Unit 10: Infinite Sequences and Series

Syllabus
2020
Section
—
Level
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10.1 Defining Convergent and Divergent Infinite Series

Syllabus
2020
Topic
10.1
Level
—

A Series Converges When Its Partial Sums Settle

For the infinite series ∑k=1∞ak\sum_{k=1}^{\infty}a_k, the nnth partial sum SnS_n adds only the first nn terms. The infinite series converges to SS exactly when the sequence S1,S2,S3,…S_1,S_2,S_3,\ldots approaches the finite real number SS.

S_n=\sum_{k=1}^{n}a_k,\qquad \sum_{k=1}^{\infty}a_k=S\iff\lim_{n\to\infty}S_n=S

  1. Write a formula for the finite partial sum SnS_n.
  2. Simplify SnS_n before taking a limit.
  3. Evaluate lim⁡n→∞Sn\lim_{n\to\infty}S_n.
  4. If the limit is a finite real number, the series converges to it; if no finite limit exists, the series diverges.

Example: since 1/[k(k+1)]=1/k−1/(k+1)1/[k(k+1)]=1/k-1/(k+1),
Sn=∑k=1n1k(k+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1.S_n=\sum_{k=1}^{n}\frac{1}{k(k+1)}=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)=1-\frac1{n+1}.
Therefore lim⁡n→∞Sn=1\lim_{n\to\infty}S_n=1, so ∑k=1∞1/[k(k+1)]\sum_{k=1}^{\infty}1/[k(k+1)] converges and its sum is 11. By contrast, for ∑k=1∞1\sum_{k=1}^{\infty}1, Sn=n→∞S_n=n\to\infty, so the series diverges.

A convergent series still contains infinitely many terms; convergence does not mean the adding stops. It means the finite partial sums approach one finite value. Writing that a divergent series has a sum of ∞\infty is not the same as convergence to a real number.

10.2 Working with Geometric Series

Syllabus
2020
Topic
10.2
Level
—

A Geometric Series Converges Only When Its Ratio Shrinks

A geometric series has the form a+ar+ar2+⋯a+ar+ar^2+\cdots, so dividing any term after the first by its preceding nonzero term gives the same common ratio rr. For a nonzero series, its terms shrink toward zero exactly when ∣r∣<1|r|<1.

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad\text{when }|r|<1;\qquad |r|\ge1\Rightarrow\text{divergence}

The finite partial sum through exponent NN is SN=a(1−rN+1)/(1−r)S_N=a(1-r^{N+1})/(1-r) for r≠1r\ne1. When ∣r∣<1|r|<1, rN+1→0r^{N+1}\to0, leaving a/(1−r)a/(1-r). If ∣r∣≥1|r|\ge1, that power does not approach zero, so the partial sums do not approach a finite value.

Example: 6−3+32−34+⋯6-3+\tfrac32-\tfrac34+\cdots is geometric because each term is the previous term multiplied by −1/2-1/2. Here a=6a=6 and r=−1/2r=-1/2. Since ∣−1/2∣<1|-1/2|<1, the series converges, and
S=61−(−1/2)=63/2=4.S=\frac{6}{1-(-1/2)}=\frac{6}{3/2}=4.
The alternating signs do not prevent convergence because the term magnitudes shrink.

Use the actual first term of the displayed series as aa. For example, ∑n=1∞3(1/2)n\sum_{n=1}^{\infty}3(1/2)^n begins with 3/23/2, not 33; either reindex from zero or use a=3/2a=3/2. Also, r=−1r=-1 alternates but does not shrink, so it diverges.

10.3 The nth Term Test for Divergence

Syllabus
2020
Topic
10.3
Level
—

If the Terms Do Not Approach Zero, the Series Diverges

If ∑an\sum a_n converges, adding the next term must eventually make almost no change to its partial sum. Because an=Sn−Sn−1a_n=S_n-S_{n-1}, convergence of the partial sums forces an→0a_n\to0. Therefore any failure of the terms to approach zero proves divergence.

\lim_{n\to\infty}a_n\ne0\text{ or does not exist}\Rightarrow\sum_{n=1}^{\infty}a_n\text{ diverges};\qquad \lim_{n\to\infty}a_n=0\Rightarrow\text{inconclusive}

Example 1: for ∑n=1∞n/(n+1)\sum_{n=1}^{\infty}n/(n+1),
lim⁡n→∞nn+1=1≠0.\lim_{n\to\infty}\frac{n}{n+1}=1\ne0.
The terms keep contributing about 11, so the nth term test proves that the series diverges. No other convergence test is needed.

