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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    i got it converges over the intervale -3<x<3

  2. anonymous
    • 5 years ago
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    am i doing it right

  3. anonymous
    • 5 years ago
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    does this stem from a previous question? what is the power series?

  4. anonymous
    • 5 years ago
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    the link i posted is the question

  5. anonymous
    • 5 years ago
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    okay it was not showing up , one sec I'll take a look

  6. anonymous
    • 5 years ago
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    thanks :) i think i am doing it wrong though

  7. anonymous
    • 5 years ago
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    yeah, i am getting different numbers

  8. anonymous
    • 5 years ago
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    what did you get though

  9. anonymous
    • 5 years ago
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    i have not check the end points yet but I got 8<x<12

  10. anonymous
    • 5 years ago
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    how did you do it though

  11. anonymous
    • 5 years ago
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    I used the ratio test

  12. anonymous
    • 5 years ago
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    that is what i did

  13. anonymous
    • 5 years ago
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    \[\frac{(x-5)^{(n+1)}}{(n+1)2^{(n+1)}}\frac{n2^n}{(x-5)^n}\]

  14. anonymous
    • 5 years ago
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    simplifying\[=(x-5)\frac{n}{2(n+1)}\]

  15. anonymous
    • 5 years ago
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    and \[lim_{n\rightarrow\infty}(x-5)\frac{n}{2(n+1)}\]\[=(x-5)\frac{1}{2}\]

  16. anonymous
    • 5 years ago
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    wait a minute

  17. anonymous
    • 5 years ago
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    yes that is what I got

  18. anonymous
    • 5 years ago
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    you are right

  19. anonymous
    • 5 years ago
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    oh

  20. anonymous
    • 5 years ago
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    sorry 3<x<7

  21. anonymous
    • 5 years ago
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    oh how

  22. anonymous
    • 5 years ago
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    \[-1<\frac{1}{2}(x-5)<1\]

  23. anonymous
    • 5 years ago
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    so\[-2<x-5<2\]

  24. anonymous
    • 5 years ago
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    ah i see

  25. anonymous
    • 5 years ago
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    adding 5\[3<x<7\]

  26. anonymous
    • 5 years ago
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    oh ok

  27. anonymous
    • 5 years ago
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    it does not converge absolutely at x=3

  28. anonymous
    • 5 years ago
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    although it does converge at x=3, just not absolutely

  29. anonymous
    • 5 years ago
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    it does not converge at all at x=7

  30. anonymous
    • 5 years ago
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    hmm yes so (3,7) or [3.7)

  31. anonymous
    • 5 years ago
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    well, it depends it converges absolutely on (3,7) and conditionally at 3

  32. anonymous
    • 5 years ago
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    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

  33. anonymous
    • 5 years ago
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    by the alternating series test

  34. anonymous
    • 5 years ago
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    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

  35. anonymous
    • 5 years ago
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    hmm yes so (3,7) or [3.7)

  36. anonymous
    • 5 years ago
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    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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