AP Calculus BC

Infinite Sequences and Series: Convergence Tests, Taylor and Maclaurin Series

Calculus BC

  • ✓By the end of this lesson students will be able to determine the convergence or divergence of infinite sequences and series.
  • ✓By the end of this lesson students will be able to apply various convergence tests (e.g., Ratio, Alternating Series, Integral) to infinite series.
  • ✓By the end of this lesson students will be able to find the Taylor and Maclaurin series for a given function.
  • ✓By the end of this lesson students will be able to determine the radius and interval of convergence for power series, including Taylor and Maclaurin series.
  • ✓By the end of this lesson students will be able to use known Maclaurin series to derive new series.

Key concepts

Infinite Sequences

An infinite sequence is an ordered list of numbers, denoted as {a_n} = a_1, a_2, a_3, ..., a_n, .... A sequence converges if the limit of its terms as n approaches infinity exists and is a finite number L (lim (n->inf) a_n = L). Otherwise, the sequence diverges.

lim (n->inf) a_n = L (for convergence)
Infinite Series

An infinite series is the sum of the terms of an infinite sequence, denoted as sum (n=1 to inf) a_n = a_1 + a_2 + a_3 + .... A series converges if the sequence of its partial sums S_N = sum (n=1 to N) a_n converges to a finite limit S (lim (N->inf) S_N = S). Otherwise, the series diverges.

S = lim (N->inf) sum (n=1 to N) a_n (for convergence)
Nth-Term Test for Divergence

If lim (n->inf) a_n does not equal 0, then the series sum (n=1 to inf) a_n diverges. If lim (n->inf) a_n = 0, the test is inconclusive; the series may converge or diverge.

If lim (n->inf) a_n != 0, then sum a_n diverges.
Integral Test

If f is a positive, continuous, and decreasing function for x >= 1, and a_n = f(n), then the series sum (n=1 to inf) a_n and the improper integral integral (1 to inf) f(x) dx either both converge or both diverge.

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P-Series Test

A p-series is a series of the form sum (n=1 to inf) 1/n^p. It converges if p > 1 and diverges if p <= 1.

sum (n=1 to inf) 1/n^p converges if p > 1, diverges if p <= 1.
Direct Comparison Test

Suppose that sum a_n and sum b_n are series with positive terms. If 0 <= a_n <= b_n for all n greater than some N: 1. If sum b_n converges, then sum a_n converges. 2. If sum a_n diverges, then sum b_n diverges.

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Limit Comparison Test

Suppose that sum a_n and sum b_n are series with positive terms. If lim (n->inf) (a_n / b_n) = L, where L is a finite, positive number (0 < L < inf), then either both series converge or both series diverge.

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Alternating Series Test (AST)

An alternating series is a series whose terms alternate in sign, typically of the form sum (n=1 to inf) (-1)^(n-1) b_n or sum (n=1 to inf) (-1)^n b_n, where b_n > 0. The series converges if both conditions are met: 1. lim (n->inf) b_n = 0. 2. b_n is a decreasing sequence (i.e., b_(n+1) <= b_n for all n greater than some N).

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Absolute and Conditional Convergence

A series sum a_n is absolutely convergent if the series of absolute values sum |a_n| converges. If sum a_n converges but sum |a_n| diverges, then the series sum a_n is conditionally convergent.

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Ratio Test

For a series sum a_n, let L = lim (n->inf) |a_(n+1) / a_n|. 1. If L < 1, the series converges absolutely. 2. If L > 1 or L = inf, the series diverges. 3. If L = 1, the test is inconclusive.

L = lim (n->inf) |a_(n+1) / a_n|
Root Test

For a series sum a_n, let L = lim (n->inf) |a_n|^(1/n). 1. If L < 1, the series converges absolutely. 2. If L > 1 or L = inf, the series diverges. 3. If L = 1, the test is inconclusive.

