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Math Help - sine sequence

  1. #1
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    sine sequence

    Hello!

    What is the best way to determine if a sequence like a_n = \frac{sin(n^2)}{n} converges or diverges?

    And how do we find its limit (the sin(n^2) on the numerator confuses me). Thanks in advance...
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  2. #2
    MHF Contributor Calculus26's Avatar
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    A good trick that sometimes works when sines and cosines are involved is the squeezing thm

    -1/n < sin(n^2)/n < 1/n

    since lim-1/n and lim 1/n converge to 0 so does sin(n^2)/n

    Makes Sense since you can think of the continuous curve sin(x)/x

    as a sine curve whose amplitude 1/x decays to 0
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  3. #3
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    So, the sequence converges to 0.

    How come you chose to squeeze sin(n^2)/n between 1/n ? Why didn't you squeeze it between say, - n^2/n and n^2/n ?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    The main reason is we have to squeeze it between 2 things which converge to the same limit. For sines and cosines their ranges are

    -1 to 1 which is what makes the trick work.

    yes -n^2/n < sin(n^2)/ n < n^2

    But one limit is neg inf and the other pos inf

    Which means we can only conclude the limit is somewhere between - inf and pos inf which means we know nothing about the limit.
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  5. #5
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    can we use Cauchy's method

    At first we check whether |sin(n^2) / n| converges or diverges ,

    int {from 1 to oo} sin(x^2)/x dx
    Sub u = x^2 => du = 2x dx
    the integral becomes 1/2 int { from 1 to oo } sin(u) / u du
    We know that int { from 0 to oo } sin(u) / u du converges to pi/2
    so this integral 1/2 int { from 1 to oo } sin(u) / u du converges

    Since the summation of |sin(n^2) / n| converge ,
    sin(n^2) / n also converges too
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  6. #6
    MHF Contributor Calculus26's Avatar
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    Remember if the corresponding series converges it must converge to 0.

    In your example
    Not only does sin(n^2) converge it must converge to 0

    You can always show a sequence converges to 0 by showing the corresponding series converges.

    However in practice I've found it not to be the best first choice.

    One notable exception is the sequence an = e^n/ (n!)
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