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AP Calculus BC 10.10 Alternating Series Error Bound Study Notes - New Syllabus

AP Calculus BC 10.10 Alternating Series Error Bound Study Notes- New syllabus

AP Calculus BC 10.10 Alternating Series Error Bound Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Applying limits may allow us to determine the finite sum of infinitely many terms.

Key Concepts: 

  • Alternating Series Error Bound

AP Calculus BC-Concise Summary Notes- All Topics

AP Calculus BC Exam Questions MCQs and FRQs

Alternating Series Error Bound

Alternating Series Error Bound

The Alternating Series Error Bound is used to estimate how close a partial sum \( S_N \) of an alternating series is to the actual sum \( S \). It gives a bound on the size of the error when approximating the infinite sum by a finite number of terms.

Theorem (Alternating Series Remainder Estimate): If \( \displaystyle \sum_{n=1}^\infty (-1)^{n+1} b_n \) satisfies:

  1. \( b_{n+1} \leq b_n \) for all \( n \) (terms decrease in magnitude), and
  2. \( \lim_{n \to \infty} b_n = 0 \),

then the error \( R_N = S – S_N \) when using \( N \) terms satisfies:

\( |R_N| \leq b_{N+1} \)

This means: The absolute error is less than or equal to the magnitude of the first omitted term.

Interpretation: If you stop adding terms at \( N \), the maximum possible error is just the next term’s magnitude.

Steps to Apply the Alternating Series Error Bound:

  1. Check that the series is alternating and satisfies the Alternating Series Test.
  2. Identify \( b_{N+1} \), the magnitude of the first omitted term.
  3. Use \( |R_N| \leq b_{N+1} \) to find the error bound.
  4. Optionally, find how many terms are needed to achieve a desired accuracy.

Example

Approximate \( \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n} \) by the sum of its first 5 terms. Find the error bound.

▶️ Answer/Explanation

Step 1 — Identify \( b_{N+1} \): Here \( b_n = \dfrac{1}{n} \). For \( N = 5 \), \( b_{N+1} = b_6 = \dfrac{1}{6} \).

Step 2 — Apply error bound: \( |R_5| \leq b_6 = \dfrac{1}{6} \approx 0.1667 \).

Conclusion: The actual sum differs from the 5-term partial sum by at most 0.1667.

Example

Find the minimum number of terms needed so that the alternating series \( \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n^2} \) is approximated within \( 0.001 \) of the true sum.

▶️ Answer/Explanation

Step 1 — Error bound requirement: We need \( b_{N+1} \leq 0.001 \) where \( b_n = \dfrac{1}{n^2} \).

Step 2 — Solve inequality: \( \dfrac{1}{(N+1)^2} \leq 0.001 \) \( (N+1)^2 \geq 1000 \) \( N+1 \geq \sqrt{1000} \approx 31.62 \) So \( N \geq 31 \).

Conclusion: At least 31 terms are required to achieve the desired accuracy.

Example

The series \( \displaystyle \sum_{n=1}^\infty \dfrac{(-1)^{n}}{n^3} \) is approximated by its first 4 terms. The maximum possible error is:

A. \( \dfrac{1}{64} \)
B. \( \dfrac{1}{125} \)
C. \( \dfrac{1}{216} \)
D. \( \dfrac{1}{16} \)

▶️ Answer/Explanation

Step 1 — Identify \( b_{N+1} \): Here \( b_n = \dfrac{1}{n^3} \), \( N = 4 \), so \( b_{N+1} = b_5 = \dfrac{1}{125} \).

Step 2 — Apply formula: \( |R_4| \leq \dfrac{1}{125} \).

Final Answer: B

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