L = lim (n->inf) |a_n|^(1/n)
Power Series

A power series is a series of the form sum (n=0 to inf) c_n (x-a)^n, where x is a variable and a is the center of the series. For a given power series, there are three possibilities for convergence: 1. The series converges only at x = a. 2. The series converges for all x (R = inf). 3. There is a positive number R (radius of convergence) such that the series converges if |x-a| < R and diverges if |x-a| > R. The interval of convergence must be checked at the endpoints x = a +/- R.

sum (n=0 to inf) c_n (x-a)^n
Taylor Series

If a function f has derivatives of all orders at x = a, then the Taylor series for f(x) centered at a is given by the sum of its derivatives evaluated at a, divided by n! and multiplied by (x-a)^n.

f(x) = sum (n=0 to inf) (f^(n)(a) / n!) (x-a)^n = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ...
Maclaurin Series

A Maclaurin series is a special case of a Taylor series where the series is centered at a = 0.

f(x) = sum (n=0 to inf) (f^(n)(0) / n!) x^n = f(0) + f'(0)x + (f''(0)/2!)x^2 + ...

Key facts to remember

  • 1A sequence {a_n} converges if lim (n->inf) a_n exists and is finite.
  • 2A series sum a_n converges if its sequence of partial sums converges.
  • 3Nth-Term Test for Divergence: If lim (n->inf) a_n != 0, then sum a_n diverges. If the limit is 0, the test is inconclusive.
  • 4P-Series sum (n=1 to inf) 1/n^p converges if p > 1 and diverges if p <= 1.
  • 5Geometric Series sum ar^(n-1) converges if |r| < 1 to a/(1-r).
  • 6The Ratio Test is generally effective for series involving factorials, powers of n, and exponential terms.
  • 7Taylor Series for f(x) centered at a: sum (n=0 to inf) (f^(n)(a) / n!) (x-a)^n.
  • 8Maclaurin Series is a Taylor series centered at a = 0.

Worked examples

Example 1

Determine if the series sum (n=1 to inf) (n^2) / (e^n) converges or diverges.

IIdentify the terms of the series: a_n = n^2 / e^n.
IISince the series involves powers of n and an exponential term, the Ratio Test is a good choice.
IIICalculate a_(n+1): a_(n+1) = (n+1)^2 / e^(n+1).
IVForm the ratio |a_(n+1) / a_n|: |((n+1)^2 / e^(n+1)) / (n^2 / e^n)| = ((n+1)^2 / e^(n+1)) * (e^n / n^2).
VSimplify the ratio: ((n+1)^2 / n^2) * (e^n / e^(n+1)) = ((n+1)/n)^2 * (1/e) = (1 + 1/n)^2 * (1/e).
VICalculate the limit L = lim (n->inf) |a_(n+1) / a_n|: L = lim (n->inf) [(1 + 1/n)^2 * (1/e)].
VIIAs n -> inf, 1/n -> 0, so (1 + 1/n)^2 -> (1+0)^2 = 1. Therefore, L = 1 * (1/e) = 1/e.
VIIICompare L to 1: Since L = 1/e approx 0.368, which is less than 1 (L < 1).
9Conclusion: By the Ratio Test, the series converges absolutely.

Answer

The series sum (n=1 to inf) (n^2) / (e^n) converges absolutely.

The Ratio Test is particularly effective for series involving factorials, powers of n, and exponential terms.

Example 2

Find the Taylor series for f(x) = ln(x) centered at a = 1.

IRecall the Taylor series formula: f(x) = sum (n=0 to inf) (f^(n)(a) / n!) (x-a)^n.
IIIdentify a = 1 and f(x) = ln(x).
IIICalculate the first few derivatives of f(x) and evaluate them at a = 1:
IVf(x) = ln(x) => f(1) = ln(1) = 0
Vf'(x) = 1/x => f'(1) = 1/1 = 1
VIf''(x) = -1/x^2 => f''(1) = -1/1^2 = -1
VIIf'''(x) = 2/x^3 => f'''(1) = 2/1^3 = 2
VIIIf''''(x) = -6/x^4 => f''''(1) = -6/1^4 = -6
9Observe the pattern for f^(n)(1) for n >= 1: f^(n)(1) = (-1)^(n-1) * (n-1)!.
10Substitute these values into the Taylor series formula:
11f(x) = f(1) + f'(1)(x-1) + (f''(1)/2!)(x-1)^2 + (f'''(1)/3!)(x-1)^3 + ...
12f(x) = 0 + 1(x-1) + (-1/2!)(x-1)^2 + (2/3!)(x-1)^3 + (-6/4!)(x-1)^4 + ...
13f(x) = (x-1) - (1/2)(x-1)^2 + (2/6)(x-1)^3 - (6/24)(x-1)^4 + ...
14f(x) = (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3 - (1/4)(x-1)^4 + ...
15Write the series in summation notation using the observed pattern:
16f(x) = sum (n=1 to inf) ((-1)^(n-1) * (n-1)! / n!) (x-1)^n
17f(x) = sum (n=1 to inf) ((-1)^(n-1) / n) (x-1)^n.

