Series (For Calculus BC Students Only) - Review the Knowledge You Need to Score High

5 Steps to a 5 AP Calculus AB & BC, 2012-2013 Edition (2011)

STEP 4. Review the Knowledge You Need to Score High

Chapter 14. Series (For Calculus BC Students Only)

IN THIS CHAPTER

Summary: In this chapter, you will learn different types of series and the many tests for determining whether they converge or diverge. The types of series include geometric series, p-Series, alternating series, power, and Taylor series. The tests for convergence include the integral test, the comparison test, the limit comparison test, and the ratio test for absolute convergence. These tests involve working with algebraic expressions and lengthy computations. It is important that you carefully work through the practice problems provided in the chapter, and check your solutions with the given explanations.

Key Ideas

Sequences and Series

Geometric Series, Harmonic Series, and p-Series

Convergence Tests

Alternating Series and Absolute Convergence

Power Series

Taylor Series

Operations on Series

Error Bounds

14.1 Sequences and Series

Main Concepts: Sequences and Series, Convergence

A sequence is a function whose domain is the non-negative integers. It can be expressed as a list of terms {a_n} = {a₁, a₂, a₃, …, a_n, …} or by a formula that defines the nth term of the sequence for any value of n. A series is the sum of the terms of a sequence {a_n}. Associated with each series is a sequence of partial sums, {s_n}, where s₁ = a₁, s₂ = a₁ + a₂, s₃ = a₁ + a₂ + a₃, and in general, s_n = a₁+ a₂ + a₃ + … + a_n.

Example 1

Find the first three partial sums of the series .

Step 1: Generate the first three terms of the sequence .

, ,

Step 2: Find the partial sums.

Example 2

Find the fifth partial sum of the series .

Step 1: Generate the first five terms of the sequence .

Step 2: The fifth partial sum is .

Convergence

The series Σa_n converges if the sequence of associated partial sums, {s_n}, converges. The limit , where S is a real number, and is the sum of series, . If and are convergent, then and .

Example 1

Determine whether the series converges or diverges. If it converges, find its sum.

Step 1: Find the first few partial sums.

Step 2: The sequence of partial sums {0.2, 0.24, 0.248, 0.2496, …} converges to 0.25, so the series converges, and its sum .

Example 2

Find the sum of the series , given that and .

Step 1:

Step 2:

14.2 Types of Series

Main Concepts: p-Series, Harmonic Series, Geometric Series, Decimal Expansion

p-Series

The p-series is a series of the form . The p-series converges when p > 1, and diverges when 0 < p ≤ 1.

Harmonic Series

The harmonic series is a p-series with p = 1. The harmonic series diverges.

Geometric Series

A geometric series is a series of the form where a ≠ 0. A geometric series converges when |r| < 1. The sum of the first n terms of a geometric series is . The sum of the series .

Example 1

Determine whether the series converges. is a geometric series with a = 1 and . Since r > 1, the series diverges.

Example 2

Find the tenth partial sum of the series .

While it is possible to extend the terms of the series and directly compute the tenth partial sum, it is quicker to recognize that this is a geometric series. The ratio of any two subsequent terms is and the first term is a = 12.

Example 3

Find the sum of the series Since is a geometric series with a = 12 and , .

Decimal Expansion

The rational number equal to the repeating decimal is the sum of the geometric series that represents the repeating decimal.

Example

Find the rational number equivalent to .

Step 1:

Step 2: is a geometric series with and . The sum of the series is .

Step 3:

14.3 Convergence Tests

Main Concepts: Integral Test, Ratio Test, Comparison Test, Limit Comparison Test

Integral Test

If a_n = f(n) where f is a continuous, positive, decreasing function on [c, ∞), then the series is convergent if and only if the improper integral exists.

Example 1

Determine whether the series converges or diverges.

Step 1: is continuous, positive, and decreasing on the interval [2, ∞).

Step 2: . The improper integral exists so converges.

Example 2

Determine whether the series converges or diverges.

Step 1: is continuous, positive, and decreasing on the interval [1, ∞).

Step 2:

Since the improper integral exists, the series converges.

Example 3

Determine whether the series converges or diverges.

Step 1: is continuous, positive, and decreasing on [1, ∞).

Step 2: . Since the improper integral does not converge, the series diverges.

Example 4

Determine whether the series converges or diverges.

Step 1: is continuous, positive, and decreasing on [1, ∞).

Step 2: Since the improper integral does not converge, the series diverges.

Ratio Test

If is a series with positive terms and , then the series converges. If the limit is greater than 1 or is ∞, the series diverges. If the limit is 1, another test must be used.

Example 1

Determine whether the series converges or diverges.

Step 1: For all n ≥ 1, is positive.

