Show that $$\sum_{k=1}^{\infty} \frac{\rho ^k}{k} \cos (k \theta) = - \frac{1}{2} \log(1 -2 \rho \cos(\theta) + \rho ^2)$$ with $\rho < 1$ .
I'm quite sure I can use that expansion $\operatorname{Log}(1-z) = \sum_{k=1}^{\infty} \frac{z^k}{k}$, but I don't know how to treat the term $\cos(k \theta)$.
Any hint?
Using $\log(1-z) = - \sum_{k=1}^{\infty} \frac{z^k}{k}$ you write $$ - \frac{1}{2} \log(1 -2 \rho \cos(\theta) + \rho ^2) \\ = - \frac{1}{2} \log((1 - \rho e^{i \theta})(1 - \rho e^{-i \theta}) \\ = - \frac{1}{2}\log(1 - \rho e^{i \theta}) - \frac{1}{2} \log(1 - \rho e^{-i \theta})) \\ = \frac{1}{2}\sum_{k=1}^{\infty} \frac{\rho ^k}{k} (e^{i k\theta} + e^{-i k \theta}) \\ = \sum_{k=1}^{\infty} \frac{\rho ^k}{k} \cos (k \theta) $$