Test of Convergens question

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]

 

If I replace x by either pi or 0 it get zero. doesn't my original inequality hold then?

 

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 ?

 

It is true because n is a positive number.

 

I get its as far now.

But What is the next logical step then??

Convert into sums ??

 

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]

 

Please see my post above......

 

What else can there be said than if n tends to infinity, then the integral converges to zero ??

 

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 ?