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Stuck with a limit comparison test problem.

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the Convergence or Divergence of [tex]\sum_{n=1}^\infty\left(sin(1/n)\right)[/tex]

    2. Relevant equations
    limit comparison test



    3. The attempt at a solution
    I don't know what to compare this series to, I tried the harmonic series to get sin(1/n)/(1/n) = nsin(1/n) which is just an infinity that oscilates. Which I believe to be divergent.
    I've been really confused about a lot of comparisons because i'm only aware of a handful of series that converge or diverge the harmonic series (1/n) another series (1/n^2) which converges to pi2/6 p: series which if p > 1 it converges if 0 < p < 1 the p series diverges, geometric series which converge according to the formula a1 / (1 - r) if 0 < |r| < 1 and diverges if |r| > 1 And the Telescoping Series which converge according to the terms that cancel out in the set of real numbers it generates.
     
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  3. Jul 8, 2011 #2

    Dick

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    sin(1/n) doesn't really oscillate, does it? I think it approaches zero monotonically. You've got the right limit comparison there. sin(x)/x approaches 1 as x->0, doesn't it?
     
  4. Jul 8, 2011 #3
    it's not monotonic because of the fact that it does oscillate, even if it does approach zero. lim[x -> 0] sin(x)/x = 0 by squeeze theorem, yes. My question is can I let u = 1/n to get sin(u)/u to let that happen? that would diverge because 0 is not positive. I'm also worried there would be a problem about the fact that I substituted in the first place.
     
    Last edited: Jul 8, 2011
  5. Jul 8, 2011 #4

    Dick

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    Um, the series sin(1/1), sin(1/2), sin(1/3)... does NOT oscillate. Other series involving sin do oscillate, but this isn't one of them. And sin(x)/x doesn't approach 0 as x->0.
     
  6. Jul 8, 2011 #5
    -1 < sin(x) < 1
    -1/x < sin(x)/x < 1/x
    as x approaches infinity it does equal zero by squeeze theorem, as x approaches zero it approaches 1, and I would be taking the limit as x approaches infinity, not zero. Isn't that what I do with limit comparison tests, it's at least what i've been doing with the other ones.
     
  7. Jul 8, 2011 #6

    Dick

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    1/n doesn't approach infinity as n->infinity. It approaches 0! That's the limit you should be caring about, isn't it??
     
  8. Jul 8, 2011 #7
    nono you're right, i agree, the answer says it should be convergent though. I keep getting divergent....is this book wrong?
     
  9. Jul 8, 2011 #8

    Dick

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    If the books says it's convergent, then the book is wrong.
     
  10. Jul 8, 2011 #9
    okay, well then I have had the right answer for wrong reasons for about 2 hours.
    thanks for helping me clear this up, but my takeaway from this would be to ask how to figure out, when using direct comparisons and limit comparisons, what to compare the series to. My professor asks the question "what does an look like?" that is a hard question to answer when I have something like ln(n)/(n+1) because the comparison has nothing to do with ln(n)
     
  11. Jul 8, 2011 #10

    Dick

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    I'd say ln(n)/(n+1) 'looks like' ln(n)/n. Because the '+1' is unimportant. But I would still compare it with 1/(n+1) because I know ln(n)>1 for n>e and I know 1/(n+1) diverges. You could also compare it with ln(n)/(n+n) and show that diverges by an integral test. There's more than one way to do these problems.
     
  12. Jul 9, 2011 #11
    I'll keep that in mind, and come back for questions. Thanks, for your help.
     
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