Textbook Question
Repeating Decimals
Express each of the numbers in Exercises 23–30 as the ratio of two integers.
3.1̅4̅2̅8̅5̅7 = 3.142857142857 ...
Ch. 10 - Infinite Sequences and Series
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 10, Problem 10.6.52
Error Estimation
In Exercises 49–52, estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series.
1 / (1 + t) = ∑ (from n = 0 to ∞) [(-1)ⁿ tⁿ],0 < t < 1
Verified step by step guidance1
Recognize that the given series is the Taylor (or Maclaurin) series expansion of the function \(\frac{1}{1+t}\) around \(t=0\), expressed as \(\sum_{n=0}^\infty (-1)^n t^n\) for \(0 < t < 1\).
Since the series is an alternating series with terms decreasing in magnitude, the Alternating Series Estimation Theorem applies. This theorem states that the magnitude of the error when approximating the sum by the first \(N\) terms is less than or equal to the magnitude of the first omitted term.
Identify the first four terms as those with indices \(n=0,1,2,3\). The next term, which is the first omitted term, corresponds to \(n=4\) and has the form \((-1)^4 t^4 = t^4\).
Therefore, the magnitude of the error in approximating the sum by the first four terms is at most \(|t^4|\), which simplifies to \(t^4\) since \(t > 0\).
To estimate the error for a specific value of \(t\) in the interval \((0,1)\), substitute that value into \(t^4\) to find the upper bound on the error magnitude.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Alternating Series and Their Convergence
An alternating series is a series whose terms alternate in sign. For the series 1/(1+t) expanded as ∑ (-1)^n t^n, the terms decrease in magnitude and approach zero when 0 < t < 1, ensuring convergence. Understanding this helps justify using partial sums to approximate the series.
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Alternating Series Test
Error Estimation in Alternating Series
The error when approximating an alternating series by its first n terms is less than or equal to the absolute value of the (n+1)th term. This property allows us to bound the magnitude of the error simply by evaluating the next term in the series.
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06:00Geometric Series
Power Series Representation and Radius of Convergence
A power series expresses a function as an infinite sum of powers of a variable. The radius of convergence defines the interval where the series converges to the function. For 1/(1+t), the series converges for |t| < 1, which is crucial for valid error estimation.
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07:36Radius of Convergence
Related Practice
Textbook Question
In Exercises 15–22, determine if the geometric series converges or diverges. If a series converges, find its sum.
1 − (2/e) + (2/e)² − (2/e)³ + (2/e)⁴ − …
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Textbook Question
In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 1 to ∞) [3ⁿ / n³]
Textbook Question
Limit Comparison Test
In Exercises 9–16, use the Limit Comparison Test to determine if each series converges or diverges.
∑ (from n=2 to ∞) 1 / ln n
(Hint: Limit Comparison with ∑ (from n=2 to ∞) (1/n))
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Textbook Question
Applying the Integral Test
Use the Integral Test to determine if the series in Exercises 1–12 converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied.
∑ (from n = 2 to ∞) ln(n²) / n
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Textbook Question
Telescoping Series
In Exercises 39–44, find a formula for the nth partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum.
∑ (from n = 1 to ∞) [ (1/n) − (1/(n + 1)) ]
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