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Homework Help: Infinite series sin(1/n)/n ?

  1. Apr 13, 2010 #1
    Infinite series sin(1/n)/n ???

    1. The problem statement, all variables and given/known data
    does the series (sin (1/n)) / sqrt ( n ) converge or diverge? (series from n = 1 to infinity...)


    2. Relevant equations



    3. The attempt at a solution
    I thought that for this we could do a comparison of sin (1/n) to a finite number. Lets say we use the number -2. -2 is less than sin(1/n) for all n. With that being said, -2/ sqrt( n ) diverges, as it is a p series with p less than 1.
    But, this series actually converges... what did I do wrong?
     
  2. jcsd
  3. Apr 13, 2010 #2
    Re: Infinite series sin(1/n)/n ???

    Suppose instead of comparing sin(1/n) to a finite number, you compared it to a function. As n->infinity, what do you know about sin(1/n) and its relationship to some function that might be easier to work with?

    (By the way, the term comparison tests only work if the terms of both series are all positive. That's why you can't use -2 as a comparison term to show that your series diverges.)
     
  4. Apr 13, 2010 #3
    Re: Infinite series sin(1/n)/n ???

    You could probbably use the fact that

    [tex]|sin\theta|<\theta[/tex]

    Edit: I think this is basically what hgfalling was suggesting. That is [tex]sin(\theta)=O(\theta), \mbox{ or } sin(\theta)=\Omega (\theta) \mbox{ as } \theta\rightarrow 0. [/tex]
     
  5. Apr 14, 2010 #4
    Re: Infinite series sin(1/n)/n ???

    You can use the comparison tests
    since sin(1/n) is positive since the angle (1/n) is in the first quadratic for n=1,2,3,....

    To test it, you could use the limit comparison test with a p-series, can you do that ?
     
  6. Apr 14, 2010 #5
    Re: Infinite series sin(1/n)/n ???

    Show it converges faster than 1/x?
     
  7. Apr 14, 2010 #6
    Re: Infinite series sin(1/n)/n ???

    as n --> infinity, 1/n ---> 0. sin(0) = 0. You can literally say that because the value at infinity is 0, it converges.
     
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