Transition between fourier transforms of periodic and aperiodic signals

Question

So, I was trying to prove that if I had a signal that exists only for $-a \leq x \leq +a$, and vanishes for the rest of $x$, then the fourier transform, $F\left(\omega\right)$, for the said function would be continuous but if I repeat the upper function with a periodicity of $2a$ so that I construct a periodic function out of it, the fourier transform will be a discrete version of $F\left(\omega\right)$. So, I start with $$F\left\{f\left(t\right)\right\} = \int_{-\infty}^{+\infty}f\left(t\right)e^{-j\omega t}\mathrm{d}t = \int_{-a}^{+a}f\left(t\right)e^{-j\omega t}\mathrm{d}t$$ Since $f\left(t\right)$ lies in the range of $-a \leq t \leq +a$ and is zero else where. Then I construct a periodic function $g\left(t\right)$ such that $$g\left(t\right) = \sum_{k=-\infty}^{+\infty}f\left(t+2ak\right)$$ Then I go for the fourier transform of $g\left(t\right)$ such that $$F\left\{g\left(t\right)\right\} = \int_{-\infty}^{+\infty}\sum_{k=-\infty}^{+\infty}f\left(t+2ak\right)e^{-jwt}\mathrm{d}t = \sum_{k=-\infty}^{+\infty}e^{j\omega 2ak}\int_{-\infty}^{+\infty}f\left(\tau\right)e^{-j\omega \tau}\mathrm{d}\tau = \sum_{k=-\infty}^{+\infty}e^{j\omega 2ak}F\left\{f\left(t\right)\right\} = F\left\{f\left(t\right)\right\}\sum_{k=-\infty}^{k=\infty}e^{j\omega 2ak}$$ Herein, $\tau = t+2ak$, is a substitution that I made, but is there some way to simplify the sum term $\sum_{k=-\infty}^{+\infty}e^{j\omega 2ak}$. I also know that if $f\left(t\right)$ is periodic, then the $\omega$ space will be perforated and $\omega = \frac{n\pi}{a}$, but if I plug that into the above sum part, then I get $$\sum_{k=-\infty}^{+\infty}e^{j2\pi nk}$$ which is obviously equal to infinity, thus my whole attempt at the proof blows up. Where am I going wrong? Please help. Please don't take me through some other route for the proof, I want to know where I am going wrong, since this is the way I am trying to construct the proof. Also, try and go a little easy on the explanation part, since I am an engineer and not a mathematician, hence thoroughly illiterate compared to the people in this forum.

Answer 1

$x:=\omega 2a$

$\sum\limits_{k=-m}^{+m}e^{jxk}=-1+\frac{e^{jx(m+1)}-1}{e^{jx}-1}+\frac{e^{-jx(m+1)}-1}{e^{-jx}-1}=e^{jxm}$

The amount is $1$, independend of $m$.

To choose $\omega :=\frac{\pi n}{a}$ means $x=2\pi n$ and it follows $\,e^{j2\pi nm}=1^m=1\,$.