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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 10.17

9–36. Comparison tests Use the Comparison Test or the Limit Comparison Test to determine whether the following series converge.


∑ (k = 1 to ∞) 1 / (k³ᐟ² + 1)

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Identify the given series: \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2} + 1} \). We want to determine if this series converges using the Comparison Test or the Limit Comparison Test.
Choose a simpler series to compare with. Since for large \( k \), \( k^{3/2} + 1 \) behaves like \( k^{3/2} \), consider the series \( \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} \), which is a p-series with \( p = \frac{3}{2} > 1 \) and is known to converge.
Apply the Comparison Test by noting that \( \frac{1}{k^{3/2} + 1} < \frac{1}{k^{3/2}} \) for all \( k \geq 1 \). Since \( \sum \frac{1}{k^{3/2}} \) converges, this suggests the original series converges by direct comparison.
Alternatively, apply the Limit Comparison Test by computing the limit \( L = \lim_{k \to \infty} \frac{\frac{1}{k^{3/2} + 1}}{\frac{1}{k^{3/2}}} = \lim_{k \to \infty} \frac{k^{3/2}}{k^{3/2} + 1} \).
Evaluate the limit \( L \). If \( L \) is a finite positive number, then both series either converge or diverge together. Since \( \sum \frac{1}{k^{3/2}} \) converges, this confirms the convergence of the original series.

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Key Concepts

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

Comparison Test

The Comparison Test determines the convergence of a series by comparing it to a second series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges. Conversely, if the terms are larger than those of a divergent series, it diverges.
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Direct Comparison Test

Limit Comparison Test

The Limit Comparison Test compares two series by taking the limit of the ratio of their terms. If this limit is a positive finite number, both series either converge or diverge together. This test is useful when direct comparison is difficult but the terms behave similarly for large indices.
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Limit Comparison Test

p-Series and its Convergence

A p-series is of the form ∑ 1/k^p, which converges if and only if p > 1. Recognizing that the given series resembles a p-series helps in applying comparison tests effectively. For example, since 3/2 > 1, the series ∑ 1/k^(3/2) converges.
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P-Series and Harmonic Series