- #1

Mathman23

- 254

- 0

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]

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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