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 10.5.25
Chapter 10, Problem 10.5.25

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


∑ (k = 1 to ∞) 1 / (2k − √k)

Verified step by step guidance
1
Identify the given series: \( \sum_{k=1}^{\infty} \frac{1}{2k - \sqrt{k}} \). We want to determine if this series converges or diverges.
To apply the Comparison Test or Limit Comparison Test, find a simpler series to compare with. Notice that for large \(k\), \(2k - \sqrt{k} \approx 2k\), so the terms behave like \( \frac{1}{2k} \).
Consider the comparison series \( \sum_{k=1}^{\infty} \frac{1}{k} \), which is a harmonic series known to diverge.
Use the Limit Comparison Test by computing \( \lim_{k \to \infty} \frac{\frac{1}{2k - \sqrt{k}}}{\frac{1}{k}} = \lim_{k \to \infty} \frac{k}{2k - \sqrt{k}} \). Simplify this limit to determine if it is a finite positive number.
If the limit is a positive finite constant, then both series either both converge or both diverge. Since the harmonic series diverges, conclude the behavior 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:
2m
Was this helpful?

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.
Recommended video:
09:25
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.
Recommended video:
07:45
Limit Comparison Test

Behavior of Series Terms and Dominant Terms

Analyzing the dominant terms in the series' general term helps simplify the expression for large indices. For example, in 1/(2k - √k), the term 2k dominates √k as k grows large, so the series behaves like 1/(2k). Understanding this helps in choosing an appropriate comparison series.
Recommended video:
05:44
Divergence Test (nth Term Test)
Related Practice
Textbook Question

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


∑ (k = 1 to ∞) (k² − 1) / (k³ + 4)

51
views
Textbook Question

72–86. Evaluating series Evaluate each series or state that it diverges.

∑ (k = 1 to ∞) ((1/3) × (5/6)ᵏ + (3/5) × (7/9)ᵏ)

53
views
Textbook Question

13–52. Limits of sequences

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


{√((1 + 1 / 2n)ⁿ)}

37
views
Textbook Question

33–38. {Use of Tech} Remainders in alternating series Determine how many terms of the following convergent series must be summed to be sure that the remainder is less than 10⁻⁴ in magnitude. Although you do not need it, the exact value of the series is given in each case.


π / 4 = ∑ (k = 0 to ∞) (−1)ᵏ / (2k + 1)

61
views
Textbook Question

55–70. More sequences

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


{(75n⁻¹ / 99ⁿ) + (5ⁿsinn / 8ⁿ)}

59
views
Textbook Question

9–30. The Ratio and Root Tests Use the Ratio Test or the Root Test to determine whether the following series converge absolutely or diverge.

∑ (from k = 1 to ∞) ((-1)ᵏ⁺¹ × k²ᵏ) / (k! × k!)

58
views