If $|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k}|<M$, is it true that $|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k^{3}}|<M$?

Question

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?

Answer 1

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 $$\left|\sum_{k=1}^{n}\frac{\sin(k\theta)}{k^3}\right|$$

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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 $1.85$.