What is the Fourier Transform of a weighted Dirac comb?

Question

I have the follwowing Dirac comb: \begin{equation*} \mu=\sum\limits_{n\in\mathbb{Z}}(-1)^{n}\delta_{(n,0)} \end{equation*} and i would like to know what $\widehat{\mu}$ ist. I know that $\widehat{\delta_{\mathbb{Z}}}= \delta_{\mathbb{Z}}$ is. But i dont think this works here. So i looked up how to Fourier Transform the normal dirac comb. First you take the Fourier series $\mu(t)=\sum\limits_{n=-\infty}^{\infty}c_{n}e^{inw_{0}t}$, where $w_{0}=\frac{2\pi}{T}$. In this case $T=2$. But at this point I'm not too sure what to do. Because I'dont know if i can take the $(-1)^{n}$ in the sum or not? And if there are in it, what ist $\mathcal{F}((-1)^{n})$. Because the next step would be to take the Fourier Transform on both sides. So that means: \begin{equation} \mathcal{F}(\mu(t))=\mathcal{F}(\sum\limits_{n=-\infty}^{\infty}c_{n}e^{inw_{0}t}) \end{equation} So my Question is , is this the right way and what to do with $(-1)^{n}$?