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Homework Help: Determine whether the series converges or diverges problem

  1. Apr 5, 2010 #1
    given the series, determine whether the series converges or diverges:
    (E is my sigma)

    E sin(4/3n)

    originally i started with the integral test and i got up to the point where
    (integral) sin (4/(3x)) dx

    but i could not come up with how to solve the integral yet. (maybe by parts)

    I was talking to a friend and he said he solved it by direct comparison or limit comparison.
    Now he said he used 1/n (harmonic p-series diverges) to relate to the problem but im stuck on how to get rid of the sin. Once i figure that out that its all easy.

    Am i able to remove the sin out of the whole problem or what? not sure. any help would be great
     
  2. jcsd
  3. Apr 5, 2010 #2

    lanedance

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    Re: Series

    how about the magnitude of the nth term - does it go to zero for large n?
     
  4. Apr 5, 2010 #3

    Dick

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    Re: Series

    Can you show that if x close enough to 0 then sin(x)>x/2?
     
  5. Apr 5, 2010 #4

    Dick

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    Re: Series

    The poster means sin(4/(3n)), not sin((4/3)*n).
     
  6. Apr 5, 2010 #5
    Re: Series

    im not sure what you mean by that
     
  7. Apr 5, 2010 #6

    Dick

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    Re: Series

    Pretty much what I said to begin with. I mean can you show that if x is positive and close to zero, that sin(x)>x/2? That's one way to 'get rid of the sin' by using a comparison test.
     
  8. Apr 5, 2010 #7
    Re: Series

    what do you mean if x get close enough to 0 than sin(x)>x/2?

    are you talking about the squeeze thrm?
    ill try to solve the integral using by parts
     
  9. Apr 5, 2010 #8
    Re: Series

    where are you getting the x?

    i only got the x from doing the integral test. if not than shouldn't it be n's?
     
  10. Apr 5, 2010 #9
    Re: Series

    Dick is suggesting that you try to compare your series with another series. Your given series has terms of the form [itex]\sin(x)[/itex] where [itex]x=\frac{3}{4n}[/itex]. How does [itex]\sin(x)[/itex] compare to [itex]\frac{x}{2}[/itex] if [itex]|x|<1[/itex]?

    If you can show [itex]\sin(x) > \frac{x}{2}[/itex], what does it say about

    [tex]\sum_{n=1}^\infty \underbrace{\sin\left(\frac{3}{4n}\right)}_{\sin(x)}
    \quad \text{vs}\quad \sum_{n=1}^\infty \underbrace{\frac{1}{2}\cdot \frac{3}{4n}}_{\frac{1}{2}x}[/tex]
     
  11. Apr 5, 2010 #10

    Dick

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    Re: Series

    Thank you for explaining it so well rs1n!
     
  12. Apr 5, 2010 #11
    Re: Series

    ok. So i understand that you substituted x= 3/4n

    if i show that sin(x) > x/2 than the series converges. is that what your saying?
     
  13. Apr 5, 2010 #12
    Re: Series

    Look up the direct comparison test, and see if you are applying it correctly. Does the series [itex]\sum_{n=1}^\infty \left(\frac{1}{2}\cdot \frac{3}{4n}\right)[/itex] converge or diverge? (What special series does it almost look like?)
     
  14. Apr 5, 2010 #13
    Re: Series

    looks like the harmonic p-series 1/n which diverges.
     
  15. Apr 5, 2010 #14
    Re: Series

    so are you saying that by comparing 1/n p-series diverges

    then (1/2* 3/4n) must also diverge

    indicating that sin(3/4n) diverges also?
     
  16. Apr 6, 2010 #15
    Re: Series

    That's the idea, but you actually need to show that those statements are true.
     
  17. Apr 6, 2010 #16
    Re: Series

    Now wouldn't it just be the same if i compared 1/n to 3/4n to diverge

    Then could I just say since 3/4n diverges then sin(3/4n) must also diverge?

    I wouldn't really know how to do the algebra for the direct comparison test or limit comparison test with the sin being there. But i can do the algebra for comparing 1/n with 3/4n
     
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