I really have no clue where to start with this one as I am used to having the argument of the trig function be (πx) which makes it easy to treat it like an alternating series.
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You didn't provide a series – Chickenmancer Apr 28 '17 at 07:04
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Apparently $\displaystyle\sum_{n\geq 1}\frac{\cos(kn)}{n}=-\log\left|2\sin\frac{k}{2}\right|$ for $k\not\in 2\pi\mathbb{Z}$ so the answer is presumably yes here too – Henry Apr 28 '17 at 07:17
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Again apparently, Wolframalpha can do the summation and the result is $(3+2 \pi)/4 \simeq 2.3208$. This doesn't tell us though why the series converges and how to arrive at that result. – Andreas Apr 28 '17 at 07:36
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1It converge by Dirichlet's test. you can verify this by verifying the partials sums of the form $\left|\sum_{x=1}^p \sin\left(\frac32 x\right)\right|$ is bounded. – achille hui Apr 28 '17 at 07:44
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Assuming that $\frac{\sin(3x/2)}{x}$ is defined at $x=0$ as $\frac{3}{2}$, it is a convergent series by Dirichlet's test, since the partial sums of the sequence $\left\{\sin\left(\frac{3}{2}n\right)\right\}_{n\geq 0}$ are bounded. Additionally, it is well known that the Fourier series $$ \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ is pointwise convergent to the function $\frac{\pi-x}{2}$ on the interval $(0,2\pi)$, hence
$$ \sum_{n\geq 0}\frac{\sin\left(\frac{3}{2}n\right)}{n} = \frac{3}{2}+\frac{\pi-\frac{3}{2}}{2} = \color{red}{\frac{3+2\pi}{4}}.$$
Jack D'Aurizio
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Could you elaborate quickly why the partial sum of $\sin(\frac32 n)$ is bounded ? :) – Zubzub Apr 28 '17 at 07:53
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@Zubzub: for short, $\sin(z)=\text{Im}(e^{iz})$, $\text{Im}$ and $\sum$ commute and $\sum_{n=1}^{N}e^{niz}$ can be computed as a geometric series. – Jack D'Aurizio Apr 28 '17 at 07:54
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By Dirichlet's test we have that $1/x$ is a decreasing sequence, going to zero. Then $\sum_{x=0}^\infty\dfrac{\sin\left(\frac{3x}{2}\right)}{x}$ will converge if $|\sum_{x=0}^\infty \sin\left(\frac{3x}{2}\right)|$ is bounded. However, this is the case. So the answer to the question is "yes".
As for the boundedness: Let $S_N = \sum_{n=0}^N \sin\left(\frac{3n}{2}\right)$. Then $$2\sin \left( \frac1{2} \right) \times S_N = \sum_{n=0}^{N} \left( \cos \left( \frac{3(n-1)}{2}\right) - \cos \left( \frac{3(n+1)}{2}\right)\right) \\ = \cos \left( -\frac3{2} \right) + \cos \left( 0 \right) - \cos\left( \frac{3(N)}{2}\right)- \cos\left( \frac{3(N+1)}{2} \right)$$
So for any $N$, $S_N$ is bounded.
Andreas
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