anonymous
  • anonymous
so here is a question about absolute convergence see if it is right
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
i got it converges over the intervale -3
anonymous
  • anonymous
am i doing it right

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anonymous
  • anonymous
does this stem from a previous question? what is the power series?
anonymous
  • anonymous
the link i posted is the question
anonymous
  • anonymous
okay it was not showing up , one sec I'll take a look
anonymous
  • anonymous
thanks :) i think i am doing it wrong though
anonymous
  • anonymous
yeah, i am getting different numbers
anonymous
  • anonymous
what did you get though
anonymous
  • anonymous
i have not check the end points yet but I got 8
anonymous
  • anonymous
how did you do it though
anonymous
  • anonymous
I used the ratio test
anonymous
  • anonymous
that is what i did
anonymous
  • anonymous
\[\frac{(x-5)^{(n+1)}}{(n+1)2^{(n+1)}}\frac{n2^n}{(x-5)^n}\]
anonymous
  • anonymous
simplifying\[=(x-5)\frac{n}{2(n+1)}\]
anonymous
  • anonymous
and \[lim_{n\rightarrow\infty}(x-5)\frac{n}{2(n+1)}\]\[=(x-5)\frac{1}{2}\]
anonymous
  • anonymous
wait a minute
anonymous
  • anonymous
yes that is what I got
anonymous
  • anonymous
you are right
anonymous
  • anonymous
oh
anonymous
  • anonymous
sorry 3
anonymous
  • anonymous
oh how
anonymous
  • anonymous
\[-1<\frac{1}{2}(x-5)<1\]
anonymous
  • anonymous
so\[-2
anonymous
  • anonymous
ah i see
anonymous
  • anonymous
adding 5\[3
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
it does not converge absolutely at x=3
anonymous
  • anonymous
although it does converge at x=3, just not absolutely
anonymous
  • anonymous
it does not converge at all at x=7
anonymous
  • anonymous
hmm yes so (3,7) or [3.7)
anonymous
  • anonymous
well, it depends it converges absolutely on (3,7) and conditionally at 3
anonymous
  • 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
  • anonymous
by the alternating series test
anonymous
  • 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
  • anonymous
hmm yes so (3,7) or [3.7)
anonymous
  • 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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