Skip to main content
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 10.6.58
Chapter 10, Problem 10.6.58

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]

Verified step by step guidance
1
Identify the given series: \( \sum_{n=1}^{\infty} \frac{3^n}{n^3} \). This is an infinite series with terms \( a_n = \frac{3^n}{n^3} \).
Recognize that the numerator \(3^n\) is an exponential function and the denominator \(n^3\) is a polynomial function. Generally, exponential growth dominates polynomial growth as \(n\) becomes large.
Consider applying the Root Test or Ratio Test, which are effective for series involving exponential terms. For the Ratio Test, compute \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
Calculate \( \frac{a_{n+1}}{a_n} = \frac{3^{n+1} / (n+1)^3}{3^n / n^3} = 3 \cdot \frac{n^3}{(n+1)^3} \). Then find the limit \( L = \lim_{n \to \infty} 3 \cdot \frac{n^3}{(n+1)^3} \).
Evaluate the limit \( L \). If \( L > 1 \), the series diverges; if \( L < 1 \), the series converges; if \( L = 1 \), the test is inconclusive. Use this conclusion to determine the behavior of the series.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

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

Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. Determining convergence means checking if the sum approaches a finite limit as the number of terms grows. If the series converges, its partial sums get closer to a specific value; if it diverges, the sums grow without bound or oscillate.
Recommended video:
06:52
Convergence of an Infinite Series

Ratio Test

The Ratio Test is used to determine the convergence of series with positive terms, especially those involving exponentials or factorials. It examines the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges.
Recommended video:
05:35
Ratio Test

Comparison of Growth Rates (Exponential vs Polynomial)

In series terms, exponential functions grow faster than any polynomial function as n approaches infinity. When the numerator grows exponentially and the denominator polynomially, the terms often do not approach zero, leading to divergence. Recognizing this helps in quickly assessing series behavior.
Recommended video:
09:29
Exponential Growth & Decay
Related Practice
Textbook Question

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]

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 = (2n + 2)! / (2n − 1)!

1
views
Textbook Question

Absolute and Conditional Convergence

Which of the series in Exercises 15–48 converge absolutely, which converge, and which diverge? Give reasons for your answers.

∑ (from n = 1 to ∞) [(-1)ⁿ⁻¹ / (n² + 2n + 1)]

Textbook Question

Suppose that aₙ > 0 and limₙ→∞ n²aₙ = 0. Prove that ∑aₙ converges.

1
views
Textbook Question

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

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))

1
views