Proving the series of partial sums of $\sin (in)$ is bounded? So I'm trying to prove some series converges, and I'm trying to show it by using the Drichlet test. 
So I need to prove that the series of partial sums $\Sigma_{k=1}^\infty \sin(ki)$ is bounded. I tried proving it by dividing and multiplying with $2\cos(\frac{i}{2})$ and then using the trigonometric identity $2\cos(\beta)\sin(\alpha)=sin(\alpha+\beta)-sin(\alpha-\beta)$, which creates a telescopic series - but I didn't manage to bound the result.
Any assistance would be great! Thanks in advance! 
 A: One way to obtain a bound is to derive a closed-from expression for the sum.  Here, this is accomplished by summing  geometric series. To that end,
$$\begin{align}
\left|\sum_{n=1}^N \sin(nx)\right| &=\left| \text{Im} \left(\sum_{n=1}^N e^{inx}  \right) \right|\\
&=\left|\text{Im} \left( \frac{e^{ix}-e^{i(N+1)x}}{1-e^{ix}} \right)\right| \\
&=\left|\frac{\sin \left(\frac{Nx}{2} \right) \sin\left(\frac{(N+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right|\\
&\le \left|\csc\left(\frac{x}{2}\right)\right|
\end{align}$$
Note:  A tighter bound can be obtained by noting 
$$\left|\sin \left(\frac{Nx}{2} \right) \sin\left(\frac{(N+1)x}{2}\right)\right|= \frac12 \left||\cos (\frac{x}{2})-\cos(N+\frac12)x \right|\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)$$
A: I'll use $x$ in place of $i$ in order to avoid confusion with the imaginary number $\sqrt{-1}$. Also, I'll assume $x$ is not an integral multiple of $2\pi$.
Since
\begin{align}\sum_{k = 1}^n \sin(kx) &= \csc(x/2) \sum_{k = 1}^n \sin(kx)\sin(x/2)\\
& = \frac{1}{2}\csc(x/2) \sum_{k = 1}^n [\cos((k - 1/2)x) - \cos((k + 1/2)x)]\\
& = \frac{1}{2}\csc(x/2) [\cos(x/2) - \cos((n+1/2)x]
\end{align}
and the cosine is bounded by $1$,
$$\left|\sum_{k = 1}^n \sin(kx)\right| \le |\csc(x/2)|.$$
Since the upper bound is independent of $n$, the sequence of partial sums of $\sum_{k = 1}^\infty \sin(kx)$ is bounded.
A: You are almost there. Since:
$$\begin{eqnarray*} 2\sin\frac{i}{2}\sum_{k=1}^{K}\sin(ki) &=& \sum_{k=1}^{K}\left(\cos\left(\left(k-\frac{1}{2}\right)i\right)-\cos\left(\left(k+\frac{1}{2}\right)i\right)\right)\\&=&\cos\frac{i}{2}-\cos\frac{(2K+1)i}{2}&\end{eqnarray*}$$
we have that for any $K$:
$$\left|\sum_{k=1}^{K}\sin(ki)\right|\leq\frac{\left|\cos\frac{i}{2}\right|+1}{2\left|\sin\frac{i}{2}\right|}$$
This bound is optimal since for any $\varepsilon > 0$ we can find a $K\in\mathbb{N}$ such that the difference between the LHS and the RHS of the last line is less than $\varepsilon$.
