- anonymous

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

- katieb

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- anonymous

i got it converges over the intervale -3

- anonymous

am i doing it right

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## More answers

- anonymous

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

- anonymous

the link i posted is the question

- anonymous

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

- anonymous

thanks :) i think i am doing it wrong though

- anonymous

yeah, i am getting different numbers

- anonymous

what did you get though

- anonymous

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

- anonymous

how did you do it though

- anonymous

I used the ratio test

- anonymous

that is what i did

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wait a minute

- anonymous

yes that is what I got

- anonymous

you are right

- anonymous

oh

- anonymous

sorry 3

- anonymous

oh how

- anonymous

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

- anonymous

so\[-2

- anonymous

ah i see

- anonymous

adding 5\[3

- anonymous

oh ok

- anonymous

it does not converge absolutely at x=3

- anonymous

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

- anonymous

it does not converge at all at x=7

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

by the alternating series test

- anonymous

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

- anonymous

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

- anonymous

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

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