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Question is :

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What i have done so far is :

I see that $\frac{\pi}{n}\rightarrow 0$ and so should be the sequence $u_n=\sin (\frac{\pi}{n})$.

i.e., $u_n$ converges to $0$ and so $(b)$ is true.

I remember that

$\sin \theta$ is same as $\theta$ for small angles

So, I Guess $\sin(\frac{\pi}{n})=\frac{\pi}{n}$ for large $n$

So, I guess $$\sum _{n -\text{large}}u_n=\sum _{n -\text{large}}\sin(\frac{\pi}{n})=\pi \sum _{n -\text{large}}\frac{1}{n}$$

But then $\sum \frac{1}{n}$ is divergent so should be the series $$\pi \sum _{n -\text{large}}\frac{1}{n}$$ i.e., $\sum u_n$ is divergent.

So, I wanted to conclude $(c)$ is true. But then for above question there is only one correct option.

So, If $c$ is true then both $(a),(d)$ should be false which is not possible according to question.

So, just with some sense I see that $(c)$ should be the only possibility for the correct option as this does not contradict anything while other would contradict.

So, $(c)$ should be false.. But, I am not sure how to see this.

please help me to see this in detail...

Thank you

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  • $\begingroup$ there should be only one correct option... So, if series is divergent it can not be convergent and so it can not be absolutely convergent... but then both $a,d$ will be false and both will be correct options where as given question has only one correct option.. $\endgroup$ – user87543 Nov 17 '13 at 9:17
  • $\begingroup$ a and d are false $\endgroup$ – Nirvanacs Nov 17 '13 at 9:21
  • $\begingroup$ @Nirdonkey could you please explain a bit... I also got the same result but i am not sure. $\endgroup$ – user87543 Nov 17 '13 at 9:22
  • $\begingroup$ "sinθ is same as θ for small angles" is true to some extent. $\endgroup$ – Nirvanacs Nov 17 '13 at 9:25
  • $\begingroup$ Yes. I see that but it is not very helpful $\endgroup$ – user87543 Nov 17 '13 at 9:26

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