Answer

The Taylor series for f(x) = ln(x) centered at a = 1 is sum (n=1 to inf) ((-1)^(n-1) / n) (x-1)^n.

The first term (n=0) for ln(x) at a=1 is 0, so the series starts from n=1. This is a common occurrence.

Example 3

Find the Maclaurin series for f(x) = x^2 * cos(3x) and state its radius of convergence.

IRecall the known Maclaurin series for cos(u): cos(u) = sum (n=0 to inf) ((-1)^n * u^(2n)) / (2n)! for all u, with R = inf.
IISubstitute u = 3x into the series for cos(u):
IIIcos(3x) = sum (n=0 to inf) ((-1)^n * (3x)^(2n)) / (2n)! = sum (n=0 to inf) ((-1)^n * 3^(2n) * x^(2n)) / (2n)!.
IVMultiply the series for cos(3x) by x^2:
Vx^2 * cos(3x) = x^2 * sum (n=0 to inf) ((-1)^n * 3^(2n) * x^(2n)) / (2n)!.
VIDistribute x^2 into the summation (since x^2 is independent of n):
VIIx^2 * cos(3x) = sum (n=0 to inf) ((-1)^n * 3^(2n) * x^2 * x^(2n)) / (2n)!.
VIIICombine the powers of x: x^2 * x^(2n) = x^(2n+2).
9The Maclaurin series for f(x) = x^2 * cos(3x) is sum (n=0 to inf) ((-1)^n * 3^(2n) * x^(2n+2)) / (2n)!.
10Determine the radius of convergence: Since the Maclaurin series for cos(u) converges for all u (R = inf), substituting 3x for u does not change the radius of convergence. The series for cos(3x) converges for all x, and multiplying by x^2 also does not change this. Thus, the radius of convergence is R = inf.

Answer

The Maclaurin series for f(x) = x^2 * cos(3x) is sum (n=0 to inf) ((-1)^n * 3^(2n) * x^(2n+2)) / (2n)!. The radius of convergence is R = inf.

Manipulating known series is often faster than calculating derivatives for a new series. The radius of convergence for a power series is generally preserved under multiplication by a polynomial or substitution of a linear function.

Common mistakes

  • ✗Confusing the convergence of a sequence (lim a_n) with the convergence of a series (lim S_N). If lim a_n = 0, the series may still diverge.
  • ✗Incorrectly applying the conditions for convergence tests, especially the Alternating Series Test (e.g., forgetting to check if b_n is decreasing).
  • ✗Failing to test the endpoints of the interval of convergence for power series. The Ratio/Root Test is inconclusive at L=1, requiring separate analysis.
  • ✗Assuming that if the Ratio or Root Test yields L=1, the series diverges. This means the test is inconclusive, and another test must be used.
  • ✗Algebraic errors when simplifying the ratio |a_(n+1)/a_n| or the root |a_n|^(1/n).

Exam tips

  • ★Memorize the conditions and conclusions for each major convergence test (Nth-Term, P-Series, Geometric, Integral, Comparison, Limit Comparison, Alternating Series, Ratio, Root).
  • ★Know the common Maclaurin series (e^x, sin(x), cos(x), 1/(1-x)) and how to manipulate them to find new series.
  • ★When finding the interval of convergence for a power series, always use the Ratio Test (or Root Test) to find the radius, then *separately* test the endpoints of the interval.
  • ★For series involving factorials or `n` in the exponent, the Ratio Test is usually the most efficient choice. For series with `n` in the base and exponent, the Root Test might be easier.

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