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Squeeze Theorem for Sequences Help?

Hi can someone help me with this calculus question?

Find the limit as n approaches infinity using the squeeze theorem for (sin(1/n))/(sin(n))

Thank you so much!!

2 Answers

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  • ?
    Lv 6
    2 months ago

    Might you perhaps have meant (sin(1/n))(sin(n))? Because then

    |sin(1/n)·sin(n)|

    = |sin(1/n)|·|sin(n)|

    ≤ |1/n|·1,

    which →0 as n→∞. Hence so will (sin(1/n))(sin(n)) by the squeeze theorem.

    Unfortunately the squeeze theorem cannot be applied to the sequence (sin(1/n))/(sin(n)) because the sequence doesn't converge. The behavior of 1/sin(n) is the culprit. Namely, 1/sin(n) is not O(n), which is what you would need for the sequence to converge. Rather, for infinitely many n you'll have |1/sin(n)| > n²/π.

    The reason has to do with the fact that π, thus also 1/π,  is irrational. As such, it's well-known (and not too hard to prove) that there are infinitely many pairs of integers m and n>0 such that

    |1/π – m/n| < 1/n²

    or equivalently

    |n – mπ| < π/n².

    So for these n and m you'll have

    |sin(n)|

    = |sin(n – mπ)| (since sin is periodic modulo π)

    ≤ |n – mπ| (since |sin(x)| ≤ |x|)

    < π/n².

    But then, once these n get sufficiently large, you'll have

    |sin(1/n) / sin(n)|

    = |sin(1/n)| / |sin(n)|

    > |sin(1/n)|·(n²/π)

    ≥ |(1/2)(1/n)|·n²/π,

    because |sin(x)| ≥ |x/2| for all sufficiently small x. Thus, for infinitely many positive integers n you'll have

    |sin(1/n) / sin(n)| > n/(2π),

    proving the sequence doesn't converge.

  • 2 months ago

    There's no reason to think that the expression converges. Let's look at what's going on with that expression.

    You have 0 <= sin(1/n) <= 1/n

    But sin(n) can get very close to 0.  You may be lead to believe that the limit will go to 0, and it does get very close. It will never be = zero because n will never be a multiple of π. But when you divide the positive numerator by a very small denominator, the answer can get large, and there will always be some n where the ratio is not close to 0.

    Consider these examples, where sin(n) gets smaller and smaller, enough to keep the ratio away from 0:

    n    sin(1/n)     sin(n)     ratio

    1    0.841470985     0.841470985     1

    3    0.327194697     0.141120008     2.318556392

    22    0.045438895     (0.008851309)    -5.133578908

    333    0.003002998     (0.008821166)    -0.34043101

    355    0.002816898     (0.000030144)    -93.44694343

    103993    0.000009616     (0.000019129)    -0.502685088

    104348    0.000009583     (0.000011015)    -0.870022884

    208341    0.000004800     0.000008114     0.591525159

    312689    0.000003198     0.000002901     1.102515423

    833719    0.000001199     0.000002313     0.518584816

    1146408    0.000000872     (0.00000059)    -1.484041415

    4272943    0.000000234     (0.00000055)    -0.425835951

    Sorry this isn't a proof. It intuitively rests on the fact that one can always find values for n where the fractional part of n/π, which is what determine the value of |sin(n)|m us going to be very small. There's no guarantee that the numerator, sin(1/n), will be much close to 0, so the ratio will bounce around. While "most" of the values of the ratio may go to 0, there will always be outliers that are bigger than any epsilon you might choose in trying to show that the limit = 0.

    Look at the graph below, which shows the values for the n=1 to 1300. For clarity, I've limited the range of values to ±0.2.  You can see how the bulk of the numbers do get close to 0, but there are always outliers. It's those outliers that are the reason that the correct answer is, it doesn't converge.

    Attachment image
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