Convergence of $\sum_{k \geq 1} e^{-tk} \cos kz$

Question

I would like to find the convergence of the series $\sum_{k \geq 1} e^{-tk} \cos kz$. Clearly, this series converge in using the comparison test or the integral. How could I get an explicit function of this convergence? Is there anyone could give me a hint how to proceed?

Answer 1

Note that $\cos kz = \mathrm{Re}(\cos kz + i \sin kz) = \mathrm{Re}(e^{ikz})$. Thus:

$$ \sum_{k \geq 1} e^{-tk} \cos kz = \sum_{k \geq 1} \mathrm{Re}(e^{-tk})\mathrm{Re}(e^{ikz}) = \sum_{k \geq 1} \mathrm{Re}(e^{k(-t + iz)}) = \mathrm{Re}\left(\sum_{k \geq 1} \left(e^{-t + iz}\right)^k\right) $$

which is a geometric series.