Is there any result about this formula?

Question

$${\sum_{k=0}^\infty \frac{\sin kx}{k}\over \sum_{k=0}^\infty \frac{\cos kx}{k}}$$ I want to use Euler formula but failed.I think it may be relevant to Fourier series so that you can transform the series to a new function.

Answer 1

Hint

Using $$\sin(t)=\Im\left(e^{i t}\right)\qquad \text{and} \qquad \cos(t)=\Re\left(e^{i t}\right)$$ you face sums looking like $$S=\sum_{k=1}^\infty \frac {y^k} k$$ Differentiate once to get $$S'=\sum_{k=1}^\infty {y^{k-1}}$$ which you know. Integrate again to get each sum.

You should end with a rather simple expression.

Answer 2

Taking account that $\displaystyle \boxed{\cos(k\, x)=\frac{e^{i\,kx}+e^{-i\,kx}}{2}}$ and $\displaystyle \boxed{\sin(k\, x)=\frac{e^{i\,kx}-e^{-i\,kx}}{2i}}$ your fraction can be written as

$$\frac{\sum_{k=1}^{\infty} \frac{\sin kx}{k}}{\sum_{k=1}^\infty \frac{\cos kx}{k}}=-i\,\frac{\sum_{k=1}^{\infty}\frac{(e^{i\,x})^k}{k}-\frac{(e^{-i\,x})^k}{k} }{\sum_{k=1}^{\infty}\frac{(e^{i\,x})^k}{k}+\frac{(e^{-i\,x})^k}{k}}$$

Now, using $\displaystyle \log(1-x)=-\sum_{k=1}^{\infty}\frac{x^k}{k}$ you arrives

$$-i\frac{-\log(1-e^{i\,x})+\log(1-e^{-i\,x})}{-\log(1-e^{i\,x})-\log(1-e^{-i\,x})}=i\frac{\log(-e^{-ix})}{\log(2-e^{i\,x}-e^{-i\,x})}=\boxed{\frac{x- \text{sign}(x)\,\pi}{\log(2)+\log(1-\cos(x))}} $$