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Prove if sin(1/x) diverges or converges

  1. Oct 17, 2013 #1
    I want to prove if sin(1/x) diverges or converges as x approaches infinity. I know the answer is diverge because the function oscillates. I want to prove it more rigorously though, for my own well being and for the exam coming up. I am having trouble with the ratio test and I rewrote the expression as
    sin(1/x)[sin^2(1/x) +cos^2(1/x)] to see if it would help but that won't because if I set b equal to sin^2(1/x) +cos^2(1/x) that is not smaller than the original expression of course so I can't use the direct comparison test. I assume I will be using the limit comparison test but I need a little kick start. Thanks.
     
  2. jcsd
  3. Oct 17, 2013 #2

    jbunniii

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    Are you sure about that? What happens to ##1/x## as ##x \rightarrow \infty##?
     
  4. Oct 17, 2013 #3

    jbunniii

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    Do you know this inequality? ##|\sin(x)| \leq |x|##, valid for all ##x##.
     
  5. Oct 18, 2013 #4
    Sin(x) is a bounded function with an upper bound of 1 and a lower bound of -1. As values of x take values close to 0 then sin(x) takes values also close to 0. So in your case when x takes really big values close to infinity what does 1/x take values close to and what does that say for sin(1/x).
     
  6. Oct 18, 2013 #5
    Both of you are saying exactly what I thought at first. I thought the function would converge because the limit of 1/x as x approaches infinity is zero. Sine of zero is zero. So I concluded that the sum converges. When I looked at the answer in the back of the book it said the series was divergent. I believed this to be true against my initial intuition when I looked at a graph of the function it does oscillate as x approaches infinity, even when I made the viewing window quite large.
    jbunniii, yes I do know that inequality. All these things are the reason I don't see the answer. I need to see clarity in this problem.
     
  7. Oct 18, 2013 #6

    D H

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    What series are you talking about?

    The original post only asks about sin(1/x) as x tends to infinity. That has a well defined limit. If you are asking about something such as ##\sum_{n=1}^{\infty} \sin(\frac 1 n)##, that's a very different question than ##\lim_{x\to \infty} \sin(\frac 1 x)##.
     
  8. Oct 19, 2013 #7
    I'm sorry. Yes i'm talking about the sum you have written above from n=1 to infinity.
     
  9. Oct 19, 2013 #8

    D H

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    You should have said so in your original post! Be specific when you ask a question. Use the proper notation. We can't read your mind.

    Just because sin(1/x) converges to zero as x→∞ (which it does) does not mean that Ʃ sin(1/n) converges (which it doesn't).
     
  10. Oct 19, 2013 #9

    Dick

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    If x is very close to zero (like 1/n is for n large) then sin(x) is very close to being x. Look at the graphs near 0. Does that suggest a way to create a comparison series that is smaller but diverges?
     
    Last edited: Oct 19, 2013
  11. Oct 21, 2013 #10
    The graph near zero is very intense. This is because the function will experiences infinitely many oscillations as it gets closer to zero correct? This problem is confusing. I can't use the alternating series test so my only option would be comparison like you said, correct? So I want a function that will be undefined at zero and most likely involves a trig function that is less than sin(1/n) that I can show the limit as n approaches infinity of this new series diverges. Do I compare it to another sine function? Since it's not an alternating series I know can't use sin(1/(n+1)). Plus, this series converges.
     
  12. Oct 21, 2013 #11

    Dick

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    1/n is near 0 for large n. That's where you are concerned with an estimate of sin. I meant look at the graphs of sin(x) and x near 0.
     
  13. Oct 26, 2013 #12
    Sorry it took so long to reply. I believe I have the answer though. If I use the limit comparison test with 1/n as my series to compare I get the correct answer. I didn't know that you could use expressions within a trigonometric function as your series to compare.
     
  14. Oct 26, 2013 #13

    Dick

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    Any test is free game as long as it works. You could also do a direct comparison test with 1/(2n) since 1/(2n)<=sin(1/n) when n is large.
     
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