Category Archives: Sequences and series

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.)

Introducing Power Series

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The posts for the next several weeks will be on topics tested only on the BC Calculus exams. Continuing with some posts on introducing power series (the Taylor and Maclaurin series)

Introducing Power Series 1 Two examples to lead off with.

Introducing Power Series 2 Looking at the graph of a power series foreshadows the idea of the interval of convergence.

Introducing Power Series 3 The Taylor Approximating Polynomial with examples of using a series to approximate.

Graphing Taylor Polynomials Graphing calculator hints.

 

 

 

 

 

Infinite Series

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The posts for the next several weeks will be on topics tested only on the BC Calculus exams. The first few weeks will be past posts on sequences and series. That will be followed, in February, by a discussion of Parametric and Vector functions, and Polar equation.

We begin with series.

Comparison Test Chart A table summarizing the convergence test required for the AP Calculus BC exam, their hypotheses and Conclusions (Updated 2-1-18)

What Conversion Test Should I Use? Part 1

What Conversion Test Should I Use? Part 2

Good Question 14 A riff on the Integral Test

Everyday series   decimals and roots

Amortization Calculating your mortgage payment – a practical and useful example of finite series

Revised 2-26-2018

 

 

 

Synthetic Summer Fun

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Today, for some summer fun, let’s look at synthetic division a/k/a synthetic substitution. I’ll assume you all know how to do that since it is a pretty common pre-calculus topic and even comes up again in calculus.

Why Does Synthetic Division Work?

An example: consider the polynomial

P(x)=2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+14x-1.

This can be written in nested form like this

P(x)=((((2x-3)x-11)x+14)x-1)

To evaluate this last expression at, say x = 2, we do the arithmetic as follows:

  1.   2 x 2 – 3 = 1
  2.   2 x 1 – 11 = –9
  3.   2 x (–9) + 14 = –4
  4.   2 x (–4) – 1 = – 9 = f(2)

Notice that this requires only multiplication and addition or subtraction, no raising to powers. More to the point, this is the same arithmetic, in the same order when you do the evaluation by synthetic division, and the work is a little easier to keep track of.

\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}

Synthetic division has another advantage: the other numbers in the second row are the coefficients of a quotient polynomial, a polynomial of one less degree that the original. So,

\displaystyle \frac{P(x)}{x-2}=2{{x}^{3}}+{{x}^{2}}-9x-4+\frac{-9}{x-2}

The Remainder Theorem and the Factor Theorem

In general, a polynomial of degree n, divided by a linear factor (x – a) gives a polynomial Q(x) of degree n – 1 and a remainder R

\displaystyle \frac{P(x)}{x-a}=Q(x)+\frac{R}{x-a}

Or

P(x)=Q(x)(x-a)+R

From here it is easy to see that P(a)=R. This is called the remainder theorem. It has a corollary called the factor theorem: If R = 0, then (x – a) is a factor of P(x).

Calculus

But wait there is more: differentiating the equation above using the product rule gives

{P}'(x)=Q(x)(1)+Q(x)(x-a)+0 and substituting x = a  gives

{P}'(a)=Q(a). The value of the quotient polynomial at a is the derivative of the original polynomial at a.

Of course, we could also rewrite the same equation as \displaystyle \frac{P(x)-P(a)}{x-a}=Q(x) . Then

\displaystyle {P}'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{P(x)-P(a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,Q(x)=Q(a)

Taylor Series

But wait, there’s even more.

A polynomial is a Maclaurin series in which all the terms after the nth term are zero. When you students are first learning how to write a Taylor series, by finding all the derivatives and substituting in the general term, a good exercise is to have them write the Taylor series for a polynomial centered away from the origin. For the example above:

P(x)=-9-2\left( x-2 \right)+19{{\left( x-2 \right)}^{2}}+13{{\left( x-2 \right)}^{3}}+2{{\left( x-2 \right)}^{4}}

Then ask them to expand the expression above and collect term etc. They should get the original polynomial again (and have some great practice expand powers of a binomial).

Can synthetic division help us? Yes, of course.  Here, is the original computation again:

\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}

If we ignore the –9 and divide the quotient numbers by 2 we get

\begin{matrix} {} & 2 & 1 & -9 & -4 \\ 2) & 2 & 5 & 1 & -2 \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(2) \\ \end{matrix}

\begin{matrix} {} & 2 & 5 & 1 \\ 2) & 2 & 9 & 19 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}'}'(2)}{2} \\ \end{matrix}

And again

\begin{matrix} {} & 2 & 9 \\ 2) & 2 & 13 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{{P}'}'}'\left( 2 \right)}{3!} \\ \end{matrix}

One more time

\begin{matrix} {} & 2 \\ 2) & 2 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}^{(4)}}\left( 2 \right)}{4!} \\ \end{matrix}

What do you see? Right, the last numbers in each computation, –9, –2, 19, 13, and 2, are the coefficients of the Taylor polynomial!

