# Estimate or evaluate the sum: $\sum_{k=1}^n \frac{\cos k x}{k}$.

Estimate or evaluate the sum: $$\sum_{k=1}^n \frac{\cos k x}{k}$$ where $x\in\mathbb R$. My approach: $$f(x)=\sum_{k=1}^n \frac{\cos k x}{k}\Rightarrow f'(x)= - \sum_{k=1}^n \sin k x$$ If we use $$\sum\limits_{k=1}^n\sin{kx}=\dfrac{1}{\sin{\frac{x}{2}}}\sum\limits_{k=1}^n \left(\sin{kx}\cdot\sin{\frac{x}{2}}\right)$$ and the identity $$\sin{\alpha}\sin{\beta}=\dfrac{1}{2}(cos(\alpha-\beta)-cos(\alpha+\beta))$$ we obtain $$f'(x)=\frac{1}{2}\cot(x/2)-\frac{\cos(nx+(x/2))}{2\sin(x/2)}$$ Thus: $$f(x)-f(a)=\frac{1}{2}\int_a^x\cot(t/2) dt-\int_a^x\frac{\cos(nt+(t/2))}{2\sin(t/2)}dt$$ If we choose initial point at $a=\pi$, we have a known value for $f(a)= \sum_{k=1}^n \frac{(-1)^k}{k}$. My question is the following. I have evaluate the sum, but do you know another method for evaluate (or estimate) this sum?

Thank you very much.

• I am not sure what do you want... do you want the estimation of the series $\sum_{k=1}^{\infty}\frac{\cos\left(kx\right)}{k}$ or another method for $\sum_{k=1}^{n}\frac{\cos\left(kx\right)}{k}$? Commented Apr 14, 2015 at 7:04
• The finite sum is non-elementary, but the infinite series can be written as a combination of logarithms and exponentials. Commented Apr 14, 2015 at 7:24
• You can show that $\sum_{k=1}^{n}\frac{\cos\left(kx\right)}{k}\leq C-\log |sin (x/2)|$. math.stackexchange.com/questions/1173560/… Commented Apr 14, 2015 at 7:34
• Sorry, there was a misunderstanding. I meant to write "sum" and not "series".
– Mark
Commented Apr 14, 2015 at 16:52

Assuming that you want to evaluate the infinite sum, note that $$\frac{\cos kx}{k} =\frac{1}{2}\left(\left(\frac{e^{ikx}}{k}\right)+\left(\frac{e^{-ikx}}{k}\right)\right) =\frac{1}{2}\left(\frac{(e^{ix})^k}{k}\right)+\frac{1}{2}\left(\frac{(e^{-ix})^k}{k}\right).$$
• Where did the $i$'s go from the second last to last line? Commented Feb 1, 2022 at 18:32