I need to evaluate the limit
$$\lim_{n\to\infty}\,\sum _{k=0}^n\frac{\cos(k\pi/4)}{2^k}$$
I see that this series is convergent because is limited by $1/2^n$ that is a convergent geometric series but I can't define what is this $\cos(n\pi /4)$ so I can't find what its ratio $r$ is. If it were $\cos(n\pi)$, then I should put $(-1)^n$ but what about this?
We can evaluate the series in closed form by using Euler's Formula to write $\cos(\pi k/4)=\text{Re}(e^{i\pi k/4})$.
Proceeding, we find that
$$\begin{align} \lim_{n\to \infty}\sum_{k=0}^n\frac{\cos(\pi k/4)}{2^k}&=\lim_{n\to \infty}\text{Re}\left(\sum_{k=0}^n\left(\frac{e^{i\pi/4}}{2}\right)^k \right)\\\\ &=\lim_{n\to \infty}\text{Re}\left(\frac{1-\frac1{2^{n+1}}e^{i\pi(n+1)/4}}{1-\frac12e^{i\pi/4}}\right)\\\\ &=\frac{4-\sqrt2}{5-2\sqrt2}\\\\ &=\frac1{17}(16+3\sqrt 2) \end{align}$$