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Alternative examples, alternating series test

  1. Oct 22, 2015 #1
    Hey guys, this one is just for funnsies. So when dealing with an alternating series test, 3 requirements must be met, :
    u(sub n) ≥ u(sub n+1) for all n ≥ N, for some integer N
    u(sub n) → 0 as n → ∞.

    So I have been coming up with examples where of these are true, and one isnt. A little bit further, im not sure if this makes sense, but is possible to find an example for each where like, one diverges and one converges, or does that not make sense?
  2. jcsd
  3. Oct 22, 2015 #2
    Come on, come up with some examples yourself first! It's not hard to find some counterexamples! Tell us what you've got already, and I'll help you fill in the parts you don't have.

    (By the way, there are not always examples possible where one diverges and one converges. I can see one obvious condition where it wouldn't be the case).
  4. Oct 23, 2015 #3
    Oh right! forgot about that, havnt slept in waaayy to long. Alright so i think the sum of ((-1)^2n)/n wouldnt alternate (rule 1) but seems to follow the rest, sum of cos(pi)/n I think would alternate and go to zero. I am thinking the one that your thinking of, which ive had a hunch on, is u(sub n) ≥ u(sub n+1) for all n ≥ N, for some integer N for diverging, but i cant really put it into words or make sense of it,its just a hunch , probably due to my lack of sleep
  5. Oct 23, 2015 #4
    The one I'm thinking of is ##u_n \rightarrow 0##. If this is false, can the series ever converge?
    Last edited: Oct 23, 2015
  6. Oct 23, 2015 #5
    Well, couldn't it converge to someone other than 0?
  7. Oct 23, 2015 #6
    also, dont forget that u (Sub n) is in absolute values, sorry forgot to put that
  8. Oct 23, 2015 #7
    Do you know the "divergence test"?
  9. Oct 23, 2015 #8
    Right, if it doesnt go to zero then it diverges because of the infinite sums. forgot about that. Yeah so i guess for that one it is impossible
  10. Oct 23, 2015 #9
    OK, so you have ##\sum 1/n## which does not alternate but does satisfy the other conditions. This series does not converge. Do you have one which does converge?
  11. Oct 23, 2015 #10
    ∑sin(n)/n I think would converge
  12. Oct 23, 2015 #11
    never mind it alternates haha
  13. Oct 23, 2015 #12
    ∑Sin(npi)/n right? because it would keep going to 0s and not alternating
  14. Oct 23, 2015 #13
    Does it alternate? I'm not so sure? Alternating means it gets positive and negative at each turn. Does this do this?
  15. Oct 23, 2015 #14
    You could probably write this easier.
  16. Oct 23, 2015 #15
    I just tried to plug it into my calculator and it denied me. My instincts tell me it doesnt. but I put down sin(3pi/4)= negative and sin(3pi/4)= still negative, and before pi its positive
  17. Oct 23, 2015 #16
  18. Oct 23, 2015 #17
    Again, doesn't seem alternating to me...
  19. Oct 23, 2015 #18
    Hang on, would ∑1/n work? because the absolute value of the next term has to be greater then the last right?
  20. Oct 23, 2015 #19
    Here is one ##\sum_n 2^{-n}##.
  21. Oct 23, 2015 #20
    No, the sequence needs to be decreasing. Every next term needs to have smaller absolute value.
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