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anonymous

  • 5 years ago

radius and interval of convergence of the series sqrt(n)*x^n as n goes from 1 to infinity?? please help..

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  1. anonymous
    • 5 years ago
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    the radius = 0, since x stands alone here :)

  2. anonymous
    • 5 years ago
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    wait

  3. anonymous
    • 5 years ago
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    Just give me a minute.

  4. anonymous
    • 5 years ago
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    ok.

  5. anonymous
    • 5 years ago
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    are you sure it's the sqrt of n and not the power of n?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    Well.. I am going to use the ratio test first to find the radius of convergence.

  8. anonymous
    • 5 years ago
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    you can't use the ratio test here anwar, since it's to the power of 1/n and not n, you'll have to use the root test here ^_^ and hey :)

  9. anonymous
    • 5 years ago
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    but since the question doesn't have any sort of exponential function, then "maybe" you can :)

  10. anonymous
    • 5 years ago
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    but it's going to be difficult

  11. anonymous
    • 5 years ago
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    Yes I can, and it won't be difficult. And hello! :)

  12. anonymous
    • 5 years ago
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    lol, root test is much easier here :)

  13. anonymous
    • 5 years ago
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    let: \[an = x^n\] so \[|an^{\frac{1}{n}} |= (|x^n)^{\frac{1}{n}} |= |x|\]

  14. anonymous
    • 5 years ago
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    and :\[0\le |x| < 1\] so the radius is = 1 in this case, I guess, correct me if I'm wrong please :)

  15. anonymous
    • 5 years ago
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    so since :\[0 \le |x| < 1\] then : \[-1 < x < 1\] is the interval of convergence ^_^

  16. anonymous
    • 5 years ago
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    I took this section yesterday morning during my calculus class lol

  17. anonymous
    • 5 years ago
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    so I might have a couple of mistakes ^_^" maybe

  18. anonymous
    • 5 years ago
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    just so its clear the series is..\[\sqrt{n}*x ^{n}\]

  19. anonymous
    • 5 years ago
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    oh >_<! I thought it was to the power of 1/n

  20. anonymous
    • 5 years ago
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    \[\left| {a_{n+1} \over a_n} \right|=\left| {\sqrt{n+1}. x^{n+1} \over \sqrt{n} x^n} \right|=\left| x \right|.\left| {\sqrt{n+1} \over \sqrt n} \right|\]

  21. anonymous
    • 5 years ago
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    then find the limit , oh my bad, I completely misunderstood the question >_<

  22. anonymous
    • 5 years ago
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    proceed with anwar's answer, it seems correct :)

  23. anonymous
    • 5 years ago
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    oh sorry for the confusion..ok yea i got till there..and then?

  24. anonymous
    • 5 years ago
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    continue anwar, you're on the right track :)

  25. anonymous
    • 5 years ago
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    So now take the limit as n goes to infinity: \[\lim_{n \rightarrow \infty}{\left| x \right|.\left| {\sqrt{n+1} \over \sqrt n} \right|}=\left| x \right|\]

  26. anonymous
    • 5 years ago
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    oh, almost the same LOL =P bleh

  27. anonymous
    • 5 years ago
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    Is the limit part clear?

  28. anonymous
    • 5 years ago
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    yess

  29. anonymous
    • 5 years ago
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    oh congrats on being a champion anwar ^_^

  30. anonymous
    • 5 years ago
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    Good. As I said before, the series is convergent when the limit value is <1. In our case, the radius of convergent is: \[R=\left| x \right|<1 \implies -1<x<1\]

  31. anonymous
    • 5 years ago
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    Thanks sstarica :)

  32. anonymous
    • 5 years ago
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    :) sure

  33. anonymous
    • 5 years ago
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    the radius of convergence = 1 and the interval is the one on top lol ^_^

  34. anonymous
    • 5 years ago
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    Now you should check the endpoints. check if the series converges at x=1 and x=-1.

  35. anonymous
    • 5 years ago
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    but i thought it had to be less than one to converge..

  36. anonymous
    • 5 years ago
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    oo nevermind. understood

  37. anonymous
    • 5 years ago
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    Well.. It converges "for sure" when it's less than 1, and diverges when it's greater than one. But the test fails when it's equal to 1. So we have to check it separately. Does that make sense?

  38. anonymous
    • 5 years ago
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    does x also have to be greater than -1?? y is that?

  39. anonymous
    • 5 years ago
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    Yeah. because we had IxI<1, that means x is between -1 and 1.

  40. anonymous
    • 5 years ago
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    can we always use the ratio test for this and if we do that isnt x always in between -1 and 1?

  41. anonymous
    • 5 years ago
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    I have an exam today, and still have a lot of material to study. So, I should get going now :(. I am sure sstarica will be here to help.

  42. anonymous
    • 5 years ago
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    oh ok then! thanks for all the help!

  43. anonymous
    • 5 years ago
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    No, you don't have always to use the ratio test. You can use other tests as well, especially the root test. It just depends on the series itself. The radius of convergence differs from series to another. GOOD LUCK!

  44. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=01LzAU__J-0

  45. anonymous
    • 5 years ago
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    thank u so much! good luck to you too!

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