Does the series $\sum_{x=0}^\infty\frac{\sin\left(\frac{3x}{2}\right)}{x}$ converge or diverge? 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.
 A: 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
$$ \sum_{n\geq 0}\frac{\sin\left(\frac{3}{2}n\right)}{n} = \frac{3}{2}+\frac{\pi-\frac{3}{2}}{2} = \color{red}{\frac{3+2\pi}{4}}.$$
A: 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.
