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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.5.18
Chapter 10, Problem 10.5.18

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 ∞) [(-1)ⁿ n² e⁻ⁿ]

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1
Identify the given series: \( \sum_{n=1}^{\infty} (-1)^n n^2 e^{-n} \). This is an alternating series because of the factor \( (-1)^n \), which causes the terms to alternate in sign.
Check the absolute value of the terms: \( a_n = n^2 e^{-n} \). Since \( e^{-n} = \frac{1}{e^n} \), the terms involve a polynomial factor \( n^2 \) multiplied by an exponential decay \( e^{-n} \).
Determine if the terms \( a_n = n^2 e^{-n} \) approach zero as \( n \to \infty \). Because exponential decay dominates polynomial growth, \( a_n \to 0 \). This is a necessary condition for convergence.
Consider the absolute convergence by examining the series \( \sum_{n=1}^{\infty} n^2 e^{-n} \). Since \( e^{-n} \) decays faster than any polynomial grows, this series converges by comparison to a convergent geometric series.
Conclude that since the series converges absolutely (the series of absolute values converges), the original alternating series \( \sum_{n=1}^{\infty} (-1)^n n^2 e^{-n} \) converges as well.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Convergence and Divergence of Infinite Series

An infinite series converges if the sum of its terms approaches a finite limit as the number of terms grows indefinitely; otherwise, it diverges. Understanding this concept is fundamental to analyzing whether a given series sums to a finite value or not.
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Convergence of an Infinite Series

Alternating Series Test

The Alternating Series Test determines convergence for series whose terms alternate in sign. If the absolute value of the terms decreases monotonically to zero, the series converges. This test is useful for series with factors like (-1)^n.
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Comparison and Limit Comparison Tests

These tests compare a given series to a known benchmark series to determine convergence or divergence. By analyzing the behavior of terms relative to simpler series, one can infer the original series' behavior, especially when exponential or polynomial terms are involved.
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Limit Comparison Test
Related Practice
Textbook Question

Finding Taylor Polynomials

In Exercises 1–10, find the Taylor polynomials of orders 0, 1, 2, and 3 generated by f at a.

f(x) = √(1 − x),a = 0

Textbook Question

Determining Convergence or Divergence

In Exercises 1–14, determine whether the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test.

∑ (from n = 1 to ∞) [(-1)ⁿ⁺¹ (1 / n^(3/2))]

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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 = 2 + (0.1)ⁿ

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Textbook Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?

∑ (from n = 1 to ∞) [ (x − 1)ⁿ / (n³ 3ⁿ) ]

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Textbook Question

In Exercises 37–42, find the series’ radius of convergence.

∑ (from n = 1 to ∞) [ n! xⁿ / nⁿ ]

Textbook Question

In Exercises 67–72, use the results of Exercises 63 and 64 to determine if each series converges or diverges.

∑(from n=2 to ∞) [(ln n)¹⁰⁰⁰ / n¹.⁰⁰¹]

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