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liliy

  • 3 years ago

absolute convergence , converge or diverge: n^2/2^n. sumation going from n=1 until infinity

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  1. anhkhoavo1210
    • 3 years ago
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    \[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]

  2. anhkhoavo1210
    • 3 years ago
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    Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]

  3. liliy
    • 3 years ago
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    how?

  4. anhkhoavo1210
    • 3 years ago
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    sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]

  5. anhkhoavo1210
    • 3 years ago
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    with \[u_n=\frac{n^2}{2^n}\] here

  6. liliy
    • 3 years ago
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    wats sr?

  7. liliy
    • 3 years ago
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    i dont understand the formula ur using

  8. anhkhoavo1210
    • 3 years ago
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    this is d'Alembert test

  9. anhkhoavo1210
    • 3 years ago
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    if this limit < 1, the series is converge

  10. liliy
    • 3 years ago
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    neverheard of that...aight

  11. anhkhoavo1210
    • 3 years ago
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    what is the test you studied?

  12. liliy
    • 3 years ago
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    we did power, ratio?

  13. anhkhoavo1210
    • 3 years ago
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    It seems to be a ratio test :-? may be differ name, display ratio test here

  14. liliy
    • 3 years ago
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    given sumation A sub n. let the ratio= lim n goes to infinity of absolute val of A sub n +1 / A sub n then if ratio<1 series Coverges Absloutely ratio>1, diverges ratio=1, try diff test

  15. anhkhoavo1210
    • 3 years ago
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    yes it is

  16. liliy
    • 3 years ago
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    so how do u write it out?

  17. anhkhoavo1210
    • 3 years ago
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    \[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]

  18. liliy
    • 3 years ago
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    howo does thatcancel out?

  19. anhkhoavo1210
    • 3 years ago
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    I don't understand what you say, but if this limit <1, you can deduce this series converges. Moreover, it is absolute converge because \[\frac{n^2}{2^n}>0\]

  20. liliy
    • 3 years ago
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    how do you get that second part of the limit?

  21. anhkhoavo1210
    • 3 years ago
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    \[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???

  22. liliy
    • 3 years ago
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    |dw:1336454779193:dw|

  23. anhkhoavo1210
    • 3 years ago
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    It means \[\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\] That's Ok?

  24. liliy
    • 3 years ago
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    ya...now wat?

  25. anhkhoavo1210
    • 3 years ago
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    you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]

  26. liliy
    • 3 years ago
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    o.m.g. yes! wow. sorry im so slow

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