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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 10.4.15
Chapter 10, Problem 10.4.15

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))

Verified step by step guidance
1
Identify the given series: \( \sum_{n=2}^{\infty} \frac{1}{\ln n} \). We want to determine if this series converges or diverges using the Limit Comparison Test.
Choose a comparison series that is easier to analyze. The hint suggests using \( \sum_{n=2}^{\infty} \frac{1}{n} \), which is a well-known divergent harmonic series.
Set up the Limit Comparison Test by computing the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{\ln n}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{\ln n} \).
Analyze the limit \( L \). Since \( n \) grows faster than \( \ln n \), this limit tends to infinity, which is a positive number (not zero or infinite in the sense of the test's conditions).
Interpret the result: Because the limit \( L \) is infinite and the comparison series \( \sum \frac{1}{n} \) diverges, the Limit Comparison Test tells us that the original series \( \sum \frac{1}{\ln n} \) also diverges.

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Key Concepts

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

Limit Comparison Test

The Limit Comparison Test is used to determine the convergence or divergence of an infinite series by comparing it to a second series with known behavior. It involves taking the limit of the ratio of the nth terms of the two series. If the limit is a positive finite number, both series either converge or diverge together.
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Limit Comparison Test

Behavior of the Harmonic Series

The harmonic series ∑ 1/n is a well-known divergent series. Understanding its divergence is crucial when using it as a comparison series in the Limit Comparison Test. Since it diverges, any series that behaves similarly (in terms of term size) will also diverge.
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P-Series and Harmonic Series

Properties of the Natural Logarithm Function

The natural logarithm function ln(n) grows slowly as n increases, but it still tends to infinity. Recognizing how 1/ln(n) compares to 1/n helps in applying the Limit Comparison Test. Since ln(n) grows slower than n, 1/ln(n) decreases more slowly than 1/n, affecting convergence.
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Properties of Functions
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

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

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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

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

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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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 = 0 to ∞) (2x)ⁿ