Does this series converge?
$$ \sum_{n=1}^{\infty} \frac{1}{\sin 1 + \sin 2 + \ldots + \sin n} $$
What test do I have to use for it?
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What tests do you know and what have you tried? – Calvin Khor Sep 10 '15 at 14:03
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First thing is to check whether the terms tend to $0$. Can you find a more tractable form for $$\sum_{k = 1}^n \sin k,?$$ – Daniel Fischer Sep 10 '15 at 14:06
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You might consider the sum of the first $43$, $665$, and $709$ terms, related to $\pi \approx \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \ldots$ – Henry Sep 10 '15 at 14:18
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We have: $$ \sin(1)+\sin(2)+\ldots+\sin(n) = \text{Im}\sum_{j=1}^{n}e^{ji} = \frac{\sin\left(\frac{n}{2}\right)\sin\left(\frac{n+1}{2}\right)}{\sin\left(\frac{1}{2}\right)}\tag{1}$$ hence: $$ \sum_{n=1}^{N}\frac{1}{\sin(1)+\sin(2)+\ldots+\sin(n)}=2\sin\left(\frac{1}{2}\right)\sum_{n=1}^{N}\frac{1}{\cos\left(\frac{1}{2}\right)-\cos\left(n+\frac{1}{2}\right)}\tag{2}$$ but since the sequence given by $a_n = \exp\left(\frac{i}{2}+in\right)$ is dense in the unit circle, the general term of our series is not even bounded, hence the series cannot be convergent. An alternative proof, following Kelenner's comment below, comes from noticing that $\left|\frac{1}{\sin(1)+\sin(2)+\ldots+\sin(n)}\right|\geq\sin\left(\frac{1}{2}\right).$
Jack D'Aurizio
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2I think that you can simplify: let $u_n$ the sum of the $\sin(k)$, you have proven that $ |u_n|\leq 1/\sin(1/2)$, so $1/u_n$ does not go to $0$.. – Kelenner Sep 10 '15 at 14:34
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@Kelenner: that's true. Not only the general term is unbounded, but it is bounded away from a neighbourhood of zero. Two good reasons for which the series cannot be convergent. – Jack D'Aurizio Sep 10 '15 at 14:59