Estimate $\sup_{n\in\mathbb N}\sum_{k=1}^n \frac{\cos kx-\cos(kx+\alpha k)}{k^2}$. I would estimate
$$\sup_{n\in\mathbb N}\sum_{k=1}^n \frac{\cos kx-\cos(kx+\alpha k)}{k^2}$$
where $\alpha>0, x\in \mathbb R$.
I do not know how to proceed. Do you have an idea? Any help please.
 A: A simple estimation can be this $$\sum_{k\leq n}\frac{\cos\left(kx\right)-\cos\left(k\left(x+\alpha\right)\right)}{k^{2}}\leq2\sum_{k\leq n}\frac{1}{k^{2}}<\frac{\pi^{2}}{3}.$$
  But if you want more precise estimation, we have$$\sum_{k\leq n}\frac{\cos\left(kx\right)}{k^{2}}-\sum_{k\leq n}\frac{\cos\left(k\left(x+\alpha\right)\right)}{k^{2}}=\sum_{k\geq1}\frac{\cos\left(kx\right)}{k^{2}}-\sum_{k\geq1}\frac{\cos\left(k\left(x+\alpha\right)\right)}{k^{2}}-\sum_{n<k}\frac{\cos\left(kx\right)}{k^{2}}+\sum_{n<k}\frac{\cos\left(k\left(x+\alpha\right)\right)}{k^{2}}$$
  and now note$$-\psi^{(1)}\left(n\right)=-\sum_{n<k}\frac{1}{k^{2}}\leq\sum_{n<k}\frac{\cos\left(kx\right)}{k^{2}}\leq\sum_{n<k}\frac{1}{k^{2}}=\psi^{(1)}\left(n\right)$$
 where $\psi^{(n)}$ is the polygamma function, and observe that $\psi^{(1)}\left(n\right)$
  is a monotonic decreasing function. So if $0\leq x\leq2\pi$,
  $0\leq x+\alpha\leq2\pi$
 , using Fourier series$$\sum_{k\geq1}\frac{\cos\left(kx\right)}{k^{2}}=\frac{\pi^{2}}{6}-\frac{\pi x}{2}+\frac{x^{2}}{4}$$
 we have $$\frac{\pi\alpha}{2}-\frac{2\alpha x+\alpha^{2}}{4}-\psi^{(1)}\left(1\right)\leq\sup_{n\in\mathbb{N}}\left(\sum_{k\leq n}\frac{\cos\left(kx\right)-\cos\left(k\left(x+\alpha\right)\right)}{k^{2}}\right)\leq\frac{\pi\alpha}{2}-\frac{2\alpha x+\alpha^{2}}{4}+\psi^{(1)}\left(1\right).$$