Step 2: Since this limit is less than 1, the series converges.

Example 2

Determine whether the series converges or diverges.

Step 1: For all n ≥ 1, is positive.

Step 2: Since this limit is greater than one, the series diverges.

Comparison Test

Suppose and are series with non-negative terms, and is known to converge. If a term-by-term comparison shows that for all n, a_n ≤ b_n, then converges. If diverges, and if for all n, a_n ≥ b_n, then diverges. Common series that may be used for comparison include the geometric series, which converges for r < 1 and diverges for r ≥ 1, and the p-series, which converges for p > 1 and diverges for p ≤ 1.

Example 1

Determine whether the series converges or diverges.

Step 1: Choose a series for comparison. The series can be compared to , the p-series with p = 2. Both series have non-negative terms.

Step 2: A term-by-term comparison shows that for all values of n.

Step 3: converges, so converges.

Example 2

Determine whether the series converges or diverges.

Step 1: The series can be compared to .

Step 2: , so for all n ≥ 1.

Step 3: diverges, also diverges.

Limit Comparison Test

If and are series with positive terms, and if where 0 < L < ∞, then either both series converge or both diverge. By choosing, for one of these, a series that is known to converge, or known to diverge, you can determine whether the other series converges or diverges. Choose a series of a similar form so that the limit expression can be simplified.

Example 1

Determine whether the series converges or diverges. The given series . Choose for comparison. Since it has a similar structure, and we know it is a p-series with p = 1, it diverges. The . The limit exists and is greater than zero; therefore, since diverges, also diverges.

Example 2

Determine whether the series converges or diverges.

Compare to the known convergent . The . and converges, also converges.

14.4 Alternating Series

Main Concepts: Alternating Series, Error Bound, Absolute Convergence

A series whose terms alternate between positive and negative are called alternating series. Alternating series have one of two forms: or . An alternating series converges if and .

Example 1

Determine whether the series converges or diverges.

Step 1:

Step 2: Note that , and in general, , since multiplying by eⁿ⁺¹ gives en > n + 1.

Step 3: , .14936, .07326, …}, so . Therefore the series converges.

Example 2

Determine whether the series converges or diverges. If it converges, find its sum.

Step 1: is a geometric series with a = 4 and . Since r < 1, the series converges.

Step 2:

Error Bound

If an alternating series converges to the sum S, then S lies between two consecutive partial sums of the series. If S is approximated by a partial sum s_n, the absolute error |S − s_n| is less than the next term of the series a_n_{+ 1}, and the sign of S − s_n is the same as the coefficient of a_n₊₁.

Example

converges to 3.2. This value is greater than s_n for n odd, and less than s_n for n even. If S is approximated by the third partial sum, s3 = 3.25, the absolute error |S − s₃| = |3.2−3.25| = |−0.05| = 0.05, which is clearly less than . The coefficient of a₄ is negative, as is S − s₃.

Absolute Convergence

If the series converges, then converges. A series that converges absolutely converges.

Example

Determine whether the series converges.

Step 1: Consider the series . For this series, s₁ = 1, , , . The sequence of partial sums, , converges to 1.5. Or, note that this is a geometric series with a = 1, ; thus it converges to .

Step 2: converges, converges.

14.5 Power Series

Main Concepts: Power Series, Radius and Interval of Convergence

A power series is a series of the form , where c₁, c₂, c₃, … are constants, and x is a variable.

Radius and Interval of Convergence

A power series centered at x = a converges only for x = a, for all real values of x, or for all x in some open interval (a − R, a + R), called the interval of convergence. The radius of convergence is R. If the series converges on (a − R, a+ R), then it diverges if x < a − R or x > a + R: but convergence or divergence much be investigated individually at x = a − R and at x = a + R.

Example 1

Find the values of x for which the series converges.

Step 1: Use the ratio test.

Step 2: The series converges absolutely, so converges for all real x.

Example 2

Find the interval of convergence for the series .

Step 1: Use the ratio test.

Step 2: The series converges absolutely when |x − 2| < 1, −1 < x − 2 < 1, or 1 < x < 3.

Step 3: When x = 1, the series becomes . Since 1 > and , this alternating series converges. When x = 3, the series becomes which is a p-series with p = 1, and therefore diverges. Therefore, the interval of convergence is [1, 3).

14.6 Taylor Series

Main Concepts: Taylor Series and MacLaurin Series, Common MacLaurin Series

Taylor Series and MacLaurin Series

A Taylor polynomial approximates the value of a function f(x) at the point x = a. If the function and all its derivatives exist at x = a, then on the interval of convergence, the Taylor series converges to f(x). The MacLaurin series is the name given to a Taylor series centered at x = 0.

Example 1

Find the Taylor polynomial of degree 3 for about the point x = 3.

