Skip to main content
Ch. 10 - Sequences and Infinite Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.8.49
Chapter 10, Problem 10.8.49

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)(⁵√k) / ⁵√(k⁷ + 1)

Verified step by step guidance
1
First, rewrite the general term of the series to better understand its behavior: the term is given by \(\frac{\sqrt[5]{k}}{\sqrt[5]{k^{7} + 1}}\).
Simplify the expression inside the fifth root in the denominator by factoring out \(k^{7}\): \(\sqrt[5]{k^{7} + 1} = \sqrt[5]{k^{7}(1 + \frac{1}{k^{7}})}\).
Use the property of roots to separate the terms: \(\sqrt[5]{k^{7}(1 + \frac{1}{k^{7}})} = \sqrt[5]{k^{7}} \cdot \sqrt[5]{1 + \frac{1}{k^{7}}} = k^{\frac{7}{5}} \cdot \sqrt[5]{1 + \frac{1}{k^{7}}}\).
Rewrite the original term as \(\frac{k^{\frac{1}{5}}}{k^{\frac{7}{5}} \cdot \sqrt[5]{1 + \frac{1}{k^{7}}}} = \frac{1}{k^{\frac{6}{5}} \cdot \sqrt[5]{1 + \frac{1}{k^{7}}}}\).
As \(k\) approaches infinity, \(\sqrt[5]{1 + \frac{1}{k^{7}}}\) approaches 1, so the term behaves like \(\frac{1}{k^{\frac{6}{5}}}\). Use the p-series test to determine convergence: since \(\frac{6}{5} > 1\), the series converges.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

Key Concepts

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

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.
Recommended video:
06:52
Convergence of an Infinite Series

Comparison Test for Series

The Comparison Test involves comparing the given series to a known benchmark series with positive terms. If the given series is smaller than a convergent series or larger than a divergent series, conclusions about its convergence or divergence can be drawn.
Recommended video:
09:25
Direct Comparison Test

Asymptotic Behavior and Simplification of Terms

Analyzing the behavior of terms for large indices (k → ∞) helps simplify complex expressions. Approximating dominant terms allows us to compare the series to simpler p-series or geometric series, facilitating the application of convergence tests.
Recommended video:
5:50
Asymptotes of Hyperbolas
Related Practice
Textbook Question

55–70. More sequences

Find the limit of the following sequences or determine that the sequence diverges.


aₙ = (−1)ⁿ ⁿ√n

87
views
Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{(3ⁿ⁺¹ + 3)⁄3ⁿ}

64
views
Textbook Question

55–70. More sequences

Find the limit of the following sequences or determine that the sequence diverges.


{tan⁻¹n / n}

76
views
Textbook Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.


{(1 + (2 / n))ⁿ}

69
views
Textbook Question

1–10. Choosing convergence tests Identify a convergence test for each series. If necessary, explain how to simplify or rewrite the series before applying the convergence test. You do not need to carry out the convergence test.

∑ (from k = 1 to ∞) (k² / (k⁴ + k³ + 1))

79
views
Textbook Question

54–69. Telescoping series

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sₙ}. Then evaluate limₙ→∞ Sₙ to obtain the value of the series or state that the series diverges.


63. ∑ (k = 1 to ∞) 1 / ((k + p)(k + p + 1)), where p is a positive integer

106
views