A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

use the limit comparison test to determine whether sigma 2^n/3^n-1 converges or diverges. I solved the problem but i am stuck at this point lim n-->inf 3^n/3^n-1

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

    I will just continue from where you're stuck, assuming all you have done so far is right.

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

    \[\lim_{n \rightarrow \infty}{3^n \over 3^n-1}=1\] (since they both grow equally).

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

    i do not understand how did u got one

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

    Factor out 3^n from the numerator and denominator.

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

    \[\frac{3^n}{3^n-1}=\frac{3^n}{3^n(1-\frac{1}{3^n})}=\frac{1}{1-\frac{1}{3^n}}\]

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

    oh okay i see it

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

    The limit as n approaches infinity is \[\frac{1}{1-0}=1\]

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

    Hey Lucas :)

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

    Hi Anwar :)

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

    i have a question in a problem like this sigma ( 1/2n-1-1/2n) what do i compare it to

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

    brb...one sec

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

    k

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

    Can you write that expression in the equation editor, because going off what you've written, your expression looks equal to -1.

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

    ?

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

    \[\sum_{1}^{\infty} (1/ 2n-1-1/2n)\]

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

    how do you write a fraction

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

    There's a box in the equation editor, or you can type, frac{}{} and put your numerator stuff in the first box, denominator stuff in the second.

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

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

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

    Okay...now I can look at it :)

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

    :)

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

    i forgot we have to use the limit comparison test here

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

    I think you'd be best to combine your fractions and use the limit comparison on the result. Your summand combines as\[\frac{1}{4n^2-2n}\]

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

    This is asymptotically equivalent to 1/(4n^2), which means you can eyeball the fact you could compare it to 1/n^2 for limit comparison.

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

    \[\frac{\frac{1}{4n^2-2n}}{\frac{1}{n^2}}=\frac{n^2}{4n^2-2n}=\frac{1}{4-2/n} \rightarrow \frac{1}{4}\]as n goes to infinity. The limit is positive and finite, so by the limit comparison test, since Sum(1/n^2) converges, your series does too.

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

    hey i got the same things i did along with u thank u for ur help

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

    you're welcome

  27. 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.