Question

so here is a question about absolute convergence see if it is right

Answer 1

http://openstudy.com/updates/attachments/4da96792d6938b0b4499a54d-avnis-1302948850351-calculus.JPG

Answer 2

i got it converges over the intervale -3<x<3

Answer 3

am i doing it right

Answer 4

does this stem from a previous question? what is the power series?

Answer 5

the link i posted is the question

Answer 6

http://openstudy.com/updates/attachments/4da96792d6938b0b4499a54d-avnis-1302948850351-calculus.JPG

Answer 7

http://openstudy.com/updates/attachments/4da96792d6938b0b4499a54d-avnis-1302948850351-calculus.JPG

Answer 8

okay it was not showing up , one sec I'll take a look

Answer 9

thanks :) i think i am doing it wrong though

Answer 10

yeah, i am getting different numbers

Answer 11

what did you get though

Answer 12

i have not check the end points yet but I got 8<x<12

Answer 13

how did you do it though

Answer 14

I used the ratio test

Answer 15

that is what i did

Answer 16

\[\frac{(x-5)^{(n+1)}}{(n+1)2^{(n+1)}}\frac{n2^n}{(x-5)^n}\]

Answer 17

simplifying\[=(x-5)\frac{n}{2(n+1)}\]

Answer 18

and \[lim_{n\rightarrow\infty}(x-5)\frac{n}{2(n+1)}\]\[=(x-5)\frac{1}{2}\]

Answer 19

wait a minute

Answer 20

yes that is what I got

Answer 21

you are right

Answer 22

oh

Answer 23

sorry 3<x<7

Answer 24

oh how

Answer 25

\[-1<\frac{1}{2}(x-5)<1\]

Answer 26

so\[-2<x-5<2\]

Answer 27

ah i see

Answer 28

adding 5\[3<x<7\]

Answer 29

oh ok

Answer 30

it does not converge absolutely at x=3

Answer 31

although it does converge at x=3, just not absolutely

Answer 32

it does not converge at all at x=7

Answer 33

hmm yes so (3,7) or [3.7)

Answer 34

well, it depends

it converges absolutely on (3,7) and conditionally at 3

Answer 35

If the question asks what is the interval of absolute convergence, I would answer (3,7)

because clearly the series that arises when x=3 converges conditionally

Answer 36

by the alternating series test

Answer 37

okay just looked at the question again

My answer would be :
converges absolutely on (3,7), conditionally at x=3 and diverges at x=7

Answer 38

hmm yes so (3,7) or [3.7)

Answer 39

The question asked "for what values of x does the series, converge absolutely? converge conditionally?, diverge?

So you must address all three questions which my answer above does