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3k+2 00 63. 5k

3k+2 00 63. 5k

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Could you show the Algebra for this? What exactly are we doing algebraicly here to get to the (3/5)^k?

The given integral therefore diverges by the Integral Test. **10.4.63** This can be written in the form \( 9 \sum_{k=1}^{\infty} \left( \frac{3}{5} \right)^k \) so it is convergent because \( \sum_{k=1}^{\infty} \left( \frac{3}{5} \right)^k \) is a convergent geometric series. In fact, its value is \( 9 \left( \frac{\frac{3}{5}}{1 - \frac{3}{5}} \right) = 9 \left( \frac{3}{2} \right) = \frac{27}{2} \). **10.4.64** Additional handwritten notes appear, showing: \[ \frac{3a}{5} \cdot 3^k = 9 \sum \left(\frac{3}{5}\right)^k \]
Transcribed Image Text:The given integral therefore diverges by the Integral Test. **10.4.63** This can be written in the form \( 9 \sum_{k=1}^{\infty} \left( \frac{3}{5} \right)^k \) so it is convergent because \( \sum_{k=1}^{\infty} \left( \frac{3}{5} \right)^k \) is a convergent geometric series. In fact, its value is \( 9 \left( \frac{\frac{3}{5}}{1 - \frac{3}{5}} \right) = 9 \left( \frac{3}{2} \right) = \frac{27}{2} \). **10.4.64** Additional handwritten notes appear, showing: \[ \frac{3a}{5} \cdot 3^k = 9 \sum \left(\frac{3}{5}\right)^k \]
The image shows a mathematical expression for a series, labeled with the number 63. It represents an infinite sum given by: \[ \sum_{k=1}^{\infty} \frac{3^{k+2}}{5^k} \] This series involves: - The summation symbol \(\sum\), indicating an infinite series starting from \(k=1\). - The term inside the summation, \(\frac{3^{k+2}}{5^k}\), where \(3^{k+2}\) is the numerator and \(5^k\) is the denominator. The goal is to find the sum of this infinite series.
Transcribed Image Text:The image shows a mathematical expression for a series, labeled with the number 63. It represents an infinite sum given by: \[ \sum_{k=1}^{\infty} \frac{3^{k+2}}{5^k} \] This series involves: - The summation symbol \(\sum\), indicating an infinite series starting from \(k=1\). - The term inside the summation, \(\frac{3^{k+2}}{5^k}\), where \(3^{k+2}\) is the numerator and \(5^k\) is the denominator. The goal is to find the sum of this infinite series.
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