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Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblema 10.8.79
Capitolo 10, Problema 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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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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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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Convergence of an Infinite Series

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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Introduction to Cotangent Graph
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