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Homework Help: Series Comparison Test

  1. Dec 9, 2017 #1
    1. The problem statement, all variables and given/known data
    ##\sum _{n=0}^{\infty }\:\sin \left(\frac{1}{n}\right)##

    2. Relevant equations


    3. The attempt at a solution
    Can I try comparison test by
    ##\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)##

    since ##\left(\frac{1}{1+n}\right)## diverges also ##sin\left(\frac{1}{n}\right)## diverges ?
     
  2. jcsd
  3. Dec 9, 2017 #2

    Ray Vickson

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    What do YOU think?
     
  4. Dec 9, 2017 #3
    Its true
     
  5. Dec 9, 2017 #4

    Ray Vickson

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    Right, and you should be able to explain why that is.
     
  6. Dec 9, 2017 #5
    I already explained it ? Like how
     
  7. Dec 9, 2017 #6

    Mark44

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    No, you didn't explain it -- you just asserted the inequality with no explanation of why it must be true.
     
  8. Dec 9, 2017 #7
    ##\left(\frac{1}{1+n}\right)<sin\left(\frac{1}{n}\right)## well I am not sure...I put some numbers and tested.

    I thought the number ##\frac{1}{n}## and its relationship with ##\sin \left(\frac{1}{n}\right)##

    Then suddenly I thought ##\frac{1}{\:n+1}##

    I dont know how to do an exact prove in this case.
     
  9. Dec 9, 2017 #8

    Ray Vickson

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    It actually does not matter for small values of ##n##. As long as you have ##\sin (1/n) > k/n## for all sufficiently large ##n## and for some positive constant ##k##, that is really all you need.

    Anyway, that is not really what I was referring to. I was referring to your assertion that ##a_n > b_n > 0## and ##\sum b_n## divergent implies that ##\sum a_n## is also divergent.
     
  10. Dec 9, 2017 #9

    vela

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    Which is that ##\sin \frac 1n < \frac 1n##.

    So you have ##\frac 1{n+1} < \frac 1n##.

    You have two quantities that are less than ##1/n##. It's not clear that you can conclude that one is less than the other, i.e., ##\frac 1{n+1} < \sin \frac 1n.##

    This can suggest that perhaps you're on the right track, but as you know, you need to show it for the general case. But it could also be that you're doing things the hard way. Perhaps a different convergence test would work here. You need to experiment with different ideas to learn what works and what doesn't and to develop your intuition in solving these problems.
     
  11. Dec 9, 2017 #10
    Yeah I know it was a long shot..theres also limit test which works but It doesnt come to my mind
     
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