for the AP Calculus BC exam click here.

To be able to use these tests the students must know the hypotheses of each test and check that they are met for the series in question. On multiple-choice questions students do not need to how their work, but on free-response questions (such as checking the endpoints of the interval of convergence of a Taylor series) they should state them and say that the series meets them.

For our example we will look at the series \displaystyle 1-\frac{1}{3}+\frac{1}{9}-\frac{1}{{27}}+\frac{1}{{81}}-+\ldots =\sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}}

Spoiler: Except for the first two tests to be considered, the other tests are far more work than is necessary for this series. The point is to show that several tests may be used for a given series, and to practice the other tests.

The Geometric Series Test is the obvious test to use here, since this is a geometric series. The common ratio is (–1/3) and since this is between –1 and 1 the series will converge.

The Alternating Series Test (the Leibniz Test) may be used as well. The series alternates signs, is decreasing in absolute value, and the limit of the nth term as n approaches infinity is 0, therefore the series converges.

The Ratio Test is used extensively with power series to find the radius of convergence, but it may be used to determine convergence as well. To use the test, we find

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{{n+1}}}} \right|}}{{\left| {{{{\left( {-\frac{1}{3}} \right)}}^{n}}} \right|}}=\frac{1}{3}  Since the limit is less than 1, we conclude the series converges.

Absolute Convergence

A series, \sum\limits_{{n=1}}^{\infty }{{{{a}_{n}}}}, is absolutely convergent if, and only if, the series \sum\limits_{{n=1}}^{\infty }{{\left| {{{a}_{n}}} \right|}} converges. In other words, if you make all the terms positive, and that series converges, then the original series also converges. If a series is absolutely convergent, then it is convergent. (A series that converges but is not absolutely convergent is said to be conditionally convergent.)

The advantage of going for absolute convergence is that we do not have to deal with the negative terms; this allows us to use other tests.

Applied to our example, if the series \sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{{n-1}}}}} converges, then our series \sum\limits_{{n=1}}^{\infty }{{{{{\left( {-\frac{1}{3}} \right)}}^{{n-1}}}}} will converge absolutely and converge.

The Geometric Series Test can be used again as above.

The Integral Test says if the improper integral \displaystyle {{\int_{1}^{\infty }{{\left( {\frac{1}{3}} \right)}}}^{x}}dx converges, then our original series will converge absolutely.

\displaystyle \int\limits_{1}^{\infty }{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\int\limits_{1}^{n}{{{{{\left( {\frac{1}{3}} \right)}}^{x}}}}dx=\underset{{n\to \infty }}{\mathop{{\lim }}}\,\left( {\frac{{{{{\left( {\frac{1}{3}} \right)}}^{n}}}}{{\ln \left( {1/3} \right)}}-\frac{{{{{\left( {\frac{1}{3}} \right)}}^{1}}}}{{\ln \left( {1/3} \right)}}} \right)=0-\frac{{1/3}}{{\ln \left( {1/3} \right)}}

\displaystyle =-\frac{{1/3}}{{\ln \left( {1/3} \right)}}>0 since ln(1/3) < 0.

The limit is finite, so our series converges absolutely, and therefore converges.

The Direct Comparison Test may also be used. We need to find a positive convergent series whose terms are term-by-term greater than the terms of our series. The geometric series \sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}} meets these two requirements. Therefore, the original series converges absolutely and converges.

The Limit Comparison Test is another possibility. Here we need a positive series that converges; we can use \sum\limits_{{n=1}}^{\infty }{{{{{\left( {\frac{1}{2}} \right)}}^{{n-1}}}}} again. We look at

\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\frac{{{{{\left( {1/3} \right)}}^{n}}}}{{{{{\left( {1/2} \right)}}^{n}}}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {\frac{2}{3}} \right)}^{n}}=0  and since the series in the denominator converges, our series converges absolutely.

So, for this example all the convergences that may be tested on the AP Calculus BC exam may be used with the single exception of the p-series Test which cannot be used with this series.

Teaching suggestions

  1. While the convergence of the series used here can be done all these ways, other series lend themselves to only one. Stress the form of the series that works with each test. For example, the Limit Comparison Test is most often used for rational expressions with the numerator of lower degree than the denominator and for expressions involving radicals of polynomials. The comparison is made with a p-series of whatever degree will make the numerator and denominator the same degree allowing the limit to be found.
  2. Most textbooks, after explaining each test and giving exercises on them, include a series of mixed exercises that require all the test covered up to that point. A good way to use this set is to assign students to state which test they would try first on each series. Discuss the opinions of the class and work any questions that students are unsure of or on which several ways are suggested.
  3. Give your students the series above, or a similar one, and have them prove its convergence using each of the convergence tests as was done above.
  4. Divide your class into groups and assign each group the series and one of the convergence tests. Ask them to use the test to prove convergence and then discuss the results as a group.

Of course, I didn’t really answer the question, did I? Check What Convergence Test Should I use Part 2

Updated February 23, 2013

More on Power Series

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Continuing with post on sequences and series

New Series from Old 1 Rewriting using substitution

New Series from Old 2 Finding series by differentiating and integrating

New Series from Old 3  Rewriting rational expressions as geometric series

Geometric Series – Far Out A look at doing a question the right way and the “wrong” way?

Error Bounds The Alternating Series Error Bound and the Lagrange Error Bound

The Lagrange Highway An example explaining error bounds

Synthetic Summer Fun Using synthetic division, the Remainder Theorem, the Factor Theorem and finding the terms of a Taylor Series (Probably more than you want to know, but possibly an enrichment idea.)