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
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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
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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