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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 10, Problem 10.2.97

Make up a geometric series ∑a rⁿ⁻¹ that converges to the number 5 if
b. a = 13/2

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1
Recall the formula for the sum of an infinite geometric series: \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio with \(|r| < 1\) for convergence.
Given that the series converges to 5, set up the equation: \(5 = \frac{a}{1 - r}\).
Substitute the given value of \(a = \frac{13}{2}\) into the equation: \(5 = \frac{\frac{13}{2}}{1 - r}\).
Solve the equation for \(r\): multiply both sides by \((1 - r)\) to get \(5(1 - r) = \frac{13}{2}\), then isolate \(r\).
Check that the value of \(r\) you find satisfies \(|r| < 1\) to ensure the series converges, then write the geometric series as \(\sum_{n=1}^\infty a r^{n-1}\) with the found \(r\) and given \(a\).

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Key Concepts

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

Geometric Series and Its General Form

A geometric series is a sum of terms where each term is found by multiplying the previous term by a constant ratio r. It is generally expressed as ∑ a rⁿ⁻¹, where a is the first term and r is the common ratio. Understanding this form is essential to manipulate and analyze the series.
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Convergence of a Geometric Series

A geometric series converges if the absolute value of the common ratio |r| is less than 1. When it converges, the sum approaches a finite limit given by S = a / (1 - r). Recognizing this condition allows us to find r when the sum and a are known.
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Solving for the Common Ratio

Given the sum S and the first term a, we can solve for the common ratio r using the formula S = a / (1 - r). Rearranging gives r = 1 - (a / S). This step is crucial to determine the ratio that makes the series converge to the desired sum.
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