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$\ds{\sum_{k = -\infty}^{\infty}\delta\pars{x - k}}$ is
even and periodic ( of period $\ds{1}$ ). Then,
\begin{align}
&\sum_{k = -\infty}^{\infty}\delta\pars{x - k} =
\sum_{n = 0}^{\infty}a_{n}\cos\pars{2\pi nx}
\\[1cm] &\
\int_{-1/2}^{1/2}\cos\pars{2\pi nx}\sum_{k = -\infty}^{\infty}\delta\pars{x - k}\dd x
\\[2mm] = &\
\sum_{m = 0}^{\infty}
a_{m}\underbrace{\int_{-1/2}^{1/2}\cos\pars{2\pi nx}\cos\pars{2\pi mx}\dd x}
_{\ds{=\ {1 + \delta_{n0} \over 2}\,\delta_{nm}}}
\\[5mm] &\
\underbrace{\int_{-1/2}^{1/2}\cos\pars{2\pi nx}\delta\pars{x}\dd x}_{\ds{=\ 1}} =
{1 + \delta_{n0} \over 2}\,a_{n}
\\[5mm] &\
\implies a_{n} = 2 - \delta_{n0}
\end{align}
$$
\implies
\bbx{\sum_{k = -\infty}^{\infty}\delta\pars{x - k} =
1 + 2\sum_{n = 1}^{\infty}\cos\pars{2\pi nx}}
$$