A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

how do you determine if a series is absolutely ,conditionally convergent or divergent for a problem like this sigma (-1)^n/2^n

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

    The absolute series is a geometric series with ratio 1/2. Hence it is absolute convergent.

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

    how would you test if it is conditionally convergent

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

    since terms go to zero and it alternates it is conditionally convergent.

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

    would you use the latenating test along with another test?

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

    I believe that this is an alternating series, and is convergent. its absolutely convergent if its still convergent using |(-1)^b/2^n| with absolute value operator otherwise its conditionally convergent

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

    if it alternates, which this does because of the \[(-1)^n\] then all you need for conditional convergence is that the terms go to zero.

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

    okay so you use the alternating test along with another test to see is it is absolutely or conditionally

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

    to show absolute convergence you just check \[\sum\frac{1}{2^n}\] which as watchmath said, is a geometric series with \[r=\frac{1}{2}\]

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

    ok i sense it is not clear from your question. absolute convergence is stronger than conditional. it is converges absolutely then it certainly converges conditionally

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

    for an alternating series all you have to check for conditional convergence is that the terms go to zero.

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

    This is an example of series that conditionally convergent \(\sum (-1)^n\frac{1}{n}\) The absolute value series is divergent since it is a harmonic series. But the original series is convergent by alternating series test. So it is always better to check it for the absolute series first.

  12. 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.