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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
Capitolo 10, Problema 10.5.17

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.


∑ (from k = 1 to ∞) (−k)³ / (3k³ + 2)

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First, write down the general term of the series: \(a_k = \frac{(-k)^3}{3k^3 + 2}\).
Simplify the term \(a_k\): since \((-k)^3 = -k^3\), rewrite \(a_k\) as \(a_k = \frac{-k^3}{3k^3 + 2}\).
Analyze the behavior of \(a_k\) as \(k \to \infty\) by dividing numerator and denominator by \(k^3\): \(a_k = \frac{-1}{3 + \frac{2}{k^3}}\).
Determine the limit of \(a_k\) as \(k \to \infty\): \(\lim_{k \to \infty} a_k = \frac{-1}{3 + 0} = -\frac{1}{3}\).
Since the limit of \(a_k\) is not zero, conclude by the Test for Divergence (also called the nth-term test) that the series \(\sum_{k=1}^\infty a_k\) diverges.

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