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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 10.6.17

11–27. Alternating Series Test Determine whether the following series converge.
∑ (k = 1 to ∞) (−1)ᵏ⁺¹ k² / (k³ + 1)

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Identify the general term of the series: \(a_k = \frac{k^2}{k^3 + 1}\) and note the alternating factor \((-1)^{k+1}\), which makes this an alternating series.
Recall the Alternating Series Test, which states that an alternating series \(\sum (-1)^{k} b_k\) converges if two conditions are met: (1) the sequence \(b_k\) is decreasing, and (2) \(\lim_{k \to \infty} b_k = 0\).
Set \(b_k = \frac{k^2}{k^3 + 1}\) (the absolute value of the terms) and check the limit as \(k\) approaches infinity: compute \(\lim_{k \to \infty} \frac{k^2}{k^3 + 1}\) by dividing numerator and denominator by \(k^3\).
Analyze whether \(b_k\) is decreasing for sufficiently large \(k\) by considering the behavior of the function \(f(k) = \frac{k^2}{k^3 + 1}\) or by comparing \(b_k\) and \(b_{k+1}\).
If both conditions of the Alternating Series Test are satisfied (limit zero and decreasing terms), conclude that the series converges; otherwise, it does not converge by this test.

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Alternating Series Test

The Alternating Series Test determines the convergence of series whose terms alternate in sign. It requires that the absolute value of the terms decreases monotonically to zero. If these conditions hold, the series converges, even if it does not converge absolutely.
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Alternating Series Test

Limit of the Terms

For a series to converge, the terms must approach zero as k approaches infinity. Evaluating the limit of the general term helps verify this condition, which is essential for applying the Alternating Series Test or any convergence test.
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Behavior of Rational Functions at Infinity

Understanding how rational functions behave as k grows large helps simplify the terms of the series. For example, k²/(k³ + 1) behaves like k²/k³ = 1/k for large k, which aids in analyzing the limit and monotonicity of the terms.
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