Does $\sum_{i=1}^n \sin(i)$ converge? Prove it. I know that $\sum_{i=1}^n \sin(i)$ doesn't converge. But I'm having trouble proving it. 
 A: If $\sum a_n$ converges, $\lim a_n=0$.
A: If a summation $\sum_{n=1}^\infty a_n$ converges, then $a_n\rightarrow 0$. But $sin(n)$ does not go to zero as $n\rightarrow \infty$.
A: The easiest way is what Calvin, Vahid and user57101 have suggested. However, you can compute $a_n = \displaystyle \sum_{k=1}^n \sin(k)$ and see for yourself how the sequence behaves for large $n$.
\begin{align}
2\sin(1/2) a_n & = 2\sum_{k=1}^n \sin(k) \sin(1/2) = \sum_{k=1}^n \left(\cos(k-1/2)-\cos(k+1/2) \right)\\
& = (\cos(1/2) - \cos(3/2)) + (\cos(3/2) - \cos(5/2)) + \cdots\\
& + (\cos(n-3/2) - \cos(n-1/2)) + (\cos(n-1/2) - \cos(n+1/2))\\
& = \cos(1/2) - \cos(n+1/2)
\end{align}
Hence,
$$a_n = \dfrac{\cos(1/2) - \cos(n+1/2)}{2 \sin(1/2)} = \dfrac12 \cot(1/2) - \dfrac{\cos(n+1/2)}{2 \sin(1/2)}$$ which oscillates about $\dfrac12 \cot(1/2)$ as $n \to \infty$ and hence it doesn't converge. However, if you want to assign a regularized value to the series, then you can assign it a value $\dfrac12 \cot(1/2)$.
A: Hint: If $\sum_{i=1}^{\infty} a_n$ converges, then $a_n$ tends to 0.
Show that since $\frac {\pi}{1}$ is not rational, then by the pigeonhole principle, for any $N$, there exists $n > N$ such that $| \sin n | > \frac {1}{2}$.
