A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

what is the limit as n approaches infinity of 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

  • This Question is Closed
  1. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do you evaluate sum(i^9,i=1..n) do you know?

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please give a little more background. What class? Is this an infinite series?

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is there anything else to the question

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    looks like a geometric series

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^yeah

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so find the ratio

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have to think in terms of integrals to solve it.

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why?

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you need a power series representation and get the ratio from that

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the ratio is 2

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you want (1/n)^10 + (2/n)^10 + ...(

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i take that back

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the ratio of consecutive terms is not constant

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is an integral

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can yuo post the original question

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah thats what I asked

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh its a riemann sum

  20. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the limit is probably infinity

  21. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its not a decreasing function

  22. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no , if its a riemann sum its area under a curve

  23. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that means it converges

  24. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i just cant see the integral

  25. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well do you see that n is finite there

  26. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    uphill, in that expression

  27. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lim (n->oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9]

  28. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    n approaches infinity

  29. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes but each n is finite

  30. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is not a series

  31. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is a rieman sum

  32. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe you need to set up an inequality

  33. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    S sub n + the limit of the integral of that is > the limit (of stuff)

  34. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well the denominator is n^9

  35. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no you dont need that

  36. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lim (n->oo) 1/n[(1/n)^9 + (2/n)^9 + (3/n)^9 + ... + (n/n)^9] = 1/n [ (1^9 + 2^9 + 3^9 + ... n^9) / n^9 ]

  37. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you need to find the sum formula for sum k^9 as for k = 1 to n

  38. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    let me try and make the problem look more neat and type it again

  39. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    give me medal

  40. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i solve it

  41. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when you pull out the /n^9 isn't it just a ratio of leading coefficients at that point? then you just have 1/n which goes to 0?

  42. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    leading coefficients?

  43. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (L'pitals more or less)

  44. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well i dont think you can apply Lhopital;s easily

  45. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whichi is why u use leading coefficients

  46. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    we need to find a formula for 1^9 + 2^9 + ... n^9

  47. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a closed form in terms of n

  48. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay. i know formulas for powers 1,2,3 but not 9

  49. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and thanks you everyone for helping me out on this one. its pretty tough

  50. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait, can you state the original problem again

  51. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure. just a minute

  52. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{0}^{1}x^{9}dx=x^{10}/10, x=0..1=1/10\]

  53. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\lim_{n \rightarrow \infty} (1/n)[(1/n)^9+(2/n)^9+(3/n)^9+...+(n/n)^9]\]

  54. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  55. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    deltax=[b-a]/n=[1-0]/n

  56. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right that works

  57. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or you can do the limit proof directly , find the sum of k^9

  58. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you very much! i finally understand it. well more than i did beforehand anyways.

  59. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.