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Infinite Sequences and Series – Unit 10

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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

Topic 10.1: Defining Convergent and Divergent Series.

Topic 10. 2: Working with Geometric Series. Including the formula for the sum of a convergent geometric series.

Topics 10.3 – 10.9 Convergence Tests

The tests listed below are tested on the BC Calculus exam. Other methods are not tested. However, teachers may include additional methods.

Topic 10.3: The nth Term Test for Divergence.

Topic 10.4: Integral Test for Convergence. See Good Question 14

Topic 10.5: Harmonic Series and p-Series. Harmonic series and alternating harmonic series, p-series.

Topic 10.6: Comparison Tests for Convergence. Comparison test and the Limit Comparison Test

Topic 10.7: Alternating Series Test for Convergence.

Topic 10.8: Ratio Test for Convergence.

Topic 10.9: Determining Absolute and Conditional Convergence. Absolute convergence implies conditional convergence.

Topics 10.10 – 10.12 Taylor Series and Error Bounds

Topic 10.10: Alternating Series Error Bound.

Topic 10.11: Finding Taylor Polynomial Approximations of a Function.

Topic 10.12: Lagrange Error Bound.

Topics 10.13 – 10.15 Power Series

Topic 10.13: Radius and Interval of Convergence of a Power Series. The Ratio Test is used almost exclusively to find the radius of convergence. Term-by-term differentiation and integration of a power series gives a series with the same center and radius of convergence. The interval may be different at the endpoints.

Topic 10.14: Finding the Taylor and Maclaurin Series of a Function. Students should memorize the Maclaurin series for \displaystyle \frac{1}{{1-x}}, sin(x), cos(x), and ex.

Topic 10.15: Representing Functions as Power Series. Finding the power series of a function by, differentiation, integration, algebraic processes, substitution, or properties of geometric series.

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:

Before sequences

Amortization Using finite series to find your mortgage payment. (Suitable for pre-calculus as well as calculus)

A Lesson on Sequences An investigation, which could be used as early as Algebra 1, showing how irrational numbers are the limit of a sequence of approximations. Also, an introduction to the Completeness Axiom. 

Everyday Series

Convergence Tests

Reference Chart

Which Convergence Test Should I Use? Part 1 Pretty much anyone you want!

Which Convergence Test Should I Use? Part 2 Specific hints and a discussion of the usefulness of absolute convergence

Good Question 14 on the Integral Test

Sequences and Series

Graphing Taylor Polynomials Graphing calculator hints

Introducing Power Series 1

Introducing Power Series 2

Introducing Power Series 3

New Series from Old 1 substitution (Be sure to look at example 3)

New Series from Old 2 Differentiation

New Series from Old 3 Series for rational functions using long division and geometric series

Geometric Series – Far Out An instructive “mistake.”

A Curiosity An unusual Maclaurin Series

Synthetic Summer Fun Synthetic division and calculus including finding the (finite)Taylor series of a polynomial.

Error Bounds

Error Bounds Error bounds in general and the alternating Series error bound, and the Lagrange error bound

The Lagrange Highway The Lagrange error bound. 

What’s the “Best” Error Bound?

Review Notes

Type 10: Sequences and Series Questions

 

 

 

 

 

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2019 CED Unit 10: Infinite Sequences and Series

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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

Power Series 2

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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.

 

 

 

 

 

 

 

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

Error Bounds

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How Good is Your Approximation?

Whenever you approximate something, you should be concerned about how good your approximation is. The error, E, of any approximation is defined to be the absolute value of the difference between the actual value and the approximation. If Tn(x) is the Taylor/Maclaurin approximation of degree n for a function f(x) then the error is E=\left| f\left( x \right)-{{T}_{n}}\left( x \right) \right|.  This post will discuss the two most common ways of getting a handle on the size of the error: the Alternating Series error bound, and the Lagrange error bound.

Both methods give you a number B that will assure you that the approximation of the function at x={{x}_{0}} in the interval of convergence is within B units of the exact value. That is,

\left( f\left( {{x}_{0}} \right)-B \right)<{{T}_{n}}\left( {{x}_{0}} \right)<\left( f\left( {{x}_{0}} \right)+B \right)

or

{{T}_{n}}\left( {{x}_{0}} \right)\in \left( f\left( {{x}_{0}} \right)-B,\ f\left( {{x}_{0}} \right)+B \right).

