If $|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k}|It's proven that $\sum_{k=1}^{n}\sin(k\theta)/k$ is uniformly bounded for all $\theta\in\mathbb{R}$ and all $n\geq 1$. So there exists a $M>0$ such that $$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k}\right|<M.$$ 
I am figuring out to prove that $\sum_{k=1}^{n}\sin(k\theta)/k^3$ is also uniformly convergent, or if possible, that
$$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k^3}\right|<M$$
also holds. How do I do that?
 A: I think this is true. On the one hand we have
$$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k^3}\right| \leq 
\sum_{k=1}^{n}\frac{1}{k^3} < \sum_{k=1}^{\infty}\frac{1}{k^3} = \zeta(3) < 1.21$$
for all values of $n$ and $\theta$. On the other hand, if we take $\theta = \pi/n$, then
$$\sum_{k=1}^{n}\frac{\sin(k\theta)}{k} = \sum_{k=1}^{n}\frac{\sin(\pi k/n)}{k}$$
Note that $\sin(\pi k /n) \geq 0$ for all of the summands, and we can use the lower bounds
$$\sin(t) \geq \begin{cases}
2t/\pi & \text{ if }t \in [0,\pi/2] \\
2 - 2t/\pi & \text{ if }t \in [\pi/2, \pi] \\
\end{cases}$$
to obtain the estimate (for even values of $n$)
$$\begin{aligned}
\sum_{k=1}^{n}\frac{\sin(\pi k /n)}{k}
&= \sum_{k=1}^{n/2}\frac{\sin(\pi k /n)}{k} +
\sum_{k=n/2 + 1}^{n}\frac{\sin(\pi k /n)}{k} \\
&\geq \sum_{k=1}^{n/2} \frac{2k/n}{k} + \sum_{k=n/2+1}^{n}\frac{2-2k/n}{k} \\
&= 2\sum_{k=n/2+1}^{n}\frac{1}{k} \\
&\geq 2\int_{n/2+1}^{n+1}\frac{dx}{x} \\
&= 2(\log(n+1) - \log(n/2+1)) \\
&= 2\log\left(\frac{2n+2}{n+2}\right) \\
&= 2\log\left(2 - \frac{2}{n+2}\right) \\
\end{aligned}$$
which exceeds $1.21$ (the upper bound computed above for the other series) for all $n > 10$, and approaches $2\log(2) > 1.38$ as $n$ grows large. Consequently, any uniform bound for 
$$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k}\right|$$
must be at least $1.38$, whereas we have established the uniform bound $1.21$ for
$$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k^3}\right|$$

Edit: I just noticed that we can get a tighter lower bound for 
$$\sum_{k=1}^{n}\frac{\sin(\pi k/n)}{k}$$
Approximate $\sin(\pi x)/(\pi x)$ from above on the interval $[0,1]$ using rectangles of width $1/n$ to obtain the bound
$$\begin{aligned}
\frac{\pi}{n} + \sum_{k=1}^{n}\frac{\sin(\pi k/n)}{k}
&= \pi\left(\frac{1}{n} + \sum_{k=1}^{n}\frac{\sin(\pi k/n)}{\pi k/n}\cdot \frac{1}{n}\right) \\
&\geq \pi \int_0^1 \frac{\sin(\pi x)}{\pi x} dx \\
& = \pi \cdot \frac{\text{Si}(\pi)}{\pi} \\
&\approx \pi \cdot 0.58949 \\
&> 1.85
\end{aligned}$$
and so
$$\sum_{k=1}^{n}\frac{\sin(\pi k/n)}{k} > 1.85 - \frac{\pi}{n}$$
Consequently, any uniform bound for $\sum_{k=1}^{n}\frac{\sin(\pi k/n)}{k}$ must be at least $1.85$.
