I was asked to prove that the following equation holds true:
$$\sum_{n=-\infty}^{+\infty} e^{-\displaystyle|t+nT|} = \frac{2}{T}\sum_{m=-\infty}^{+\infty}\frac{1}{1+\displaystyle\left(\frac{2\pi m}{T}\right)^2}e^{-\displaystyle\frac{i2\pi m}{T}t}.$$
I observed that $f(t) = \sum_{n=-\infty}^{+\infty} e^{-|t+nT|}$ is periodic with period $T$. Then, I can find its Fourier series:
$$f(t) = \sum_{a=-\infty}^{+\infty}c_a e^{\frac{i2\pi a}{T}t},$$
where
$$c_a = \frac{1}{T}\int_{-T/2}^{T/2} f(t) e^{-\displaystyle\frac{i2\pi a}{T}t}dt = \frac{1}{T}\int_{-T/2}^{T/2} \sum_{n=-\infty}^{+\infty} e^{-|t+nT|} e^{-\displaystyle\frac{i2\pi a}{T}t}dt.$$
I suppose that I can switch series and integral, thus obtaining:
$$c_a = \sum_{n=-\infty}^{+\infty} \frac{1}{T}\int_{-T/2}^{T/2} e^{-|t+nT|} e^{-\displaystyle\frac{i2\pi a}{T}t}dt.$$
Furthermore, I noticed that:
$$e^{-|t+nT|} = \begin{cases} e^{-t-nT} & \text{if}~ n \geq 1 \\ e^{-|t|} & \text{if}~ n =0 \\ e^{+t+nT} & \text{if}~ n \leq -1 \end{cases},$$
in the set $\left[-\frac{T}{2}, +\frac{T}{2}\right].$
Therefore, I can rewrite $c_a$ as follows:
$$c_a = \frac{1}{T}\left( \sum_{n=1}^{+\infty} e^{-nT}\int_{-T/2}^{T/2} e^{-t} e^{-\frac{i2\pi a}{T}t}dt + \sum_{n=-\infty}^{-1} e^{nT}\int_{-T/2}^{T/2} e^{t} e^{-\frac{i2\pi a}{T}t}dt + \int_{-T/2}^{T/2} e^{-|t|} e^{-\frac{i2\pi a}{T}t}dt\right).$$
Additionally, $\sum_{n=1}^{+\infty} e^{-nT} = \sum_{n=-\infty}^{-1} e^{nT} = \frac{1}{1 + e^{-T}}$, and hence:
$$c_a = \frac{1}{T}\left(\frac{1}{1+e^{-T}}\int_{-T/2}^{T/2} (e^{-t}+e^{t}) e^{-\frac{i2\pi a}{T}t}dt + \int_{-T/2}^{T/2} e^{-|t|} e^{-\frac{i2\pi a}{T}t}dt\right).$$
Here I'm stuck. I've tried a lot of time, but I'm not able to prove the equation at the beginning of this post.
You stumbled here: $$\sum_{n=1}^{\infty}e^{-nT}=\sum_{n=-\infty}^{-1}e^{nT}=\frac{e^{-T}}{1-e^{-T}}=\frac{e^{-T/2}}{e^{T/2}-e^{-T/2}}$$ Then just do the integrals: $$\int_{-T/2}^{T/2}e^{-t}e^{-\frac{2\pi ia}Tt}dt=\left.\frac{e^{-t}e^{-\frac{2\pi ia}Tt}}{-1-\frac{2\pi ia}T}\right|_{-T/2}^{T/2}=\frac{(-1)^a\left(e^{T/2}-e^{-T/2}\right)}{1+\frac{2\pi ia}T}$$ $$\int_{-T/2}^{T/2}e^{t}e^{-\frac{2\pi ia}Tt}dt=\left.\frac{e^{t}e^{-\frac{2\pi ia}Tt}}{1-\frac{2\pi ia}T}\right|_{-T/2}^{T/2}=\frac{(-1)^a\left(e^{T/2}-e^{-T/2}\right)}{1-\frac{2\pi ia}T}$$ $$\begin{align}\int_{-T/2}^{T/2}e^{-|t|}e^{-\frac{2\pi ia}Tt}dt&=\int_{-T/2}^0e^te^{-\frac{2\pi ia}Tt}dt+\int_0^{T/2}e^{-t}e^{-\frac{2\pi ia}Tt}dt\\ &=\left.\frac{e^{t}e^{-\frac{2\pi ia}Tt}}{1-\frac{2\pi ia}T}\right|_{-T/2}^0+\left.\frac{e^{-t}e^{-\frac{2\pi ia}Tt}}{-1-\frac{2\pi ia}T}\right|_0^{T/2}\\ &=\frac{1-(-1)^ae^{-T/2}}{1-\frac{2\pi ia}T}+\frac{1-(-1)^ae^{-T/2}}{1+\frac{2\pi ia}T}\end{align}$$ So there you have it: $$\begin{align}c_a&=\frac1T\left\{\frac{e^{-T/2}}{e^{T/2}-e^{-T/2}}\frac{(-1)^a\left(e^{T/2}-e^{-T/2}\right)}{1+\frac{2\pi ia}T}+\frac{e^{-T/2}}{e^{T/2}-e^{-T/2}}\frac{(-1)^a\left(e^{T/2}-e^{-T/2}\right)}{1-\frac{2\pi ia}T}\right.\\ &\,\,\,\,\left.+\frac{1-(-1)^ae^{-T/2}}{1-\frac{2\pi ia}T}+\frac{1-(-1)^ae^{-T/2}}{1+\frac{2\pi ia}T}\right\}\\ &=\frac1T\left\{\frac1{1-\frac{2\pi ia}T}+\frac1{1+\frac{2\pi ia}T}\right\}=\frac2{T\left(1+\left(\frac{2\pi a}T\right)^2\right)}\end{align}$$