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anonymous

  • 5 years ago

What is the radius and interval of convergence of the following series: (3x-4)^(n) --------- n=1 and is going to infinity (4n)^(n)

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  1. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty} (3x-4)^n / (4n)^n\]

  2. anonymous
    • 5 years ago
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    Usually the main method is to perform the ratio test on the series, and when you have the limit in terms of x, and use the geometric series boundaries -- that the absolute value of the ratio must be less than or equal to one -- and apply it to that limit (which is the ratio).

  3. Yuki
    • 5 years ago
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    @:Canyoudothis... I like he way how you drew the series lol

  4. anonymous
    • 5 years ago
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    ...is my help a bit vague? xD

  5. anonymous
    • 5 years ago
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    thanks yuki

  6. anonymous
    • 5 years ago
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    @quantummodulus yeah its kinda vague

  7. Yuki
    • 5 years ago
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    first you want to perform the ratio test

  8. anonymous
    • 5 years ago
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    Okay, first perform the ratio test and get the final value of the limit in terms of x.

  9. anonymous
    • 5 years ago
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    help me

  10. anonymous
    • 5 years ago
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    I dont know the answer to the ratio test :(

  11. anonymous
    • 5 years ago
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    Do you know how to perform it?

  12. Yuki
    • 5 years ago
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    the nth term is \[(3x-4)^n/(4n)^n\]

  13. anonymous
    • 5 years ago
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    is it a(n+1)/ a(n)

  14. anonymous
    • 5 years ago
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    Okay, now apply that to each n in your original series, dividing by the original series itself.

  15. anonymous
    • 5 years ago
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    ...it might take a while (it's not a very clean process, usually) but once you do it should be in terms of x.

  16. anonymous
    • 5 years ago
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    so would i get (4n)^(n) / (3x-4)^(n) ?

  17. anonymous
    • 5 years ago
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    oh no, hold on

  18. anonymous
    • 5 years ago
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    (3x-4)^(n+1) (4n)^(n) ------------ * ---------- ? (4n+1)^(n+1) (3x-4)^(n)

  19. Yuki
    • 5 years ago
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    exactly

  20. anonymous
    • 5 years ago
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    soooooo? um now what haha

  21. Yuki
    • 5 years ago
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    (3x-4)^n+1 can be represented as (3x-4)*(3x-4)^n

  22. anonymous
    • 5 years ago
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    so the one on the top and the other one would cancel?

  23. Yuki
    • 5 years ago
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    yep

  24. anonymous
    • 5 years ago
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    (3x-4)(4n)^(n) ------------ (4n+1)^(n+1)

  25. Yuki
    • 5 years ago
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    now that can be written in the form\[\lim_{n \rightarrow \infty}(3x-4) * (4n)^2/(4n+1)^2\]

  26. anonymous
    • 5 years ago
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    okay thanks for the help :) ttyl

  27. Yuki
    • 5 years ago
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    since you will get |3x-4| < 1, all you have to do is to solve for the inequality. however, the endpoints are always something you have to check by hand, so you need to plug in the two endpoints to the original series to see if it converges or not, ok? let me know if you need more help.

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