I was interested in evaluating $$ \sum_{n=1}^{\infty} \frac{\cos n}{n+k} $$ I saw with computation that many of such series converge. Is there a general result? I've tried using Taylor expansion of cosine but I've realized it is meaningless since it is a good approximation only close to the origin. I was also wondering if the convergence is absolute... Any hint?
So far I have tried to use complex analysis but I guess I don't have enough background to show that
$$ \sum_{n=1}^{\infty} \frac{e^{in}}{n} = -\log (1-e^i) $$ Why is that?
I will show you a formal manipulation, leaving convergence details to check to you. $$\begin{eqnarray*} \sum_{n\geq 1}\frac{\cos n}{n+k}&=&\text{Re}\sum_{n\geq 1}\frac{e^{in}}{n+k}\\&=&\text{Re}\left(e^{-ik}\sum_{n\geq k+1}\frac{e^{in}}{n}\right)\\&=&\text{Re}\left[e^{-ik}\left(-\log(1-e^i)-\sum_{n=1}^{k}\frac{e^{in}}{n}\right)\right]\\&=&\cos(k)\left(-\log(2\sin(1/2))-\sum_{n=1}^{k}\frac{\cos n}{n}\right)-\sin(k)\left(\frac{\pi-1}{2}-\sum_{n=1}^{k}\frac{\sin n}{n}\right).\end{eqnarray*}$$