If you really want to dive this home and have some more summer fun here’s the start of a proof (at least for n = 4). Let

P(x)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{3}}+{{c}_{1}}x+{{c}_{0}} and divide this by a:

\begin{matrix} {} & {{c}_{4}} & {{c}_{3}} & {{c}_{2}} & {{c}_{1}} & {{c}_{0}} \\ a) & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{a}}a+{{c}_{1}} & {{c}_{4}}{{a}^{4}}+{{c}_{3}}{{a}^{3}}+{{c}_{a}}{{a}^{2}}+{{c}_{1}}a+{{c}_{0}}=P(a) \\ \end{matrix}

Again

\begin{matrix} {} & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{2}}a+{{c}_{1}} \\ a) & {{c}_{4}} & 2{{c}_{4}}a+{{c}_{3}} & 3{{c}_{4}}{{a}^{2}}+2{{c}_{3}}a+{{c}_{2}} & 4{{c}_{4}}{{a}^{3}}+3{{c}_{3}}{{a}^{2}}+2{{c}_{2}}a+{{c}_{1}} \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(a) \\ \end{matrix}

And I’ll leave the rest to you.  Really, why should I have all the fun?

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Sequences and Series (Type 10 for BC only)

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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.

On the free-response section there is usually one full question devoted to sequences and series. This question usually involves writing a Taylor or Maclaurin polynomial for a series.

Students should be familiar with and able to write several terms and the general term of a 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.
  • 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), ex 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 post of  February 22, 2013.
  • Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).
  • 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.

 

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.

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

Next Post

Friday April 7, 2017 The Domain of the solution of a differential equation.

 

 

Geometric Series – Far Out

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One of the great things – at least I like it – about the Taylor series of a function is that it is unique. There is only one Taylor series for any function centered at a given point, what that means is that any way you get it, it’s right.

Faced with writing the power series for, say, \displaystyle f\left( x \right)=\frac{3x}{1-2x}, instead of cranking out a bunch of derivatives, we can say this looks a lot like the formula for the sum of a geometric series,

\displaystyle \sum\limits_{k=1}^{\infty }{a{{r}^{k-1}}}=\frac{a}{1-r}. Taking  a = 3x and r = 2x, the series is

\displaystyle \frac{3x}{1-2x}=3x+6{{x}^{2}}+12{{x}^{3}}+24{{x}^{4}}+\cdots =3\cdot \sum\limits_{k=1}^{\infty }{{{2}^{k-1}}{{x}^{k}}}.

Furthermore, since a geometric series converges only when \left| r \right|<1, the interval of convergence for this series is \left| 2x \right|<1 or -\tfrac{1}{2}<x<\tfrac{1}{2} and we don’t even have to check the endpoints.

There are other choices as well.  We could write \displaystyle f\left( x \right)=3x{{\left( 1-2x \right)}^{-1}} and then expand the binomial using the binomial theorem. Or we could use the technique of long division of polynomials to divide 3x by (1 – 2x) – leaving the divisor as written here.

This works even in more complicated situations. Let \displaystyle g\left( x \right)=\frac{3x}{{{x}^{2}}-4}. Begin by dividing each term by –4. This gives \displaystyle g\left( x \right)=\frac{-\tfrac{3}{4}x}{1-\tfrac{1}{4}{{x}^{2}}}. Then treating this as a geometric series

\displaystyle g\left( x \right)=\sum\limits_{k=1}^{\infty }{-\tfrac{3}{4}x{{\left( \tfrac{1}{4}{{x}^{2}} \right)}^{k-1}}=-\frac{3}{4}x-\frac{3}{16}{{x}^{3}}-\frac{3}{64}{{x}^{5}}-\frac{3}{256}{{x}^{7}}-\cdots }

The interval of convergence is \displaystyle \left| \tfrac{1}{4}{{x}^{2}} \right|<1, or –2 < x < 2

Now the fun part

I once heard of a student making one of those great “mistakes.” For the series above, she divided by (–x2) and found that \displaystyle g\left( x \right)=\frac{-\frac{3}{x}}{1-\frac{4}{{{x}^{2}}}} and then wrote:

\displaystyle g\left( x \right)=\sum\limits_{k=1}^{\infty }{\left( -\frac{3}{x} \right){{\left( \frac{4}{{{x}^{2}}} \right)}^{k-1}}=-\frac{3}{x}-\frac{12}{{{x}^{3}}}-\frac{48}{{{x}^{5}}}-\frac{192}{{{x}^{7}}}-\cdots }

So, what’s wrong with that?

Nothing actually.

Okay, it’s not a Taylor Series since a Taylor series is allowed only non-negative exponents, but it’s still a geometric series. Let’s take a look at its interval of convergence: \displaystyle \left| \frac{4}{{{x}^{2}}} \right|<1, or \displaystyle \left| \frac{{{x}^{2}}}{4} \right|>1,  or the union of x>2 and x<-2, Whoa, that’s different and not even an interval.

The graph will make things clear (as usual):

geom-series

The original function graphed as a rational expression is shown in black. The Taylor polynomial (4 terms) is shown in blue; it approximates the function well between –2 and 2 as we should expect. The red graph is the student’s series (4 terms) and it is a good approximation of the series outside of the interval (–2, 2), far outside! Way Cool!

Of course, this kind of series is not studied in beginning calculus. It may make a good topic for a report or project for someone in your class.