Textbook Question
Use the Integral Test to show that the series
∑ (from n=0 to ∞) e^(−n²)
converges.
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 10.5.23
Determining Convergence or Divergence
In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑(from n=1 to ∞) [(2 + (−1)ⁿ) / 1.25ⁿ]
Verified step by step guidance1
Identify the general term of the series: \(a_n = \frac{2 + (-1)^n}{1.25^n}\).
Recognize that the denominator \$1.25^n\( is an exponential term with base greater than 1, which tends to grow as \)n$ increases.
Consider the behavior of the numerator \(2 + (-1)^n\), which oscillates between \(2 + 1 = 3\) (for even \(n\)) and \(2 - 1 = 1\) (for odd \(n\)), so it remains bounded.
Since the numerator is bounded and the denominator grows exponentially, compare the series to a geometric series with ratio \(r = \frac{1}{1.25} < 1\).
Conclude that because the terms behave like a geometric series with ratio less than 1, the series converges by the Comparison Test or by recognizing it as a sum of geometric series.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Geometric Series and Its Convergence
A geometric series has the form ∑ arⁿ, where r is the common ratio. It converges if |r| < 1 and diverges otherwise. Understanding this helps analyze series with terms involving exponential expressions like 1.25ⁿ.
Recommended video:
06:00
Geometric Series
Alternating Series and the Alternating Series Test
An alternating series has terms that alternate in sign, often involving (−1)ⁿ. The Alternating Series Test states such a series converges if the absolute value of terms decreases monotonically to zero. This is useful when the series includes (−1)ⁿ factors.
Recommended video:
10:54Alternating Series Test
Comparison and Limit Comparison Tests
These tests compare a given series to a known benchmark series to determine convergence or divergence. The Limit Comparison Test uses the limit of the ratio of terms to decide if both series behave similarly, aiding in analyzing complex series.
Recommended video:
07:45Limit Comparison Test
Related Practice
Textbook Question
In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 0 to ∞) [(2n + 3)(2ⁿ + 3) / (3ⁿ + 2)]
Textbook Question
In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 2 to ∞) [(3n)! / (n!)³]
Textbook Question
Determining Convergence or Divergence
Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.
∑ (from n=1 to ∞) (1 + cos n) / n²
Textbook Question
Convergence and Divergence
Which of the sequences {aₙ} in Exercises 31–100 converge, and which diverge? Find the limit of each convergent sequence.
aₙ = (n + 3) / (n² + 5n + 6)
1
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Textbook Question
Find the value of b for which
1 + eᵇ + e²ᵇ + e³ᵇ + ⋯ = 9.