AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/10/10-10)
Unit 10 · Topic 10.10
BC only10.10 Alternating Series Error Bound
BC only. If an alternating series passes the alternating series test and you stop after some number of terms, the error is no bigger than the size of the first term you left out. You can also tell whether your estimate is too high or too low.
Key terms
- error bound
- remainder
- partial sum
- first omitted term
The error bound
This whole unit is BC only. Suppose Σ (−1)ⁿ⁺¹ bₙ satisfies the alternating series test (bₙ > 0, decreasing, bₙ → 0), and its sum is S. If you approximate S with the partial sum Sₙ, then
|S − Sₙ| ≤ bₙ₊₁.
In words: the error is at most the absolute value of the first term you didn't use.
Why it works
From the number-line picture in 10.7: the partial sums hop back and forth, and the true sum is always between any two consecutive partial sums. Sₙ and Sₙ₊₁ differ by exactly bₙ₊₁, so S is within bₙ₊₁ of Sₙ.
Over or under?
The sign of the first omitted term tells you which side the true sum is on. If the first term left out is negative, adding it would bring the estimate down, so your partial sum is an overestimate. If the first term left out is positive, your partial sum is an underestimate.
Equivalently: if the last term you included was positive, you overshot; if it was negative, you undershot.
Putting both facts together gives an interval. The true sum lies between Sₙ and Sₙ₊₁. For the series in the first example below, S is between S₄ = 197/216 − 1/64 ≈ 0.8964 and S₃ ≈ 0.9120.
With Taylor series
Many Taylor series alternate when you plug in a number. The series for cos x at x = 0.5 is 1 − (0.5)²/2! + (0.5)⁴/4! − … . Using the first two terms gives cos(0.5) ≈ 1 − 0.125 = 0.875. The terms decrease in size to 0, so the error is at most the next term, (0.5)⁴/4! = 0.0625/24 ≈ 0.0026. Since that omitted term is positive, 0.875 is an underestimate. (The true value is about 0.87758.)
Using it in reverse
To guarantee an error less than some tolerance, find the first n with bₙ₊₁ < tolerance, then use Sₙ. Be careful about off-by-one errors: the bound uses the first omitted term, which is term n + 1 when you've added n terms (starting at n = 1).
This bound applies only to alternating series that pass the alternating series test. For Taylor polynomial approximations, it can apply when the series for the function alternates. Otherwise, use the Lagrange error bound (10.12).
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Error bound for a partial sum
Approximate S = Σ (n = 1 to ∞) (−1)ⁿ⁺¹/n³ using the first three terms. Give an error bound, and say whether the approximation is too high or too low.
Show the solutionHide the solution
- Step 1: The series passes the alternating series test: 1/n³ is positive, decreasing and approaches 0.
- Step 2: S₃ = 1 − 1/8 + 1/27 = 197/216 ≈ 0.91204.
- Step 3: First omitted term: (−1)⁵/4³ = −1/64. Error ≤ 1/64 = 0.015625.
- Step 4: The first omitted term is negative, so S < S₃: the estimate is too high.
- Step 5: (For reference, S ≈ 0.90154, so the actual error is about 0.0105, within the bound.)
Answer: S ≈ 197/216 ≈ 0.912, with error at most 1/64; it's an overestimate.
- Example 2
Trap: off by one when choosing n
How many terms of Σ (n = 1 to ∞) (−1)ⁿ⁺¹/n² are needed to guarantee an error less than 0.001?
Show the solutionHide the solution
- Step 1: If you use n terms, the error is at most bₙ₊₁ = 1/(n + 1)².
- Step 2: You need 1/(n + 1)² < 0.001, so (n + 1)² > 1000, so n + 1 ≥ 32 (since 31² = 961 is too small and 32² = 1024 works).
- Step 3: So n = 31 terms.
- Step 4: The trap: solving 1/n² < 0.001 gives n = 32, which is one more than needed. It's not wrong to use more terms, but it isn't the smallest number.
Answer: 31 terms (error ≤ 1/32² = 1/1024 < 0.001).
Common mistakes
- Using the last included term as the bound instead of the first omitted term.
- Applying the bound to a series that isn't alternating, or whose terms don't decrease.
- Getting over/under backwards. Look at the sign of the first omitted term.
- Off-by-one errors when the index starts at 0.
On the exam
- A common BC free-response part: use the first few nonzero terms of a Taylor series (often alternating) to approximate a value, then show the error is less than some number using the alternating series error bound.
- State that the series alternates with terms decreasing in size to 0, then name the first omitted term and its absolute value.
Connected topics
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Check yourself
4 questions on 10.10 Alternating Series Error Bound. Pick an answer to see if you got it, and why.
Let S = Σ (n = 1 to ∞) (−1)ⁿ⁺¹/(n² · 2ⁿ), and let Sₙ be the sum of the first n terms. What is the least value of n for which the alternating series error bound guarantees |S − Sₙ| < 0.0001?
Let S = Σ (n = 1 to ∞) (−1)ⁿ⁺¹/n³ = 1 − 1/8 + 1/27 − 1/64 + ⋯. The sum of the first two terms, S₂ = 7/8, is used to approximate S. Which of the following is true?
The Maclaurin polynomial 1 − x²/2 is used to approximate cos(0.5). Using the alternating series error bound, what is the bound on the error of this approximation?
The function g is defined for all real x by the Maclaurin series g(x) = 1 − x²/2! + x⁴/3! − x⁶/4! + ⋯ + (−1)ⁿ x²ⁿ/(n + 1)! + ⋯.
Described function
The sum of the first three terms of the series is used to approximate g(0.9). What is the alternating series error bound for this approximation?
0 of 4 answered