Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

liliy Group Title

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

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sum_{n=1}^{\infty} {\frac{n^2}{2^n}}\]

    • 2 years ago
  2. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Check it with the following test : \[\lim_{n\to \infty} {\frac{u^{n+1}}{u^n}}\]

    • 2 years ago
  3. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how?

    • 2 years ago
  4. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    sr, it is \[\lim_{n\to \infty} {\frac{u_{n+1}}{u_n}}\]

    • 2 years ago
  5. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    with \[u_n=\frac{n^2}{2^n}\] here

    • 2 years ago
  6. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    wats sr?

    • 2 years ago
  7. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont understand the formula ur using

    • 2 years ago
  8. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    this is d'Alembert test

    • 2 years ago
  9. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    if this limit < 1, the series is converge

    • 2 years ago
  10. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    neverheard of that...aight

    • 2 years ago
  11. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    what is the test you studied?

    • 2 years ago
  12. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    we did power, ratio?

    • 2 years ago
  13. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It seems to be a ratio test :-? may be differ name, display ratio test here

    • 2 years ago
  14. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • 2 years ago
  15. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes it is

    • 2 years ago
  16. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so how do u write it out?

    • 2 years ago
  17. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\lim_{n\to \infty} {\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}}=\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}<1\]

    • 2 years ago
  18. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    howo does thatcancel out?

    • 2 years ago
  19. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  20. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    how do you get that second part of the limit?

    • 2 years ago
  21. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\lim_{n\to \infty} {\frac{(n+1)^2}{2n^2}}=\frac{1}{2}\]???

    • 2 years ago
  22. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1336454779193:dw|

    • 2 years ago
  23. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It means \[\frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\] That's Ok?

    • 2 years ago
  24. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ya...now wat?

    • 2 years ago
  25. anhkhoavo1210 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you see \[\frac{2^n}{2^{n+1}}=\frac{1}{2}\]

    • 2 years ago
  26. liliy Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    o.m.g. yes! wow. sorry im so slow

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.