Step 1: Differentiate: , , .

Step 2: Evaluate: , , , .

Step 3:

Step 4:

Example 2

A function f(x) is approximated by the third order Taylor series 1 + 2(x − 1) − (x − 1)² + (x − 1)³ centered at x = 1. Find f′(1) and f′″(1).

Step 1: Compare to the given polynomial: , , , and .

Step 2: f′(1) = 2 · 1! = 2 and f′″(1) = 1 · 3! = 6.

Example 3

Find the MacLaurin polynomial of degree 4 that approximates f(x) = ln(1 + x).

Step 1: Differentiate: , , , .

Step 2: Evaluate: f(0) = 0, f′(0) = 1, f″(0) = −1, f′″(0) = 2, f⁽⁴⁾(0) = −6.

Step 3:

Example 4

Find the Taylor series for the function f(x) = e^−x about the point x = ln 2.

Step 1: f⁽ⁿ⁾(x) = e^−x when n is even and f⁽ⁿ⁾(x) = − e^−x when n is odd.

Step 2: Evaluate when n is even and when n is odd.

Step 3:

Example 5

Find the MacLaurin series for the function f(x) = xe^x.

Step 1: Investigating the first few derivatives of f(x) = xe^x shows that f⁽ⁿ⁾(x) = xe^x + ne^x.

Step 2: Evaluating f⁽ⁿ⁾(x) = xe^x + ne^x at x = 0 gives f⁽ⁿ⁾(0) = n.

Step 3:

Common MacLaurin Series

MacLaurin Series for the Functions e^x, sin x, cos x, and

Familiarity with these common MacLaurin series will simplify many problems.

14.7 Operations on Series

Main Concepts: Substitution, Differentiation and Integration, Error Bounds

Substitution

New series can be generated by making an appropriate substitution in a known series.

Example 1

Find the MacLaurin series for .

Step 1: Begin with the known series .

Step 2: Substitute − x² for x.

Example 2

Find the first four nonzero terms of the MacLaurin series for f(x) = cos(2x).

Step 1: Begin with the known series

Step 2: Substitute 2x for x.

Differentiation and Integration

If a function f(x) is represented by a Taylor series with a non-zero radius of convergence, the derivative f′(x) can be found by differentiating the series term by term. If the series is integrated term-by-term, the resulting series converges to . In either case, the radius of convergence is identical to that of the original series.

Example 1

Differentiate the MacLaurin series for f(x) = ln(x + 1) to find the Taylor series expansion for .

Step 1:

Step 2:

Example 2

Find the MacLaurin series for .

Step 1: We know that .

Step 2: Differentiate: .

Step 3: Multiply by − 1.

Example 3

Use MacLaurin series to approximate the integral to three decimal place accuracy.

Step 1: Substitute x² for x in the MacLaurin series representing sin x.

Step 2:

Step 3: For this alternating series, the absolute error for the nth partial sum is less than the n + 1 term so . We want three decimal place accuracy, so we need or 2000 ≤ (4n + 4)(2n + 2)! This occurs for n ≥ 2.

Step 4: Taking the sum , .

Error Bounds

The remainder, R_n(x), for a Taylor series is the difference between the actual value of the function f(x) and the nth partial sum that approximates the function. If the function f(x) can be differentiated n + 1 times on an interval containing x₀, and if |f⁽ⁿ^{+ 1)}(x)| ≤ M for all x in that interval, then for all x in the interval.

Example 1

Approximate accurate to three decimal places.

Step 1: Substitute for x in the MacLaurin series representation for e^x.

Step 2: For three decimal place accuracy, we want to find the value of n for which the remainder is less than or equal to 0.0005. Choose x₀ = 1 in the interval (0, 1). All derivatives of e^x are equal to e^x, and therefore |f⁽ⁿ^{+ 1)}(x)| ≤ e, so M = e. and when n ≥ 4.

Step 3:

Example 2

Estimate sin 4° accurate to five decimal places.

Step 1: radians. Substitute for x in the MacLaurin series that represents

Step 2: For five decimal place accuracy, we must find the value of n for which the absolute error is less than or equal to 5 × 10⁻⁶. For all x, |f⁽ⁿ^{+ 1)}(x)| ≤ 1. Choose x₀ = 0. . The absolute error is less than or equal to 5 × 10⁻⁶ for n ≥ 3.

Step 3:

14.8 Rapid Review

1. Find the sum of the series .

Answer: This is a geometric series with first term 81 and a ratio of , so .

2. Determine whether the series converges or diverges.

Answer: converges by ratio test.

3. Determine whether the series converges or diverges.

Answer: for is a p-series with p < 1, so it diverges.

and diverges by comparison to , so the series diverges.

