which is right Fourier transform of $e^{-t} 1_{t > -\epsilon}\sum_{n=-\infty}^\infty \delta (t-n)$ $$\mathcal{F}(e^{-t} 1_{t > -\epsilon})=\frac{1}{j\omega+1}$$
$$\mathcal{F}(\sum_{n=-\infty}^\infty \delta (t-n))=2\pi\sum_{n=-\infty}^{+\infty}\delta(\omega-2n\pi)$$
so with convolution:
$$\mathcal{F}(e^{-t} 1_{t > -\epsilon}\sum_{n=-\infty}^\infty \delta (t-n))=\frac{1}{2\pi} \mathcal{F}(e^{-t} 1_{t > -\epsilon})*\mathcal{F}(\sum_{n=-\infty}^\infty \delta (t-n))=\sum_{n=-\infty}^\infty \frac{1}{1+j\omega+2j\pi n}$$
with different method:
\begin{align}
\mathcal{F}(e^{-t} 1_{t > -\epsilon}\sum_{n=\infty}^\infty \delta (t-n)) &=\mathcal{F}(1+e^{-1}\delta(t-1)+e^{-2}\delta(t-2+)+e^{-3}\delta(t-3)+e^{-4}\delta(t-4)) +\cdot \\
&=1+e^{-1}e^{-j\omega}+e^{-2}e^{-j2\omega}+e^{-3}e^{-j3\omega}+e^{-4}e^{-j4\omega} +\cdot\\
&=\frac{1-e^{n(-1-j\omega)}}{1-e^{-1-j\omega}} \\
&=\frac{1}{1-e^{-1-j\omega}}
\end{align}
it's easily found 
$$\sum_{n=-\infty}^\infty \frac{1}{1+j\omega+2j\pi n}\not=\frac{1}{1-e^{-1-j\omega}}$$
So which one is the right Fourier Transform? 
 A: Let $f(z) = e^z-1$. It is analytic on the whole complex plane (entire), its zeros are at $z = 2 i k \pi$, of order $1$, and $f'(2 i k \pi) = 1$. 
Note that $g(z) = \sum_{k=-\infty}^\infty \frac{1}{z-2 i k \pi}$, understood as $\lim_{K \to \infty} \sum_{k=-K}^K \frac{1}{z-2 i k \pi}$, converges absolutely for any $z \not\in \{2 i k \pi\}$, and is meromorphic, hence $h(z) = \frac{1}{f(z)} -g(z)$ is meromorphic, and since it has no poles, it is entire. 
Finally, $h(z) = \frac{1}{f(z)} -g(z)$ is bounded on the imaginary line, and since both $f(z), g(z) \to 0$ as $Re(z) \to \pm\infty$, we get that $h(z)$ is a bounded entire function, so that $h(z) = C$ by the Liouville theorem. See below for the value of $C = \color{red}{-\frac{1}{2}}$, and not $C=0$ as I thought.
Hence :
$$\frac{1}{f(z) } = \frac{1}{e^z-1} = -\frac{1}{2}+\sum_{k=-\infty}^\infty \frac{1}{z-2 i k \pi}$$
$$\frac{1}{1-e^{-1-i\omega}} = \frac{-1}{f(-1-i \omega)} = -\frac{1}{2}+\sum_{k=-\infty}^\infty \frac{1}{2 i k \pi+1+i \omega}$$

EDIT : the flaw was that once we know by Liouville's theorem that $h(z) = \frac{1}{f(z)} - g(z) = C$ is constant, we can compute $C$ by looking at $h(0) = \lim_{z \to 0} h(z)$ : 
$$\textstyle g(z) = \frac{1}{z} + \sum_{k=1}^\infty \left(\frac{1}{z-2 i \pi k}+\frac{1}{z+2 i \pi k}\right) = \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2+4 \pi^2 k^2}$$ so that  $$\lim_{z \to 0} g(z) - \frac{1}{z} = \lim_{z \to 0} 2z\sum_{k=1}^\infty \frac{1}{4 \pi^2 k^2} = 0$$ and $$\lim_{z \to 0} \frac{1}{f(z)} - \frac{1}{z} = \lim_{z \to 0} \frac{1}{e^z-1}-\frac{1}{z} = \lim_{z \to 0} \frac{z-(e^z-1)}{z(e^z-1)} = \lim_{z \to 0} \frac{z-(z+\frac{z^2}{2}+o(z^2))}{z(z+o(z))} = -\frac{1}{2}$$ hence $$C = h(0) = \lim_{z \to 0} \frac{1}{f(z)}-\frac{1}{z}-(g(z)-\frac{1}{z}) = -\frac{1}{2}$$
