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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 10.3.7
Chapter 10, Problem 10.3.7

Find a formula for the nth partial sum Sₙ of
∑ k = 1 to ∞[(1/(k + 3)) − (1/(k + 4))]
Use your formula to find the sum of the first 36 terms of the series.

Verified step by step guidance
1
Recognize that the series is a telescoping series of the form \( \sum_{k=1}^\infty \left( \frac{1}{k+3} - \frac{1}{k+4} \right) \). This means many terms will cancel out when we write out the partial sums explicitly.
Write the nth partial sum \( S_n \) as \( S_n = \sum_{k=1}^n \left( \frac{1}{k+3} - \frac{1}{k+4} \right) \).
Expand the sum to see the cancellation pattern: \( S_n = \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \cdots + \left( \frac{1}{n+3} - \frac{1}{n+4} \right) \). Notice that most intermediate terms cancel out.
After cancellation, the partial sum simplifies to \( S_n = \frac{1}{4} - \frac{1}{n+4} \). This is the formula for the nth partial sum.
To find the sum of the first 36 terms, substitute \( n = 36 \) into the formula: \( S_{36} = \frac{1}{4} - \frac{1}{36+4} \). You can then simplify this expression to get the numerical value.

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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 where many terms cancel out when the partial sums are expanded, leaving only a few terms from the beginning and end. This simplification makes it easier to find a formula for the nth partial sum by identifying the pattern of cancellation.
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Partial Sums

The nth partial sum, Sₙ, is the sum of the first n terms of a series. Finding a formula for Sₙ helps analyze the behavior of the series and compute sums for specific values of n without adding each term individually.
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Evaluating Finite Sums

Once a formula for the nth partial sum is found, it can be used to calculate the sum of a finite number of terms by substituting the desired n value. This process avoids direct summation and leverages the simplified expression derived from the telescoping property.
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