$\sum_{n=1}^\infty\frac{\cos({2nt})}{n2^n} =$? for any $t\in\Bbb R$
Since $$\cos(2nt) = \frac{e^{2nit} + e^{-2nit}}{2}$$ we have
\begin{align}\sum_{n = 1}^\infty \frac{\cos(2nt)}{n2^n} &= \frac{1}{2}\left(\sum_{n = 1}^\infty \frac{(e^{it})^{2n}}{n2^n} + \frac{(e^{-it})^{2n}}{n2^n}\right)\\ &= -\frac{1}{2}\left(\log\left(1 - \frac{e^{2it}}{2}\right) + \log\left(1 - \frac{e^{-2it}}{2}\right)\right)\\ &= -\frac{1}{2}\log\left|1 - \frac{e^{2it}}{2}\right|^2\\ &= -\frac{1}{2}\log\left(\frac{5}{4} - \cos(2t)\right).\\ \end{align}