# 2019 CED Unit 10: Infinite Sequences and Series

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.

### 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

# Which Convergence Test Should I Use? Part 1

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

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.)

# Geometric Series – Far Out

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} 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):

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.

# Amortization

This is an example of how sequences and series are used in “real life.”  It could be used in a calculus class or an advanced math class.

When you borrow money to buy a house or a car you are making a kind of loan called a mortgage. Paying off the loan is called amortizing the loan.

The amount you borrow is called the principal. You agree to pay a fixed payment at regular intervals, usually monthly. The payment includes interest on the amount owed during the previous month plus a bit more to decrease the principal. Each month your payment decreases the remaining principal, so in the next month a little less goes to interest and a little more goes to paying off the principal. Let’s say you borrow A dollars (the principal) and agree to pay a monthly interest rate of r%, for a period of n months.

How much is the fixed (constant) monthly payment, P, so that after n months the loan and interest will be paid off?

After one month you make a payment and you now owe ${{A}_{1}}$ = the original amount plus the interest on this amount minus the payment, P.

${{A}_{1}}=A+Ar-P=A\left( 1+r \right)-P$

After the second payment is made you owe ${{A}_{2}}$ =  ${{A}_{1}}$, plus the interest ${{A}_{1}r}$, minus the payment, P.

${{A}_{2}}=\left( A\left( 1+r \right)-P \right)+\left( A\left( 1+r \right)-P \right)r-P=A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$

And likewise after the third month:

${{A}_{3}}=\left( A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P \right)+\left( A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P \right)r-P$

${{A}_{3}}=A{{\left( 1+r \right)}^{3}}-P{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$

And after the fourth month a pattern emerges:

${{A}_{4}}=A{{\left( 1+r \right)}^{4}}-P{{\left( 1+r \right)}^{3}}-P{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$

${{A}_{4}}=A{{\left( 1+r \right)}^{4}}-P\sum\limits_{k=0}^{3}{{{\left( 1+r \right)}^{k}}}$

And so on so that after n months the amount you owe is:

${{A}_{n}}=A{{\left( 1+r \right)}^{n}}-P\sum\limits_{k=0}^{n-1}{{{\left( 1+r \right)}^{k}}}$

The sigma expression represents a finite geometric series. The sum of such a series with a first term of 1 and a common ratio of R sum is given by

$\displaystyle {{S}_{n}}=\frac{1-{{R}^{n}}}{1-R}$.

Therefore

$\displaystyle {{A}_{n}}=A{{\left( 1+r \right)}^{n}}-P\left( \frac{1-{{\left( 1+r \right)}^{n}}}{1-\left( 1+r \right)} \right)=A{{\left( 1+r \right)}^{n}}-P\left( \frac{{{\left( 1+r \right)}^{n}}-1}{r} \right)$

But after n months you have paid off the loan and you owe An = 0

$\displaystyle 0=A{{\left( 1+r \right)}^{n}}-P\left( \frac{{{\left( 1+r \right)}^{n}}-1}{r} \right)$

This equation may be solved for P the monthly payment:

$\displaystyle P=\frac{Ar{{\left( 1+r \right)}^{n}}}{{{\left( 1+r \right)}^{n}}-1}$.

The equation above is a formula for finding the payment on a mortgage. For example, to borrow \$25,000 for a car at 3% per year (r = 0.0025 per month) for 5 years (n = 60) the monthly payment is

$\displaystyle \frac{25,000\left( 0.0025 \right){{\left( 1+0.0025 \right)}^{60}}}{{{\left( 1+0.0025 \right)}^{60}}-1}=\449.22$.

Of course, in “real life” someone looks up the amount on the internet and we all believe them. But now you can do it yourself and have some math fun at the same time.

# New Series from Old 3

Rational Functions and a “mistake”

A geometric series is one in which each term is found by multiplying the preceding term by the same number or expression. This number is called the common ratio, r. Geometric series converge if, and only if, $\left| r \right|<1$. If a geometric series converges, then the sum of the (infinite number of) terms is $\displaystyle \frac{{{a}_{1}}}{1-r}$ where a1 is the first term.

We can use this to write series for rational expressions.

