Determine the radius of convergence and the interval of convergence for each power series. sum_{n=1}^inftyfrac{(-2)^nx^n}{sqrt[4]{n}}

Maiclubk

Maiclubk

Answered question

2021-02-10

Determine the radius of convergence and the interval of convergence for each power series.
n=1(2)nxnn4

Answer & Explanation

AGRFTr

AGRFTr

Skilled2021-02-11Added 95 answers

Given series is: n=1anxn=n=1(2)nxnn4
By Applying ratio test , we have
limn|an+1an|=limn|(2)n+1xn+1n4n+14(2)nxn|
=limn|(2)xn4n4(1+1n)14|
=limn|(2)x(1+1n)14|
=|2x|
=2|x| for radius of convergence 2|x|=1
|x|=12
so r=12 is the radius of convergence.
Now, when x=12, given series becomes n=1(2)n(12)nn4
n=1(1)nan=n=1(1)nn4
as limnan=0 and seq. an is decreasing, so by Leibnitz alternating test, this series is convergent at x=12
Now, when x=12, given series becomes n=1(2)n(12)nn4
=n=11n4
By, p-test this series is convergent.
so, interval of convergence is [12,12]
Jeffrey Jordon

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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