4. Determine whether the series converges or diverges.

Answer: and diverges, so diverges by limit comparison wit .

5. Determine whether the series converges or diverges.

6. Find the interval to convergence for the series .

Answer: when − 1 < x + 1 < 1 or − 2 ≤ x < 0, the series converges on (−2, 0). When x = −2, the series becomes and , this alternating series converges. When x = 0, the series becomes , which is a p-series with and therefore diverges. Thus, the interval of convergence is [−2, 0).

7. Approximate the function with a fourth degree Taylor polynomial centered at x = 3.

8. Find the MacLaurin series for the function f(x) = e^−x and determine its interval of convergence.

Answer: , substitute − x to find . The ratio , so the series converges on the interval (− ∞, ∞).

14.9 Practice Problems

Determine whether each series converges or diverges.

9. Find the sum of the geometric series .

10. If the sum of the alternating series is approximated by s₅₀, find the maximum absolute error.

Find the interval of convergence for each series.

11.

12.

13. The Taylor series representation of ln x, centered at x = a.

Approximate each function with a fourth degree Taylor polynomial centered at the given value of x.

14. f(x) = at x = 1.

15. f(x) = cos πx at .

16. f(x) = ln x at x = e.

Find the MacLaurin series for each function and determine its interval of convergence.

17.

18.

19. Estimate sin 9° accurate to three decimal places.

20. Find the rational number equivalent to 1..

14.10 Cumulative Review Problems

21. The movement of an object in the plane is defined by x(t) = ln t, y(t) = t². Find the speed of the object at the moment when the acceleration is a(t) = ⟨− 1, 2⟩.

22. Find the slope of the tangent line to the curve r = 5 cos 3θ when .

23.

24.

25.

14.11 Solutions to Practice Problems

1. is a geometric series with an initial term of one and a ratio of . Since the ratio is less than one, the series converges, and .

2. By the ratio test, , therefore converges.

3. Consider the integral . Integrate by parts, with u = x, du = dx, du = e^−x dx, and v = −e^−x. . Therefore, . Since the improper integral converges, converges.

4. Use the limit comparison test, comparing to the series , which is known to diverge. Divide . The limit , and diverges, so the series diverges.

5. Use the ratio test, , so the series converges absolutely.

6. Use the ratio test. , therefore, converges.

8. Compare to the p-series with p = 2, . This series, , is term-by-term less than or equal to the p-series with p = 2. Since that p-series converges, converges.

9. The sum of the geometric series is .

10. For the alternating series approximated by s₅₀, the maximum absolute error |R_n| < a_n_{+ 1}, so .

11. Examine the ratio of successive terms. . Since , the series will converge when |x| < 1 or − 1 < x < 1. When x = 1, the series becomes . This series is term-by-term smaller than the p-series with p = 2; therefore the series converges. When x = − 1, the series becomes , which also converges. Therefore, the interval of convergence is [− 1, 1 ].

12. The ratio is and so the series will converge when |3x| < 1. This tells you that − 1 < 3x < 1 and . When , the series becomes , which is a convergent p-series. When , the series becomes , a convergent alternating series. Therefore the interval of convergence is .

13. Represent ln x by a Taylor series. Investigate the first few terms by finding and evaluating the derivatives and generating the first few terms.

f(a) = ln a, , , , so ln x can be represented by the Using the ratio test, . The series converges when , that is, . Solving the inequality, you find − a < x − a < a or 0 < x < 2a. When x = 0, the series becomes .

Since diverges, this series diverges as well. When x = 2a, the series becomes . This alternating series converges; therefore, the interval of convergence is (0, 2a].

14. Calculate the derivatives and evaluate at x = 1. f(x) = and f(1) = e. f′(x) = 2x and f′(1) = 2e. f″(x) = 4x² + 2 and f″(1) = 6e. f‴(x) = 8x³ + 12x and f‴(1) = 20e. f⁽⁴⁾(x) = 16x⁴ + 48x² + 12x and f⁽⁴⁾(1) = 76e. Then the function f(x) = can be approximated by . Simplifying .

15. . Find the derivatives and evaluate at . , , , and . Then f(x) = cos π x around can be approximated by .

16.

17. Calculate the derivatives and evaluate at f⁽ⁿ⁾(0) = n!, so the MacLaurin series . The series converges to . The series converges on (−1, 1). When x = 1, the series becomes which diverges. When x = − 1, the series becomes , which diverges. Therefore, the interval of convergence is (−1, 1).

18. Begin with the known series and replace x with − x². Then . The series converges to when . The series converges on (−1, 1). When x = − 1, the series becomes , which diverges. When x = −1, the series becomes , which diverges. Therefore, the interval of convergence is (− 1, 1).