Proving the series of partial sums of $\sin (in)$ is bounded?

Solution 1:

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

Solution 2:

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.

Solution 3:

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 $\varepsilon$.