- #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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