Example 2: for ∑n=1∞1/n\sum_{n=1}^{\infty}1/n, the term limit is 00. The nth term test is inconclusive: it does not say the series converges. A different test is required to determine its behavior.

The test is designed to prove divergence, not convergence. The statement “a convergent series has an→0a_n\to0” is true, but its converse is false: terms approaching zero is necessary, not sufficient, for the accumulated sum to approach a finite value.

10.4 Integral Test for Convergence

Syllabus
2020
Topic
10.4
Level
—

Match a Positive Decreasing Series to an Improper Integral

Suppose an=f(n)a_n=f(n) and, for all x≥Nx\ge N, the function ff is continuous, positive, and decreasing. Then the terms can be compared with adjacent strips under the graph, so the infinite series and corresponding improper integral either both converge or both diverge.

\sum_{n=N}^{\infty}a_n\text{ converges}\iff\int_N^{\infty}f(x),dx\text{ converges},\qquad a_n=f(n)

  1. Choose f(x)f(x) so that f(n)=anf(n)=a_n.
  2. State where ff is continuous, positive, and decreasing.
  3. Rewrite ∫N∞f(x) dx\int_N^{\infty}f(x)\,dx as a limit and evaluate it.
  4. Transfer only the convergence or divergence conclusion to the series.

Example: for ∑n=1∞1/(n2+1)\sum_{n=1}^{\infty}1/(n^2+1), let f(x)=1/(x2+1)f(x)=1/(x^2+1). For x≥1x\ge1, ff is continuous and positive, and f′(x)=−2x/(x2+1)2<0f'(x)=-2x/(x^2+1)^2<0, so it is decreasing. Also,
∫1∞dxx2+1=lim⁡b→∞[arctan⁡x]1b=π2−π4=π4.\int_1^{\infty}\frac{dx}{x^2+1}=\lim_{b\to\infty}[\arctan x]_1^b=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.
The improper integral converges, so the series converges by the integral test.

The value π/4\pi/4 in the example is the integral's value, not the sum of the series. The integral test determines convergence behavior; it does not normally calculate the exact series sum. If positivity or decreasing behavior fails, this test has not been justified.

10.5 Harmonic Series and p-Series

Syllabus
2020
Topic
10.5
Level
—

LIM-7.A—Determine whether a series converges or diverges—Topic 10.5

  • LIM-7.A Determine whether a series converges or diverges.
  • LIM-7.A.7 In addition to geometric series, common series of numbers include the harmonic series, the alternating harmonic series, and p-series.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.6 Comparison Tests for Convergence

Syllabus
2020
Topic
10.6
Level
—

LIM-7.A—Determine whether a series converges or diverges—Topic 10.6

  • LIM-7.A Determine whether a series converges or diverges.
  • LIM-7.A.8 The comparison test is a method to determine whether a series converges or diverges.
  • LIM-7.A.9 The limit comparison test is a method to determine whether a series converges or diverges.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.7 Alternating Series Test for Convergence

Syllabus
2020
Topic
10.7
Level
—

LIM-7.A—Determine whether a series converges or diverges—Topic 10.7

  • LIM-7.A Determine whether a series converges or diverges.
  • LIM-7.A.10 The alternating series test is a method to determine whether an alternating series converges.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.8 Ratio Test for Convergence

Syllabus
2020
Topic
10.8
Level
—

LIM-7.A—Determine whether a series converges or diverges—Topic 10.8

  • LIM-7.A Determine whether a series converges or diverges.
  • LIM-7.A.11 The ratio test is a method to determine whether a series of numbers converges or diverges.
    • Exclusion statement: The nth term test for divergence, and the integral test, comparison test, limit comparison test, alternating series test, and ratio test for convergence are assessed on the AP Calculus BC Exam. Other methods are not assessed on the exam. However, teachers may include additional methods in the course, if time permits.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.9 Determining Absolute or Conditional Convergence

