Proving the relation: $\sum_{n\in\Bbb Z} \frac{2a}{a^2+4 \pi^2 n^2} = \sum_{n\in\Bbb Z} e^{-a \left\lvert n \right\rvert}$

Question

I came across this problem in an exercise for Fourier analysis. I tried solving just $e^{−a|n|}$ to get the Fourier transform of a similar form as seen on the LHS because it looked familiar. But in that case in the denominator, I get $f^2$ (where $f$ is the frequency as given in the equation below) instead of $n^2$ (I'm integrating with respect to n). I used the following relation:

Fourier transform of $x(n)$: $$X(f)=\int_{-\infty}^\infty x(n)e^{-2\pi jfn}dn$$

Solving this gives: $$\frac{2a}{a^2+4 \pi^2 f^2}$$

I think my approach is wrong here.