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Comparison test for series convergence (trig function)

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data

    Use a comparison test to determine whether this series converges:

    [tex] \sum_{x=1}^{\infty }\sin ^2(\frac{1}{x}) [/tex]


    2. Relevant equations


    3. The attempt at a solution

    At small values of x:

    [tex] \sin x\approx x [/tex]
    [tex] a_{x}=\sin \frac{1}{x} [/tex]
    [tex] b_{x}=\frac{1}{x} [/tex]
    [tex] \lim \frac{a_{x}}{b_{x}}=\frac{\sin \frac{1}{x}}{\frac{1}{x}}=1 [/tex]

    Since 1/x diverges so does sin(1/x).

    Can I use this same method to solve the question above? ( i.e. sin^2(1/x) )
     
  2. jcsd
  3. Jan 21, 2017 #2
    Why not just try out that same method? What can you compare sin2(1/x) to?
     
  4. Jan 21, 2017 #3
    I guess what I'm not sure about is if this is true:

    [tex] \sin^2 x\approx x^{2} [/tex]

    Can I just make that assumption?
     
  5. Jan 21, 2017 #4
    No, reckless assumption often leads to errors. Do you recall that special trigonometric limit that involves sin? It involves a limit where x tends to 0 but with a clever use of substitution it may help you.
     
  6. Jan 21, 2017 #5
    Hmm, what about this, if:

    [tex] \sin x\approx x [/tex]

    then

    [tex] \sin^2 x\approx x^{2} [/tex]

    Since I'm just squaring both sides it should still be true right?

    Then I can go:

    [tex] a_{x}=\sin ^2\frac{1}{x} [/tex]
    [tex] b_{x}=\frac{1}{x^2} [/tex]

    It does work out this way, but what function did you have in mind to compare it to?
     
  7. Jan 21, 2017 #6
    You shouldn't use approximations as they don't have a place in mathematical problems of this nature and they don't mean anything without specifying a tolerance for error. sin(pi) is approximately equal to pi like 5 is approximately equal to 2

    The limit I was referring to is lim x --> 0 (sinx/x) = 1. Try and figure out its application to this problem.
     
  8. Jan 21, 2017 #7
    [tex] \sin x\leqslant 1 [/tex]
    [tex] \frac{\sin x}{x^{2}}\leqslant \frac{1}{x^{2}} [/tex]
    [tex] \frac{\sin^2 x}{x^{2}}\leqslant \sin(\frac{1}{x^{2}}) [/tex]

    Am I on the right track here?

    Sorry, I'm not really sure how sin(x)/x relates to sin^2(1/x) in terms of which is greater than the other. How can I figure that out?
     
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