Example 1. The series for $\displaystyle \frac{1}{1+{{x}^{2}}}$ that we assumed in the last post can be rewritten as $\displaystyle \frac{1}{1-\left( -{{x}^{2}} \right)}$.   This has the same form as the sum of the geometric series so we can write it as a geometric series with a1 = 1 and r = –x2 . The result is

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

Example 2: $\displaystyle \frac{6x}{2+x}=\frac{2x}{1-\left( -\tfrac{x}{2} \right)}$. Letting ${{a}_{1}}=3x$ and $r=-\tfrac{x}{2}$ we have

$\displaystyle \frac{6x}{2+x}=\frac{3x}{1-\left( -\tfrac{x}{2} \right)}=3x+3x\left( -\tfrac{x}{2} \right)+3x{{\left( -\tfrac{x}{2} \right)}^{2}}+3x{{\left( -\tfrac{x}{2} \right)}^{3}}+\cdots$

$\displaystyle \frac{6x}{2+x}=3x-\tfrac{3}{2}{{x}^{2}}+\tfrac{3}{4}{{x}^{3}}-\tfrac{3}{8}{{x}^{4}}+\tfrac{3}{16}{{x}^{5}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n-1}}}{{{2}^{n-1}}}3{{x}^{n}}+\cdots$

The interval of convergence is $\left| -\tfrac{x}{2} \right|<1$  or $-2.

Many power series for rational functions can be obtained in this way.

An instructive “mistake.”

I have heard of students making an interesting mistake with this kind of problem. Instead of dividing by 2, they divide by x and arrive at $\displaystyle \frac{6x}{2+x}=\frac{6}{1-\left( -\tfrac{2}{x} \right)}$ and then write the series as

$\displaystyle \frac{6x}{2+x}=\frac{6}{1-\left( -\tfrac{2}{x} \right)}=6+6\left( -\tfrac{2}{x} \right)+6{{\left( -\tfrac{2}{x} \right)}^{2}}+6{{\left( -\tfrac{2}{x} \right)}^{3}}+\cdots$

$\displaystyle \frac{6x}{2+x}=6-\frac{12}{x}+\frac{24}{{{x}^{2}}}-\frac{48}{{{x}^{3}}}+\frac{96}{{{x}^{4}}}-\cdots$

With an interval of convergence of $\left| -\tfrac{2}{x} \right|<1$ or $x<-2\text{ or }x>2$. Now this is not a Taylor series since the powers are in the denominators, but it is nevertheless interesting. Let’s look at the graphs.

The rational expression is the black graph and partly hidden by the other graphs where they converge. The blue graph is the 5th degree Taylor Polynomial and its interval of convergence is the white strip in the center of the graph. The red graph is the “mistake” (also 5th degree) with the two blue regions as its interval of convergence.  Both series fit the rational function but only in their own interval of convergence.

# New Series from Old 2

Differentiating and integrating a known series can help you find other series. Since $\frac{d}{dx}\sin \left( x \right)=\cos \left( x \right)$ we can find the series for cos(x) this way

$\frac{d}{dx}\sin \left( x \right)=\frac{d}{dx}(x-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-\frac{{{x}^{7}}}{7!}+\cdots +\frac{{{\left( -1 \right)}^{n-1}}{{x}^{2n-1}}}{\left( 2n-1 \right)!}+\cdots )$

$\frac{d}{dx}\sin \left( x \right)=1-\frac{3{{x}^{2}}}{3!}+\frac{5{{x}^{4}}}{5!}-\frac{7{{x}^{6}}}{7!}+\cdots \frac{{{\left( -1 \right)}^{n-1}}\left( 2n-1 \right){{x}^{2n}}}{\left( 2n-1 \right)!}+\cdots$

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

Of course, we could also have integrated the series for sin(x) to get the series for –cos(x) and then changed the signs.

In our next post we will find that

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

Recall that $\frac{d}{dx}\arctan \left( x \right)=\frac{1}{1+{{x}^{2}}}$, so we can integrate the series above to find the series for arctan(x).

$\arctan \left( x \right)=\int_{{}}^{{}}{(1-{{x}^{2}}+{{x}^{4}}-{{x}^{6}}+\cdots +{{\left( -1 \right)}^{2n-1}}{{x}^{2n-2}}+\cdots )dx}$

$\arctan \left( x \right)=C+x-\tfrac{1}{3}{{x}^{3}}+\tfrac{1}{5}{{x}^{5}}-\tfrac{1}{7}{{x}^{7}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n+1}}}{2n-1}{{x}^{2n-1}}+\cdots$

Since $\arctan \left( 0 \right)=0$ it follows that the constant  of integration $C=0$ so

$\arctan \left( x \right)=x-\tfrac{1}{3}{{x}^{3}}+\tfrac{1}{5}{{x}^{5}}-\tfrac{1}{7}{{x}^{7}}+\cdots +\tfrac{{{\left( -1 \right)}^{2n+1}}}{2n-1}{{x}^{2n-1}}+\cdots$

When differentiating or integrating the interval of convergence cannot get any larger, but it can get smaller. If the endpoints are included before differentiating or integrating then you must check to see if they are included in the new series.

Next post: Rational functions and geometric series.