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

megaray26

Does this series converge or diverge?

  • one year ago
  • one year ago

  • This Question is Closed
  1. megaray26
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sum_{n = 1}^{\infty} \frac{ k \ln k }{ (k + 1)^3 }\]

    • one year ago
  2. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    there is no n in it, si its just a big ole constant

    • one year ago
  3. megaray26
    Best Response
    You've already chosen the best response.
    Medals 0

    oops i meant k = 1

    • one year ago
  4. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    :) my gut says taht the bottom is always bigger than the top and I would want to say it converges eventually. but I got no tests in mind to prove that

    • one year ago
  5. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    have you tried a ratio test?

    • one year ago
  6. megaray26
    Best Response
    You've already chosen the best response.
    Medals 0

    Well i have, but it doesn't look like it will lead to an easier solution. It just seemed to make the problem harder, unless there is some trick that I am missing while doing so.

    • one year ago
  7. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    http://www.wolframalpha.com/input/?i=sum+k*ln%28k%29%2F%28k%2B1%29%5E3%2C+k%3D1..inf says a comparison test has it converging

    • one year ago
  8. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    think we would compare it to 1/k^3 ? or 1/(k+1)^3?

    • one year ago
  9. megaray26
    Best Response
    You've already chosen the best response.
    Medals 0

    What i had thought of using was comparing to k / k^3, or 1/ k^2, and maybe try a limit comparison test, but that didn't exactly work out. Also, for those, I'm not exactly sure if the series is less than 1 / k^3, which is what I need for convergence, but I may be wrong.

    • one year ago
  10. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    a calculator would determine the bigger or less than stuff maybe \[\frac{1}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0\] \[1-k \ln k>0\] \[1>k \ln k\]

    • one year ago
  11. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    if we do a k up top we might be able to factor out a k first, but most of this is more art than science to me

    • one year ago
  12. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{k}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0\] \[k-k \ln k>0\] \[k(1- \ln k)>0\] might be able to work out something with that

    • one year ago
  13. amistre64
    Best Response
    You've already chosen the best response.
    Medals 1

    when k=1, 1(1-0)>0 when k=e, e(1-1)=0 http://www.wolframalpha.com/input/?i=y%3D1%2F%28x%2B1%29%5E3%2C+y%3Dx*ln%28x%29%2F%28x%2B1%29%5E3 hmm, the comparison seems to be smaller in both cases than the function we are looking at

    • one year ago
  14. exraven
    Best Response
    You've already chosen the best response.
    Medals 0

    the limit comparison test works, try\[1/k^{\frac{3}{2}}\]you will get\[\lim_{k \rightarrow \infty} \frac{k^{\frac{5}{2}}\ln k}{(k + 1)^3} = 0\]therefore the series converges

    • one year 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.