Syllabus
2020
Topic
10.9
Level
—

LIM-7.A—Determine whether a series converges or diverges—Topic 10.9

  • LIM-7.A Determine whether a series converges or diverges.
  • LIM-7.A.12 A series may be absolutely convergent, conditionally convergent, or divergent.
  • LIM-7.A.13 If a series converges absolutely, then it converges.
  • LIM-7.A.14 If a series converges absolutely, then any series obtained from it by regrouping or rearranging the terms has the same value.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.10 Alternating Series Error Bound

Syllabus
2020
Topic
10.10
Level
—

LIM-7.B—Approximate the sum of a series

  • LIM-7.B Approximate the sum of a series.
  • LIM-7.B.1 If an alternating series converges by the alternating series test, then the alternating series error bound can be used to bound how far a partial sum is from the value of the infinite series.
  • Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

10.11 Finding Taylor Polynomial Approximations of Functions

Syllabus
2020
Topic
10.11
Level
—

LIM-8.A—Represent a function at a point as a Taylor polynomial

  • LIM-8.A Represent a function at a point as a Taylor polynomial.
  • LIM-8.A.1 The coefficient of the nth degree term in a Taylor polynomial for a function f centered at x = a is f⁽ⁿ⁾(a)/n!.
  • LIM-8.A.2 In many cases, as the degree of a Taylor polynomial increases, the nth degree polynomial will approach the original function over some interval.
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

LIM-8.B—Approximate function values using a Taylor polynomial

  • LIM-8.B Approximate function values using a Taylor polynomial.
  • LIM-8.B.1 Taylor polynomials for a function f centered at x = a can be used to approximate function values of f near x = a.
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

10.12 Lagrange Error Bound

Syllabus
2020
Topic
10.12
Level
—

LIM-8.C—Determine the error bound associated with a Taylor polynomial approximation

  • LIM-8.C Determine the error bound associated with a Taylor polynomial approximation.
  • LIM-8.C.1 The Lagrange error bound can be used to determine a maximum interval for the error of a Taylor polynomial approximation to a function.
  • LIM-8.C.2 In some situations, the alternating series error bound can be used to bound the error of a Taylor polynomial approximation to the value of a function.
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

10.13 Radius and Interval of Convergence of Power Series

Syllabus
2020
Topic
10.13
Level
—

LIM-8.D—Determine the radius of convergence and interval of convergence for a power series

  • LIM-8.D Determine the radius of convergence and interval of convergence for a power series.
  • LIM-8.D.1 A power series is a series of the form Σ(n=0 to ∞) aₙ(x − r)ⁿ, where n is a non-negative integer, {aₙ} is a sequence of real numbers, and r is a real number.
  • LIM-8.D.2 If a power series converges, it either converges at a single point or has an interval of convergence.
  • LIM-8.D.3 The ratio test can be used to determine the radius of convergence of a power series.
  • LIM-8.D.4 The radius of convergence of a power series can be used to identify an open interval on which the series converges, but it is necessary to test both endpoints of the interval to determine the interval of convergence.
  • LIM-8.D.5 If a power series has a positive radius of convergence, then the power series is the Taylor series of the function to which it converges over the open interval.
  • LIM-8.D.6 The radius of convergence of a power series obtained by term-by-term differentiation or term-by-term integration is the same as the radius of convergence of the original power series.
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

10.14 Finding Taylor or Maclaurin Series for a Function

Syllabus
2020
Topic
10.14
Level
—

LIM-8.E—Represent a function as a Taylor series or a Maclaurin series

  • LIM-8.E Represent a function as a Taylor series or a Maclaurin series.
  • LIM-8.E.1 A Taylor polynomial for f(x) is a partial sum of the Taylor series for f(x).
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

LIM-8.F—Interpret Taylor series and Maclaurin series

  • LIM-8.F Interpret Taylor series and Maclaurin series.
  • LIM-8.F.1 The Maclaurin series for 1/(1 − x) is a geometric series.
  • LIM-8.F.2 The Maclaurin series for sin x, cos x, and eˣ provides the foundation for constructing the Maclaurin series for other functions.
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.

10.15 Representing Functions as Power Series

Syllabus
2020
Topic
10.15
Level
—

LIM-8.G—Represent a given function as a power series

  • LIM-8.G Represent a given function as a power series.
  • LIM-8.G.1 Using a known series, a power series for a given function can be derived using operations such as term-by-term differentiation or term-by-term integration, and by various methods (e.g., algebraic processes, substitutions, or using properties of geometric series).
  • Enduring understanding LIM-8: Power series allow us to represent associated functions on an appropriate interval.