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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
All textbooksBriggs 3rd EditionCh. 10 - Sequences and Infinite SeriesProblem 10.3.65
Chapter 10, Problem 10.3.65

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.


65. ∑ (k = 1 to ∞) (1 / √(k + 1) – 1 / √(k + 3))

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Identify the general term of the series: \( a_k = \frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}} \). This is a telescoping series because each term involves the difference of two expressions with consecutive indices shifted by 2.
Write the nth partial sum \( S_n \) as the sum of the first n terms: \[ S_n = \sum_{k=1}^n \left( \frac{1}{\sqrt{k+1}} - \frac{1}{\sqrt{k+3}} \right) \].
Expand the sum to observe cancellation: \[ S_n = \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}} \right) + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}} \right) + \cdots + \left( \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+3}} \right) \].
Group the positive and negative terms to see which terms cancel out. Notice that many intermediate terms will cancel, leaving only a few terms from the beginning and the end of the sum.
Express \( S_n \) in a simplified form involving only the first few terms and the last few terms that do not cancel. Then, evaluate \( \lim_{n \to \infty} S_n \) by analyzing the behavior of the remaining terms as \( n \) approaches infinity to determine if the series converges or diverges.

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Key Concepts

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

Telescoping Series

A telescoping series is a series whose partial sums simplify because many terms cancel out. Typically, each term can be written as the difference of two expressions, causing intermediate terms to vanish when summed. This simplification helps find a closed-form expression for the nth partial sum.
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Partial Sums and Their Limits

The partial sum Sₙ of a series is the sum of its first n terms. Evaluating the limit of Sₙ as n approaches infinity determines whether the series converges or diverges. If the limit exists and is finite, the series converges to that value.
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Manipulating Radical Expressions

Working with terms involving square roots often requires rationalizing or rewriting expressions to simplify differences. Recognizing patterns in radicals can help express terms in a form suitable for telescoping, making it easier to identify cancellations in the series.
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Related Practice
Textbook Question

The first ten terms of the sequence {(1 + 1/10ⁿ)^10ⁿ}∞ ₙ₌₁ are rounded to 8 digits right of the decimal point (see table). Make a conjecture about the limit of the sequence.

n an

1 2.59374246

2 2.70481383

3 2.71692393

4 2.71814593

5 2.71826824

6 2.71828047

7 2.71828169

8 2.71828179

9 2.71828204

10 2.71828203

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Textbook Question

61–66. Sequences of partial sums For the following infinite series, find the first four terms of the sequence of partial sums. Then make a conjecture about the value of the infinite series or state that the series diverges.


4 + 0.9 + 0.09 + 0.009 + ⋯  

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Textbook Question

45–48. {Use of Tech} Explicit formulas for sequences Consider the formulas for the following sequences {aₙ}ₙ₌₁∞

 Make a table with at least ten terms and determine a plausible limit of the sequence or state that the sequence diverges.

aₙ = ⁿ² + n ;n = 1, 2, 3, …

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Textbook Question

32–49. Choose your test Use the test of your choice to determine whether the following series converge absolutely, converge conditionally, or diverge.

∑ (from k = 1 to ∞) 2ᵏ k! / kᵏ

Textbook Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.

∑ (from j = 2 to ∞)1 / (j ln¹⁰j)

Textbook Question

13–52. Limits of sequences

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


{√(n² + 1) − n} 

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