I have a goal of arriving to the fourier transform of dirac-comb $\omega_T(t)$ so i calculated the fourier series $\omega_T(t) = \frac{1}{T} + \frac{2}{T} \sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t)$ i think the next step to obtaining the fourier transform is evaluating the series, isnt it?
so $F(\omega_T(t))(z) = F(\frac{1}{T}(1 + 2 \sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t)))(z) = \frac{1}{T} F(1)(z) + 2 F(\sum_{n=1}^\infty \cos(n \frac{2 \pi}{T} t))(z) = \frac{1}{T} \omega(z) + \frac{1}{T}\sum_{n=1}^\infty F(2\cos(n \frac{2 \pi}{T} t))(z)$ do i have the right to switch the transform and the infinite series?
You can switch $F$ and the infinite series because the series converges in the space of tempered distributions $\mathscr S’(\mathbb R)$ and $F$ is continuous and linear from $\mathscr S’(\mathbb R)$ to itself.
Why is the series converging in $\mathscr S’(\mathbb R)$? Well, it has to be checked of course. Take $f\in \mathscr S(\mathbb R)$, consider the quantities $$ \int_{\mathbb R}\cos(2\pi n t/T)f(t)dt. $$ Since the Fourier transform of cosine functions are sums of two Dirac deltas, the above quantity is the same as $$ \frac{F(f)(2\pi n t/T)+ F(f)(-2\pi n t/T)}{2} $$ and since the Fourier transform preserves $\mathscr S(\mathbb R)$, $F(f)$ decays fast at infinity, so that in particular the series
$$ \sum_n\int_{\mathbb R}\cos(2\pi n t/T)f(t)dt. $$
is well-defined and converges. This shows that the Fourier series you wrote is well defined and converges in $f\in \mathscr S(\mathbb R)$, so exchanging series and Fourier transform is justified from what I said above.
With this argument of testing against a Schwartz function $f\in \mathscr S(\mathbb R)$, you might as well just compute the Fourier transform of the Dirac comb directly from the definition of Fourier transform (it’s basically the same argument computationally speaking).