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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 10.6.72
Chapter 10, Problem 10.6.72

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 1 to ∞) tan(n^(1/n))

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1
First, analyze the general term of the series: \(a_n = \tan\left(n^{\frac{1}{n}}\right)\). To determine convergence or divergence, we need to understand the behavior of \(a_n\) as \(n\) approaches infinity.
Evaluate the limit of the inside of the tangent function: \(\lim_{n \to \infty} n^{\frac{1}{n}}\). Recall that \(n^{\frac{1}{n}} = e^{\frac{\ln n}{n}}\). Since \(\frac{\ln n}{n} \to 0\) as \(n \to \infty\), this limit approaches \(e^0 = 1\).
Use the limit found to find \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \tan\left(n^{\frac{1}{n}}\right) = \tan(1)\). Since \(\tan(1)\) is a finite nonzero number, the terms \(a_n\) do not approach zero.
Recall the Divergence Test (also called the nth-term test for divergence): if \(\lim_{n \to \infty} a_n \neq 0\), then the series \(\sum a_n\) diverges. Since \(\lim_{n \to \infty} a_n = \tan(1) \neq 0\), the series diverges.
Therefore, conclude that the series \(\sum_{n=1}^\infty \tan\left(n^{\frac{1}{n}}\right)\) diverges by the Divergence Test.

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Key Concepts

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

Behavior of the General Term in a Series

To determine convergence, first analyze the limit of the general term as n approaches infinity. If the term does not approach zero, the series diverges by the Test for Divergence. Understanding the behavior of tan(n^(1/n)) as n grows large is crucial here.
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Geometric Series

Limit Comparison and Standard Convergence Tests

Various tests like the Comparison Test, Limit Comparison Test, or the Divergence Test help determine series convergence. Choosing an appropriate test depends on the form of the terms; for example, comparing tan(n^(1/n)) to a simpler function can clarify convergence behavior.
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Limit Comparison Test

Properties and Limits of the Function n^(1/n)

The expression n^(1/n) approaches 1 as n approaches infinity. Understanding this limit helps simplify tan(n^(1/n)) to approximately tan(1) for large n, which informs whether the terms tend to zero or not, a key step in analyzing series convergence.
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Properties of Functions
Related Practice
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 / (1 + 2² + 3² + ⋯ + n²)

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=3 to ∞) [2n² / n²ⁿ]

Textbook Question

Find the first four nonzero terms in the Maclaurin series for the functions in Exercises 31–38.

(tan⁻¹x)²

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 ∞) [n! / (2n + 1)!]

Textbook Question

In Exercises 35–40, find the first three nonzero terms of the Maclaurin series for each function.

f(x) = (sin x) ln(1 + x)

Textbook Question

In Exercises 81–86, find the values of x for which the given geometric series converges.

Also, find the sum of the series (as a function of x) for those values of x.

∑ (from n = 0 to ∞) [ (−1/2)ⁿ (x − 3)ⁿ ]