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Test of Convergens question

  • Thread starter Mathman23
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  • #1
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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]
 
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Answers and Replies

  • #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.
 
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  • #3
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If I replace x by either pi or 0 it get zero. doesn't my original inequality hold then?
 
  • #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!
 
  • #5
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Just the x in the denominator, not both x's. Your 'inequality' says that zero is between two positive numbers!
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 ?
 
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  • #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]
 
  • #7
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It is true because n is a positive number.
 
  • #8
Dick
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It's true because 0<=x<=pi. And n is positive.
 
  • #9
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I get its as far now.

But What is the next logical step then??

Convert into sums ??
 
  • #10
Dick
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Do the outside integrals. They are easy now.
 
  • #11
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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]
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]
 
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  • #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.
 
  • #13
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Please see my post above......
 
  • #14
Dick
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Better. That's what you want to prove, yes?
 
  • #15
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Better. That's what you want to prove, yes?
What else can there be said than if n tends to infinity, then the integral converges to zero ??
 
  • #16
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Better. That's what you want to prove, yes?
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 ?
 
  • #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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