Skip to main content
Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 10 - Infinite Sequences and SeriesProblem 10.4.8
Chapter 10, Problem 10.4.8

Direct Comparison Test
In Exercises 1–8, use the Direct Comparison Test to determine if each series converges or diverges.
∑ (from n=1 to ∞) (√n + 1) / (√(n² + 3))

Verified step by step guidance
1
Identify the given series: \( \sum_{n=1}^{\infty} \frac{\sqrt{n} + 1}{\sqrt{n^2 + 3}} \). We want to determine if it converges or diverges using the Direct Comparison Test.
Simplify the general term to understand its behavior for large \( n \). Notice that \( \sqrt{n^2 + 3} \) behaves like \( n \) as \( n \to \infty \), so the term roughly behaves like \( \frac{\sqrt{n} + 1}{n} \).
Split the term into two parts: \( \frac{\sqrt{n}}{n} + \frac{1}{n} = \frac{1}{\sqrt{n}} + \frac{1}{n} \). This helps us compare with known series.
Choose appropriate comparison series: compare \( \frac{1}{\sqrt{n}} \) with \( \sum \frac{1}{n^{1/2}} \) and \( \frac{1}{n} \) with the harmonic series \( \sum \frac{1}{n} \). Recall that \( \sum \frac{1}{n^{p}} \) converges if and only if \( p > 1 \).
Use the Direct Comparison Test by comparing the original term to these simpler terms. Since \( \frac{1}{\sqrt{n}} \) diverges and dominates the behavior, conclude about the convergence or divergence of the original series accordingly.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m

Key Concepts

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

Direct Comparison Test

The Direct Comparison Test determines the convergence or divergence of a series by comparing it to another series with known behavior. If the terms of the given series are less than or equal to the terms of a convergent series, it also converges. Conversely, if the terms are greater than or equal to those of a divergent series, it diverges.
Recommended video:
Guided course
09:25
Direct Comparison Test

Behavior of Series Terms for Large n

Analyzing the behavior of the series terms as n approaches infinity helps simplify complex expressions. For example, dominant terms in numerator and denominator guide the comparison to simpler series like p-series or geometric series, facilitating the application of convergence tests.
Recommended video:
Guided course
06:45
Intro to Series: Partial Sums

p-Series and Their Convergence

A p-series has the form ∑ 1/n^p and converges if p > 1, diverging otherwise. Recognizing or approximating a given series to a p-series allows for straightforward conclusions about convergence, which is essential when using the Direct Comparison Test.
Recommended video:
Guided course
04:30
P-Series and Harmonic Series
Related Practice
Textbook Question

Theory and Examples

Suppose that a₁, a₂, a₃, …, aₙ are positive numbers satisfying the following conditions:

i) a₁ ≥ a₂ ≥ a₃ ≥ …;


ii) the series a₂ + a₄ + a₈ + a₁₆ + … diverges.

Show that the series


a₁/1 + a₂/2 + a₃/3 + …


diverges.

1
views
Textbook Question

Absolute and Conditional Convergence

Which of the series in Exercises 15–48 converge absolutely, which converge, and which diverge? Give reasons for your answers.

∑ (from n = 1 to ∞) [(-1)ⁿ (√(n² + n) − n)]

1
views
Textbook Question

Find the sum of each series in Exercises 45–52.

∑ (from n = 1 to ∞) [ (2n + 1) / (n²(n + 1)²) ]

Textbook Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?

∑ (from n = 0 to ∞) [ n xⁿ / (4ⁿ (n² + 1)) ]

1
views
Textbook Question

In Exercises 125–134, determine whether the sequence is monotonic, whether it is bounded, and whether it converges.

aₙ = (4ⁿ⁺¹ + 3ⁿ) / 4ⁿ

1
views
Textbook Question

30. b. By differentiating the series in part (a) term by term, show that

Σ(from n=1 to ∞) n / (n + 1)! = 1.