Convergence test of the series $\sum\sin100n$ I need to prove that $$\sum_{n=1}^{\infty} {\sin{100n}} \; \text{diverges}$$ I think the best way to do it is to show that $\lim_{n\to \infty}{\sin{100n}}\not=0$. But how do I prove it?

$$\begin{align} \cos(100(k+1-1/2)) - \cos(100(k-1/2)) &= -2\sin\left(\frac{100(2k)}{2}\right)\sin{\frac{101}{2}} \\
\cos(100(k+1/2)) - \cos(100(k-1/2)) &= -2\sin(100k)\sin(50.5) \\
\cos(100(n+1/2)) - \cos(50) &= -2\sin(50.5)\sum_{i=1}^{n}\sin(100i) \\
\frac{\cos(50) - \cos(100(n+1))}{2\sin(50.5)} &= \sum^{n}_{i=1}\sin(100i)\end{align}$$
There. Try to answer it from here.
