A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Hitaro9

  • one year ago

Let ak >=0. Let the series ak diverge. Prove that the series ak/(1+ak) diverges

  • This Question is Closed
  1. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    There's nothing super simple like a comparison that can be made as far as I can see. How would one go about proving this sort of question? Do you get in to definitions of series divergence or?

  2. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\sum_{k=1}^{\infty}a _{k}\div(a _{k}+1)\]

  3. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1442920394346:dw|

  4. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hey, last problem. Thanks for your help on the last one @IrishBoy123. I'm just not sure what the general strategy is here.

  5. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    compare like terms in each series

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I think you may simply use the limit test : \(\lim\limits_{k\to\infty} \dfrac{a_k}{1+a_k} \ne 0 \implies \sum\limits_{k=0}^{\infty}\dfrac{a_k}{1+a_k} \text{ diverges}\)

  7. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Huh. How do we know ak / 1+ak doesn't go to zero? Couldn't ak go to zero and not converge (ie, 1/k)? Or do I have something off here?

  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, tad confused here too we do not know that \(\lim\limits_{k\to\infty} a_k \ne 0\) as \(a_k\) might have failed that test and still diverged, ie it is poss that \(\lim\limits_{k\to\infty} a_k = 0\) plus \(\dfrac{\dfrac{a_k}{1+a_k}}{a_k} = \dfrac{1}{1 + a_k}\) and \(a_k >=0\) so the comparison test draws a blank

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

    $$ \sum \frac{\frac 1 n }{1 + \frac 1 n }= \sum \frac {1}{n+1 }= \infty $$

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    If \(\lim\limits_{k\to\infty} a_k = 0\), then limit comparison test does the job : \(\lim\limits_{k\to\infty} \dfrac{\dfrac{a_k}{1+a_k}}{a_k} = 1\) therefore both the series, \(\sum a_k \) and \(\sum\dfrac{a_k}{1+a_k}\) will have same fate.

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    looks im just repeating what irishboy has said

  12. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    no, i chickened out of your second conclusion. can't remember why now :-)

  13. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So why does 1/1+ak -> that it diverges exactly? Sorry if this is obvious.

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    hmm question is not so clear... could you elaborate a bit

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

    limit comparison test: http://prntscr.com/8j65pv

  16. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh wait right. Thank you Jayz. Okay then that makes sense.

  17. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So basically we have two cases, one for which ak -> 0 and one for otherwise? If I understand correctly?

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

    :) no problem. ganeshie and irishboy did the hard work

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

    oh you mean if a_k converged, then yes

  20. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay yeah, my question was poorly phrased. I think I understand now. I'll go through and make it more wordy with respects to the limit test

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

    We get two statements for free using limit comparison. 1. Let ak >=0. Let the series ak diverge. Then the series ak/(1+ak) diverges 2. 1. Let ak >=0. Let the series ak converge. Then the series ak/(1+ak) converges

  22. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks all of you for the help. I appreciate it.

  23. hitaro9
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Alright, I gotta get to class, thanks again for all your help. You're all amazing people.

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

    Just to go over the argument one more time to make this proof airtight. We are given ak>=0 and ∑ak diverges. Show that ∑ak / (1 + ak) diverges. proof: there are two cases lim ak ≠0 or lim ak = 0 1. If lim ak ≠0 , then since ak>=0 we have lim ak/(1 + ak) ≠0 and thus ∑ak / (1 + ak) diverges by divergence test. 2. If lim ak = 0 then lim [ ak / (1+ ak) ] / ak = lim ( 1 / (1 + ak) = 1 / (1 + 0) = 1 . by limit comparison test ∑ak / (1 + ak) diverges since ∑ak diverges

  25. 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.