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Given the series

$$\sum_{n=1}^{\infty}(-1)^n\sin\left(\frac{n}{\pi}\right)$$

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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4 Answers 4

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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You are working modulo $2\pi$, so you need to think of $\frac{n}{\pi}$ modulo $2\pi \mathbb{Z}.$

Since $1/\pi^2$ is irrational, there are sequences of natural numbers $n_k$ such that the fractional part of $\frac{n_k}{2\pi^2}$ tends, for instance, to $1/3$ (any number such that $sin\ a\neq 0$ will do). This means that $|sin(n_k/\pi)|$ tends to a nonzero number as $k\to \infty$.

See this question (or look up Weyl equidistribution for a better result) for the density result used: Multiples of an irrational number forming a dense subset

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