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Homework Help: Help needed badly,convergence of series

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data
    given this find if it converge or diverge.

    [tex]\sum^{n=0}_{infinity} sin(\frac{1}{n^2})[/tex]

    sin (1/n^2) as n goes from zero to infinity.


    2. Relevant equations



    3. The attempt at a solution

    i tried

    sin(1/n^2) <= sin(1/n) ~ 1/n

    by squeeze theorem

    0 <= sin(1/n^2) <= 1/n

    as the limit goes to infinity the 1/n goes to zero thus sin(1/n^2) goes to zero, does that mean it converge
     
  2. jcsd
  3. Apr 19, 2010 #2

    Mark44

    Staff: Mentor

    Your LaTeX code was almost right. Click on my version to see what I did.
    [tex]\sum_{n=1}^{\infty} sin(\frac{1}{n^2})[/tex]

    BTW, the index n really should start at 1, not 0, since the expression being summed is undefined at n = 0.

    In a series
    [tex]\sum_{n=0}^{\infty} a_n[/tex]
    If
    [tex]\lim_{n \to \infty} a_n~=~0[/tex]
    then you really don't know anything about the series. For example, lim an = 0 for both of the following series, but the first one diverges and the second converges.

    [tex]\sum_{n=0}^{\infty} \frac{1}{n}[/tex]

    [tex]\sum_{n=0}^{\infty} \frac{1}{n^2}[/tex]
     
  4. Apr 19, 2010 #3

    Mark44

    Staff: Mentor

    This was a good idea - sin(1/n) ~ 1/n
    It's not right on the money, but it's a start.

    The limit comparison test is good to know.
     
  5. Apr 19, 2010 #4
    would it be valid to compare it with
    [tex]
    \sum_{n=1}^{\infty} sin(\frac{1}{n^2})
    [/tex]

    since sine if always less than one (book is telling me its a non negative series)

    and very close to zero (when n is large) its almost like
    [tex]
    \sum_{n=1}^{\infty} sin(\frac{1}{n^2})
    [/tex]

    therefore
    [tex]
    \sum_{n=1}^{\infty} sin(\frac{1}{n^2})
    [/tex] coverage thus sine should
     
  6. Apr 19, 2010 #5

    lanedance

    User Avatar
    Homework Helper

    all your latex are the same?

    [tex] sin(\frac{1}{n^2}) < sin(\frac{1}{n}) [/tex]
    for all n>1 seemed like a good start

    then as [itex] sin(\frac{1}{n}) \approx \frac{1}{n} [/itex] for large n can you think of anything similar in form to [itex] \frac{1}{n} [/itex] but such that [itex] sin(\frac{1}{n}) [/itex] is less than it for all n?
     
  7. Apr 19, 2010 #6

    Mark44

    Staff: Mentor

    No, because you don't want to compare your series with itself. Why would you want to do that?
    ????
     
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