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Given the series


I need to test for convergence/divergence. I think the divergent test might work here. I could see that the $\lim_{n\rightarrow\infty}(-1)^n\sin(\frac{n}{\pi})$ might not exist, so the series is divergent. But I still need a solid proof here.

Any help is appreciated. Thanks.

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3 Answers 3

up vote 2 down vote accepted

I guess the standard argument should work. If $S_n=\sum_{k=0}^na_k$ converges then $a_k\rightarrow 0$. The necessary condition is not satisfied.

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sorry, didn't try to re-post the same answer. I didn't see this when I was typing. I'll give you an upvote. –  Rod Nov 9 '13 at 19:10

Apply $n^{th}$ test, which states, if $\lim_{\ n\to\infty} a_n\neq0$ then $\sum_{n=0}^{\infty}a_n$ diverges.

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First, you want to think about $$\sin{\left(\frac{n}{\pi}\right)}.$$ This is similar to $\sin{(n)}$ except the period (of ocsillation) is different. We know that the limit of $\sin{(n)}$, as $n\rightarrow \infty$, does not exist since it oscillates indefinitely. Can you apply a similar argument to $\sin{(n/\pi)}$?

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