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?

We know that every subgroup of $\mathbb{R}$ is either discrete/cyclic or dense.
Consider the subgroup $G=\left\{100n+2m\pi:n,m\in\mathbb{Z}\right\}$. A simple argument  shows that this subgroup is not cyclic (if $\gamma$ was a generator, then $\gamma$ would have to be rational because $100\in G$, but would also have to be irrational because $2\pi\in G$). Thus, $G$ is dense in $\mathbb{R}$.
Therefore, there exists a sequence of distinct elements of $G$ with $\pi/4<|g_k|<\pi/2$, say $g_k=100n_k+2\pi m_k$, and changing $g_k$ by $-g_k$ if necessary, we may assume $n_k>0$. Clearly, the set $\left\{n_k:k=1,2,\ldots\right\}$ is infinite (if not, then the $|g_k|$ would go to $\infty$). But
$$\sin(100n_k)=\sin g_k>\sin\pi/4$$
so there exists infinitely many indices $n_k\in\mathbb{N}$ such that $\sin(100n_k)>\sin\pi/4$, so $\sin(100n)$ does not converge to $0$>
