How to show that $x_n = \sum_{k=1}^{n} \frac{\cos(k+1)x - \cos kx}{k}$ converges using Cauchy convergence theorem?

Question

I asked a question. It is solved by using Dirichlet's test.

Is it possible to show that $x_n = \sum_{k=1}^{n} \frac{\cos(k+1)x - \cos kx}{k}$ converges using Cauchy convergence theorem? Thank you very much.

Answer 1

Write:

$$\frac{\cos(k+1)x-\cos(kx)}{k}=\frac{\cos((k+1)x)}{k+1}-\frac{\cos(kx)}{k}+\frac{\cos((k+1)x)}{k(k+1)}$$ Then if $n\geq m$ $$\sum_{k=m}^n \left(\frac{\cos((k+1)x)-\cos(kx)}{k}\right)=\frac{\cos((n+1)x)}{n+1}-\frac{\cos(mx)}{m}+\sum_{m}^n\frac{\cos((k+1)x)}{k(k+1)}$$

We have $$\left|\frac{\cos((n+1)x)}{n+1}-\frac{\cos(mx)}{m}\right|\leq \frac{1}{n+1}+\frac{1}{m}$$ and $$\left|\sum_m^n\frac{\cos((k+1)x)}{k(k+1)}\right|\leq \sum_{m}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{1}{m}-\frac{1}{n+1}$$

Hence $$\left|\sum_{k=m}^n \left(\frac{\cos(k+1)x-\cos(kx)}{k}\right)\right|\leq \frac{2}{m}$$ and it is easy to finish.