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Determine if the series Σ(-1)" 5n 1/2 + 3 converges 3n² - 3 n=1 absolutely, converges conditionally, or diverges, and justify your answer. The Series ✗(-1)" 5n / 2 +3 3n² - 3 n=1 converges by the since the absolute value of the terms Consider the absolute value of 5N 1/1 2 + 3 the series: 3n² - 3 n=1 by the to the p- series with p = which since a p-series will く if and only if n=1 5n³½ + 3 3m² - 3 and ☑ n=1 Since (−1)n 5n1/2 + 3 3n² - 3 Σ(-1)" n=1 5n 1/2 + 3 3n² - 3 > >

Determine if the series Σ(-1)" 5n 1/2 + 3 converges 3n² - 3 n=1 absolutely, converges conditionally, or diverges, and justify your answer. The Series ✗(-1)" 5n / 2 +3 3n² - 3 n=1 converges by the since the absolute value of the terms Consider the absolute value of 5N 1/1 2 + 3 the series: 3n² - 3 n=1 by the to the p- series with p = which since a p-series will く if and only if n=1 5n³½ + 3 3m² - 3 and ☑ n=1 Since (−1)n 5n1/2 + 3 3n² - 3 Σ(-1)" n=1 5n 1/2 + 3 3n² - 3 > >

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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Determine if the series Σ(-1)" 5n 1/2 + 3 converges 3n² - 3 n=1 absolutely, converges conditionally, or diverges, and justify your answer. The Series ✗(-1)" 5n / 2 +3 3n² - 3 n=1 converges by the since the absolute value of the terms Consider the absolute value of 5N 1/1 2 + 3 the series: 3n² - 3 n=1 by the to the p- series with p = which since a p-series will く if and only if n=1 5n³½ + 3 3m² - 3 and ☑ n=1 Since (−1)n 5n1/2 + 3 3n² - 3 Σ(-1)" n=1 5n 1/2 + 3 3n² - 3 > >
Transcribed Image Text:Determine if the series Σ(-1)" 5n 1/2 + 3 converges 3n² - 3 n=1 absolutely, converges conditionally, or diverges, and justify your answer. The Series ✗(-1)" 5n / 2 +3 3n² - 3 n=1 converges by the since the absolute value of the terms Consider the absolute value of 5N 1/1 2 + 3 the series: 3n² - 3 n=1 by the to the p- series with p = which since a p-series will く if and only if n=1 5n³½ + 3 3m² - 3 and ☑ n=1 Since (−1)n 5n1/2 + 3 3n² - 3 Σ(-1)" n=1 5n 1/2 + 3 3n² - 3 > >
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