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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 10, Problem 10.8.79

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)tan⁻¹(1 / √k)

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1
Identify the series given: \( \sum_{k=1}^{\infty} \tan^{-1}\left( \frac{1}{\sqrt{k}} \right) \). We want to determine if this infinite series converges or diverges.
Recall that for large \( k \), \( \tan^{-1}(x) \) behaves approximately like \( x \) when \( x \) is close to zero. Since \( \frac{1}{\sqrt{k}} \to 0 \) as \( k \to \infty \), we can compare \( \tan^{-1}\left( \frac{1}{\sqrt{k}} \right) \) to \( \frac{1}{\sqrt{k}} \).
Use the Comparison Test or Limit Comparison Test by comparing the given series to the series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \), which is a p-series with \( p = \frac{1}{2} \). Recall that a p-series \( \sum \frac{1}{k^p} \) converges if and only if \( p > 1 \).
Since \( p = \frac{1}{2} < 1 \), the series \( \sum \frac{1}{\sqrt{k}} \) diverges. Therefore, if the terms of our original series behave like \( \frac{1}{\sqrt{k}} \) for large \( k \), the original series will also diverge by comparison.
Conclude that the series \( \sum_{k=1}^{\infty} \tan^{-1}\left( \frac{1}{\sqrt{k}} \right) \) diverges because its terms do not decrease fast enough to produce a convergent sum.

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Key Concepts

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

Convergence of Infinite Series

An infinite series converges if the sequence of its partial sums approaches a finite limit. Understanding convergence is essential to determine whether the sum of infinitely many terms results in a finite value or diverges to infinity or oscillates.
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Comparison Test and Limit Comparison Test

These tests help determine convergence by comparing the given series to a known benchmark series. If terms of the series behave similarly to a convergent or divergent series, the original series shares the same behavior, simplifying the analysis.
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Limit Comparison Test

Behavior of arctan(x) for Small Arguments

For small values of x, arctan(x) approximates x because arctan(x) ~ x as x → 0. This approximation allows us to compare the given series terms tan⁻¹(1/√k) to simpler terms like 1/√k, aiding in applying convergence tests.
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