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!
Best Answer
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 $$\left|\sin \left(\frac{Nx}{2} \right) \sin\left(\frac{(N+1)x}{2}\right)\right|= \frac12 \left||\cos (\frac{x}{2})-\cos(N+\frac12)x \right|\le \frac12\left(1+\left|\cos (\frac{x}{2})\right|\right)$$