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verges, and n+2 b) n-1 8.

verges, and n+2 b) n-1 8.

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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For part b  where it says a=(2/3)^3 , why did we raise this to the 3rd power?

**Calculus II: Review Test 3** **6. Determine whether the series converges, and if so find its sum.** a) \(\sum_{n=1}^{\infty} \left(-\frac{3}{4}\right)^{n-1}\) - **Geometric Series**: - First term (\(a\)) = 1 - Common ratio (\(r\)) = \(-\frac{3}{4}\) - Sum = \(\frac{1}{1 - \left(-\frac{3}{4}\right)} = \frac{4}{7}\) b) \(\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^{n+2}\) - **Geometric Series**: - First term (\(a\)) = \(\left(\frac{2}{3}\right)^3\) - Common ratio (\(r\)) = \(\frac{2}{3}\) - Sum = \(\frac{\left(\frac{2}{3}\right)^3}{1 - \frac{2}{3}} = \frac{8}{9}\) c) The problem continues, but the detailed explanation and transcription for part (c) is not visible in the image above.
Transcribed Image Text:**Calculus II: Review Test 3** **6. Determine whether the series converges, and if so find its sum.** a) \(\sum_{n=1}^{\infty} \left(-\frac{3}{4}\right)^{n-1}\) - **Geometric Series**: - First term (\(a\)) = 1 - Common ratio (\(r\)) = \(-\frac{3}{4}\) - Sum = \(\frac{1}{1 - \left(-\frac{3}{4}\right)} = \frac{4}{7}\) b) \(\sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^{n+2}\) - **Geometric Series**: - First term (\(a\)) = \(\left(\frac{2}{3}\right)^3\) - Common ratio (\(r\)) = \(\frac{2}{3}\) - Sum = \(\frac{\left(\frac{2}{3}\right)^3}{1 - \frac{2}{3}} = \frac{8}{9}\) c) The problem continues, but the detailed explanation and transcription for part (c) is not visible in the image above.
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