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Test of Convergens question

  1. Mar 13, 2007 #1
    Hi

    I would like to show the following:

    [tex]\frac{2}{(n+1)\pi} \leq \int_{0}^{\pi} \frac{sin(x)}{n\pi+x} dx \leq \frac{2}{n\pi}[/tex]

    To show that this is true for all [tex]n \geq 1[/tex]

    I then rewrite the integral in the terms of n and by choosing [tex]x = (n+1) \cdot \pi[/tex]

    Which then gives inturn [tex]\frac{1}{n+1} \leq 0 \leq \frac{1}{n}[/tex]

    Since then n tends towards infinity then, the above must be true

    [tex]\mathop {\lim }\limits_{n \rightarrow \infty} b_n = 0 [/tex] and

    [tex]\mathop {\lim }\limits_{n \rightarrow \infty} b_{n+1} = 0 [/tex].

    where [tex]b_n = \frac{1}{n}[/tex]

    [tex]b_{n+1} = \frac{1}{n+1}[/tex]

    Am I on the right track here???

    Best regards
    MM23

    p.s. Or is it more appropiate to claim that since

    since [tex]b_{n+1}[/tex] converges according to the alternating series test.

    then [tex]b_{n+1} \leq a_n \leq b_{n}[/tex]
     
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2

    Dick

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    How can you 'choose x' in an integral? That's why your next inequality looks so silly. Think about it. To establish your first inequality, just ask yourself i) what happens if I replace the x in the denominator with 0 and ii) if I replace it with pi? Does the integral get larger or smaller? In each of the two cases. Then work the replaced integrals.
     
    Last edited: Mar 13, 2007
  4. Mar 13, 2007 #3
    If I replace x by either pi or 0 it get zero. doesn't my original inequality hold then?
     
  5. Mar 13, 2007 #4

    Dick

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    Just the x in the denominator, not both x's. Your 'inequality' says that zero is between two positive numbers!
     
  6. Mar 13, 2007 #5
    Oh then

    [tex]\frac{2}{(n+1)\pi} \leq \int_{0}^{\pi} \frac{sin(x)}{n\pi+x} dx \leq \frac{2}{n\pi}[/tex]

    by replacing x by zero in denominator I get:

    [tex]\frac{2}{(n+1)\pi} \leq \frac{sin(x)}{n\pi} \leq \frac{2}{n\pi}[/tex]

    This part of the inequality true since [tex]\frac{2}{(n+1)\pi} \leq \frac{sin(x)}{n\pi}[/tex]

    Then by taking [tex]\pi[/tex] to replace x

    [tex]\frac{sin(x)}{\pi(n+1)} \leq \frac{2}{n\pi}[/tex]

    which again is true by by alternating series test.

    Am I on the right track now ?
     
    Last edited: Mar 13, 2007
  7. Mar 13, 2007 #6

    Dick

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    Starting with the inequality you want to prove sounds like a bad idea. And how did you make the integral just disappear??? Start here. Why is this true?

    [tex] \int_{0}^{\pi} \frac{sin(x)}{n\pi+\pi} dx \leq \int_{0}^{\pi} \frac{sin(x)}{n\pi+x} dx \leq \int_{0}^{\pi} \frac{sin(x)}{n\pi} dx[/tex]
     
  8. Mar 13, 2007 #7
    It is true because n is a positive number.
     
  9. Mar 13, 2007 #8

    Dick

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    It's true because 0<=x<=pi. And n is positive.
     
  10. Mar 13, 2007 #9
    I get its as far now.

    But What is the next logical step then??

    Convert into sums ??
     
  11. Mar 13, 2007 #10

    Dick

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    Do the outside integrals. They are easy now.
     
  12. Mar 13, 2007 #11
    By doing the outer integral I get:

    I then get [tex]\frac{2}{(n+1) \pi} \leq \int_{0}^{\pi} \frac{sin(x)}{n\pi+x} dx \leq \frac{2}{n \pi} [/tex]
     
    Last edited: Mar 13, 2007
  13. Mar 13, 2007 #12

    Dick

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    Leave the inner integral alone. It's part of what you want to prove. And the outer ones didn't go so well either. Integral of sin(x) between 0 and pi is 2. The denominators are constant.
     
  14. Mar 13, 2007 #13
    Please see my post above......
     
  15. Mar 13, 2007 #14

    Dick

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    Better. That's what you want to prove, yes?
     
  16. Mar 13, 2007 #15
    What else can there be said than if n tends to infinity, then the integral converges to zero ??
     
  17. Mar 13, 2007 #16
    What else can there be said than if n tends to infinity, then the integral becomes smaller and smaller.

    This makes the inequality true. Doesn't it ?
     
  18. Mar 13, 2007 #17

    Dick

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    Except that you didn't really need that lower bound. The integral is clearly positive. And, no, the inequality makes the convergence true, not vice versa.
     
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