Stop for a moment and consider what that means: f\left( {{x}_{0}} \right)-B and f\left( {{x}_{0}} \right)+B   are the endpoints of an interval around the actual value and the approximation will lie in this interval. Ideally, B is a small (positive) number.

Alternating Series

If a series \sum\limits_{n=1}^{\infty }{{{a}_{n}}} alternates signs, decreases in absolute value and \underset{n\to \infty }{\mathop{\lim }}\,\left| {{a}_{n}} \right|=0 then the series will converge. The terms of the partial sums of the series will jump back and forth around the value to which the series converges. That is, if one partial sum is larger than the value, the next will be smaller, and the next larger, etc. The error is the difference between any partial sum and the limiting value, but by adding an additional term the next partial sum will go past the actual value. Thus, for a series that meets the conditions of the alternating series test the error is less than the absolute value of the first omitted term:

\displaystyle E=\left| \sum\limits_{k=1}^{\infty }{{{a}_{k}}}-\sum\limits_{k=1}^{n}{{{a}_{k}}} \right|<\left| {{a}_{n+1}} \right|.

Example: \sin (0.2)\approx (0.2)-\frac{{{(0.2)}^{3}}}{3!}=0.1986666667 The absolute value of the first omitted term is \left| \frac{{{(0.2)}^{5}}}{5!} \right|=0.26666\bar{6}\times {{10}^{-6}}. So our estimate should be between \sin (0.2)\pm 0.266666\times {{10}^{-6}} (that is, between 0.1986666641 and 0.1986719975), which it is. Of course, working with more complicated series, we usually do not know what the actual value is (or we wouldn’t be approximating). So an error bound like 0.26666\bar{6}\times {{10}^{-6}} assures us that our estimate is correct to at least 5 decimal places.

The Lagrange Error Bound

Taylor’s Theorem: If f is a function with derivatives through order n + 1 on an interval I containing a, then, for each x in I , there exists a number c between x and a such that

\displaystyle f\left( x \right)=\sum\limits_{k=1}^{n}{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}+\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}}

The number \displaystyle R=\frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}} is called the remainder.

The equation above says that if you can find the correct c the function is exactly equal to Tn(x) + R. Notice the form of the remainder is the same as the other terms, except it is evaluated at the mysterious c. The trouble is we almost never can find the c without knowing the exact value of f(x), but; if we knew that, there would be no need to approximate. However, often without knowing the exact values of c, we can still approximate the value of the remainder and thereby, know how close the polynomial Tn(x) approximates the value of f(x) for values in x in the interval, i.

Corollary – Lagrange Error Bound. 

\displaystyle \left| \frac{{{f}^{\left( n+1 \right)}}\left( c \right)}{\left( n+1 \right)!}{{\left( x-a \right)}^{n+1}} \right|\le \left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right)\frac{{{\left| x-a \right|}^{n+1}}}{\left( n+1 \right)!}

The number \displaystyle \left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right)\frac{{{\left| x-c \right|}^{n+1}}}{\left( n+1 \right)!}\ge \left| R \right| is called the Lagrange Error Bound. The expression \left( \text{max}\left| {{f}^{\left( n+1 \right)}}\left( x \right) \right| \right) means the maximum absolute value of the (n + 1) derivative on the interval between the value of x and c. The corollary says that this number is larger than the amount we need to add (or subtract) from our estimate to make it exact. This is the bound on the error. It requires us to, in effect, substitute the maximum value of the n + 1 derivative on the interval from a to x for {{f}^{(n+1)}}\left( x \right). This will give us a number equal to or larger than the remainder and hence a bound on the error.

Example: Using the same example sin(0.2) with 2 terms. The fifth derivative of \sin (x) is -\cos (x) so the Lagrange error bound is \displaystyle \left| -\cos (0.2) \right|\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}, but if we know the cos(0.2) there are a lot easier ways to find the sine. This is a common problem, so we will pretend we don’t know cos(0.2), but whatever it is its absolute value is no more than 1. So the number \left( 1 \right)\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}=2.6666\bar{6}\times {{10}^{-6}} will be larger than the Lagrange error bound, and our estimate will be correct to at least 5 decimal places.

This “trick” is fairly common. If we cannot find the number we need, we can use a value that gives us a larger number and still get a good handle on the error in our approximation.

FYI: \displaystyle \left| -\cos (0.2) \right|\frac{\left| {{\left( 0.2-0 \right)}^{5}} \right|}{5!}\approx 2.61351\times {{10}^{-6}}

Corrected: February 3, 2015, June 17, 2022