(1). Does the following series converge absolutely, converge conditionally or diverge?(a). Σ^_o(5)".(b). Σ=1(-1)*+1(;).(c). Σ*, 4*.Σι (-4)"(d). Σ*=1(cos(nn)).(e). Σ^=1(−1)"+1(32)".3r(f). Σ^=(nn))".

Answered: (1). Does the following series converge…

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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Related questions

Question

(1). Does the following series converge absolutely, converge conditionally or diverge?
(a). Σ^_o(5)".
(b). Σ=1(-1)*+1(;).
(c). Σ*, 4*.
Σι (-4)"
(d). Σ*=1(cos(nn)).
(e). Σ^=1(−1)"+1(32)".
3r
(f). Σ^=(nn))".

Expert Solution

Step 1: In this step, we obtain the two subparts.

According to our guidelines, we can answer only the first question or first three subparts of a question, the rest can be reposted.

(a) Given that, the series

$sum from n equals 0 to infinity of space open parentheses negative fraction numerator 4 over denominator 5 space i end fraction close parentheses to the power of n space equals space sum from n equals 0 to infinity of space a subscript n space space left parenthesis say right parenthesis comma therefore comma space space space straight a subscript straight n space equals space space open parentheses negative fraction numerator 4 over denominator 5 space straight i end fraction close parentheses to the power of straight n open vertical bar straight a subscript straight n close vertical bar space equals space open vertical bar space open parentheses negative fraction numerator 4 over denominator 5 space straight i end fraction close parentheses to the power of straight n close vertical bar space equals open parentheses space 4 over 5 close parentheses to the power of straight n comma thus comma space limit as straight n rightwards arrow infinity of open vertical bar straight a subscript straight n over straight a subscript straight n plus 1 end subscript close vertical bar space equals space limit as straight n rightwards arrow infinity of space open parentheses space 4 over 5 close parentheses to the power of straight n space open parentheses space 5 over 4 close parentheses to the power of straight n plus 1 end exponent space equals space 5 over 4 space greater than 1$

by ratio test, the series $stack sum space with n equals 0 below and infinity on top open vertical bar a subscript n close vertical bar space$ is convergent. Hence, the orignal series $sum from n equals 0 to infinity of space a subscript n$ is absolutely convergent.

(b) Given that, the series

$stack sum space with n equals 1 below and infinity on top open parentheses negative 1 close parentheses to the power of n plus 1 end exponent open parentheses 1 over n close parentheses space equals space stack sum space with n equals 1 below and infinity on top a subscript n space left parenthesis s a y right parenthesis therefore comma space space a subscript n space equals space open parentheses negative 1 close parentheses to the power of straight n plus 1 end exponent open parentheses 1 over straight n close parentheses$ ,

by alternating series test, the series is convergent but the series $space stack sum space with n equals 1 below and infinity on top open vertical bar a subscript n close vertical bar$ is not convergent by p-series test. Thus, the series is conditionally convergent.