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Math Help - Proving a trigonometric sequence converges: {n(sin(1/n))}

  1. #1
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    Question Proving a trigonometric sequence converges: {n(sin(1/n))}

    We're studying for a test in Calc II, and are stumped on what must be a relatively simple problem:

    Determine if the following sequence is convergent or divergent:
     \lbrace n \sin (\frac{1}{n}) \rbrace

    Our best attempts to solve this have been for not.

    We've tried the Squeeze Theorem, saying that since we're dealing with all positive terms (as n starts at 0 and goes to infinity) we can squeeze it like so:

     0 \leq n \sin (\frac{1}{n}) \leq 1

    But that doesn't squeeze to anything in particular. In this case we know the answer is supposed to be that it converges with  \lim_{n \to \infty} \lbrace a_{n} \rbrace = 1 .

    We can't use L'Hospital's Rule either (if I'm not mistaken).

    Any thoughts? How to we prove it's convergent with a lim = 1?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by laughfactory View Post
    We're studying for a test in Calc II, and are stumped on what must be a relatively simple problem:

    Determine if the following sequence is convergent or divergent:
     \lbrace n \sin (\frac{1}{n}) \rbrace

    Our best attempts to solve this have been for not.

    We've tried the Squeeze Theorem, saying that since we're dealing with all positive terms (as n starts at 0 and goes to infinity) we can squeeze it like so:

     0 \leq n \sin (\frac{1}{n}) \leq 1

    But that doesn't squeeze to anything in particular. In this case we know the answer is supposed to be that it converges with  \lim_{n \to \infty} \lbrace a_{n} \rbrace = 1 .

    We can't use L'Hospital's Rule either (if I'm not mistaken).

    Any thoughts? How to we prove it's convergent with a lim = 1?
    Use the substitution n=\frac{1}{m}. As n\to\infty, m\to0, so now you have \lim_{m\to0}\frac{\sin m}{m}=1.
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  3. #3
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    Perfect! Worked like a charm. Thanks a bunch!
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    Use the substitution n=\frac{1}{m}. As n\to\infty, m\to0, so now you have \lim_{m\to0}\frac{\sin m}{m}=1.
    Edit: Nevermind horrible question.
    Would this be a "decent" way of how to approaches 1?

    \lim_{n\to\infty}nsin(\frac{1}{n})

    \lim_{n\to\infty}sin(\frac{n}{n})

    \lim_{n\to\infty}sin(1) = 1
    Last edited by RockHard; November 22nd 2009 at 12:21 PM.
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  5. #5
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    nsin(\frac{1}{n}) \neq sin(\frac{n}{n})...

    And also, sin(1) is definitely not equal to 1 :P
    Last edited by Defunkt; November 22nd 2009 at 02:19 PM.
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  6. #6
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    RockHard, good attempt but it doesn't quite work that way. The problem is that when we look at a limit problem like this we have to a evaluate first if we can just take the limit right away, or two if we need to rewrite the problem before we can take the limit, or three if we need to use a rule like L'Hospital's Rule (which involves taking the derivative of f(x) over the derivative of g(x)).

    Here we have a scenario which is unique because as we try to take the limit right away we end up with an indeterminate form with  \lim_{n \to \infty} n \sin (1/n) = \infty * 0 and we can't tell who wins, infinity or zero. Thus we consider algebraic manipulations or L'Hospital's Rule but nothing works except for substitution.

    The problem with what you did is that the first equation  \lim_{n \to \infty} n \sin(\frac{1}{n}) \neq \lim_{n \to \infty} \sin(\frac{n}{n}) . The second is essentially  \lim_{n \to \infty} \sin(1) \approx .8414 , not 1.

    Substitution works because it maintains all the mathematical elements of the first problem, it just allows us to get ourselves out of a tricky situation.

    Thus, let  m = \frac{1}{n} , note that as n approaches infinity m approaches 0... Likewise, solving for n...  n = \frac{1}{m} Plugging both (which are equivalent) into our initial equation yields  \frac{1}{m} * \sin(m) = \lim_{m \to 0} \frac{\sin(m)}{m} this, being essentially f(x) over g(x) allows us to use L'Hospital's Rule. We take the derivative of top and bottom...  \lim_{m \to 0} \frac{cos(m)}{1} = 1 because as m goes to 0, cos(0) goes to 1.
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  7. #7
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    Gee, This is why I hate limits sometimes, You never know what to do half the time, I would have never though of using that, how do you know what to use?
    Last edited by mr fantastic; November 23rd 2009 at 02:59 AM.
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  8. #8
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    lol. I'm with ya. Limits problems are a pain. My experience indicates the following approach:

    1)Can the limit be taken immediately (without resulting in an indeterminate form)?

    2)If not, with algebraic manipulation (say, dividing the top by the highest power of the bottom) can it be simplified into a form which, when the limit is taken, yields a result that is not indeterminate?

    3) If after considering all possible algebraic manipulations you still would get an indeterminate answer then use L'Hospital's Rule and, if possible after taking the derivative of the top and bottom, evaluate the limit.

    4) If after using L'Hospital's Rule you still have an indeterminate form, use L'Hospital's Rule again until you get an equation that does not yield an indeterminate answer.

    Hopefully this helps. It's kind of hard to describe my approach, but I think that's about how I work on them.
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