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Determining Convergence of ((sin(n))^4)/(1+n^2)

  1. Apr 2, 2012 #1
    Suppose I want to determine the convergence of ((sin(n))^4)/(1+n^2) using limit comparison test. I divide it by 1/(1+n^2). All that remains is (sin(n))^4. Now as the limit goes to infinty, the range of values (sin(n))^4 can give is 0 to 1. Now it gives many more values above zero then at zero, so is that why we declare it a convergent series?
  2. jcsd
  3. Apr 3, 2012 #2
    As you point out, (sin(n))^4 is always between 0 and 1.


    0 <= ((sin(n))^4)/(1+n^2) < 1/(1+n^2) < 1/n^2 for all n>0.

    1/n^2 converges. So we're home and dry.
  4. Apr 3, 2012 #3

    Well, you almost have it : [itex]\frac{\sin^4(n)}{1+n^2}\leq\frac{1}{n^2}[/itex] , which is clearly convergent.

  5. Apr 3, 2012 #4
    Lots of Thanks.
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