# 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.

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 assessed 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

Nine of nine. We continue our look at the 2021 free-response questions. We will look at ways to adapt, expand, and explore this question to help students better understand it and look at other questions that can be asked based on a similar stem.

## 2021 BC 6

This is a Sequence and Series (Type 10) question. Typically the topic of the last question on the BC exam, it tests the concepts in Unit 10 of the current Calculus Course and Exam Description. This year the previous question, 2021 BC 5, asked students to write a Taylor Polynomial. This question covers other related topics: convergence tests, radius of convergence, and the error bound.

There is a nuance here. In past years students were not asked to give the conditions for a convergence test and were expected to determine which test to use for themselves. I think the idea here, and perhaps going forward (?), is to make sure the students have considered the conditions necessary to use a test. This is in keeping with other questions where the hypotheses of the theorem students were using had to be checked (Cf. recent L’Hospital’s Rule questions).

The Convergence Test Chart  and the posts “Which Convergence Test Should I use?” Part 1 and Part 2 may be helpful.

The stem for 2021 BC 6 is:

Part (a): Students were asked to give the conditions for the integral test and use it to determine if a different series, $\sum\limits_{{n=0}}^{\infty }{{\frac{1}{{{{e}^{n}}}}}}$ converges.

Discussion and ideas for adapting this question:

• Be sure your students know the conditions necessary for each convergence test. Phrase your questions as this one is phrased – at least sometimes.
• Ask students to state the conditions for any convergence they use.
• Discuss which tests (often plural) can be used for each series you study.
• Make sure students can decide for themselves which test to use in case next year’s questions do not tell them.
• Ask what other test(s) may be used with this series (Hint: the series is geometric). This is a question to ask for any series you study.

Part (b): Students are told to use the series from part (a) with the limit comparison test to show that the given series converges absolutely when x = 1. Again, students were asked to use a specific test. Notice that even if a student could not do part (a), they were not shut out of part (b).

Discussion and ideas for adapting this question:

• Since you cannot count on being told which test to use for comparison, be sure to discuss how to decide which test(s) can be used with each series. Again see “Which Convergence Test Should I use?” Part 1 and Part 2.
• Show students that proving absolute convergence is often a good way to eliminate the need for dealing with alternating series and other series with negative signs.

Part (c): Students were asked for the radius of convergence of the series. A standard question done by using the Ratio test.

Discussion and ideas for adapting this question:

The only extension here is to determine the interval of convergence, by checking the endpoints.

Part (d): Students were asked for the alternating series error bound using the first two terms to approximate the value of g(1). Even though there are only two error bounds students are expected to be able to compute (the other is the Lagrange error bound), students were again told which one to use. The result was not expected to be expressed as a decimal.

Discussion and ideas for adapting this question:

• First, have students check that the conditions for using the alternating series error bound are met.
• Increase the number of terms to be used.
• Ask students to find the Lagrange error bound and compare the results.

This post on the series question concludes the series of posts (pun intended) considering how to expand and adapt the 2021 AP Calculus free-response questions. I hope you found them helpful.

As always, I happy to hear your ideas for other ways to use this question. Please share your thoughts and ideas.

# A Curiosity

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.

# Good Question 16

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.