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Convergence of a Series

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Ʃ[itex]^{∞}_{n=1}[/itex] [itex]\frac{sin(1/n}{\sqrt{n}}[/itex]

    2. Relevant equations


    3. The attempt at a solution
    lim[itex]_{n→∞}[/itex] [itex]\frac{sin(1/n}{\sqrt{n}}[/itex]= lim[itex]_{n→∞}[/itex] [itex]\frac{-2cos(1/n)}{n^{3/2}}[/itex] with l'hospital rule = 0

    since lim[itex]_{n→∞}[/itex]=0 therefore the series is convergent

    Do you think I did this right?
     
  2. jcsd
  3. Nov 24, 2011 #2

    HallsofIvy

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    I can't imagine why you would use L'Hopital for that: sin(2/n) lies between -1 and 1 for all n while the numerator gets larger and larger. Of course, the limit is 0.


    But that doesn't help you. [itex]a_n[/itex] going to 0 is a necessary condition for convergence of the series [itex]\sum a_n[/itex], it is not a sufficient condition.

    If [itex]a_n[/itex] does NOT go to 0, then [itex]\sum a_n[/itex] does not converge but there exist series in which [itex]a_n[/itex] converges to 0 but the series [itex]\sum a_n[/itex] does not converge.

    For example, [itex]\sum 1/n[/itex] clearly has 1/n going to 0 but the series does not converge.
     
  4. Nov 24, 2011 #3
    Consider doing a limit comparison test. What well known limit involving sine might help here?
     
  5. Nov 24, 2011 #4
    Thank you for quick reply. Okay now I understand that the proof is not sufficient. Can I instead use the fact that you said -1 < sin(1/n) < 1 and the graph shows this:

    Graph attached.

    can I say since [itex]\frac{sin(1/n}{\sqrt{n}}[/itex] < [itex]\frac{1}{\sqrt{n}}[/itex] then use the p-series to show convergence or divergence.
    or I can think of continuing what I was doing and prove that n^(3/2) converges then by definiton -2con(1/n) should also converge.
     

    Attached Files:

  6. Nov 24, 2011 #5

    Ray Vickson

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    You are doing way too much work. When n is large, 1/n is small and sin(1/n) is 1/n + O(1/n^2), so the nth term is 1/n^(3/2) + O(1/n^(5/2)). Convergence is then obvious.

    RGV
     
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