Does $\sum\limits_{x=1}^\infty\sin(x)$ converge?
Solution 1:
The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.
Solution 2:
Hint: The partial sums have an explicit form, because there are the imaginary part of some geometric series.
$$ \sum_{k=1}^n e^{ik} = \frac {1-e^{in}}{1-e^i} = \frac {e^{-in/2}-e^{in/2}}{e^{-i/2}-e^{i/2}} \frac{e^{in/2}}{e^{i/2}} = \frac{\sin \frac n2}{\sin \frac 12} e^{i(n-1)/2} $$ so the $n$th partial sum is $$ \frac{1}{\sin \frac 12} \sin \frac n2\sin{\frac{n-1}2} = \frac{1}{2\sin \frac 12} \left(\cos \frac 12 - \cos \frac{2n-1}4\right) $$
From this, you can explicitly see what values are taken by these partial sums.
Solution 3:
If $\sin(n) \to 0$, then $|\cos(n)| \to 1$, so $\sum \cos(n)$ diverges, but $\sin(n+1)= \sin(n)\cos(1) + \cos(n)\sin(1)$, so $\sum \sin(1) \cos(n) = \sum \sin(n+1) - \sum \cos(1)\sin(n)$.