Let us say we have:
$$\frac{1}{T_s}\cdot \sum _{n=-\infty }^{\infty }\delta \left(\omega -\frac{2\pi n}{T_s}\right)\:\ast X\left(\omega \right)$$
I had a small problem here since there is a series already, should it be equal to: $$\frac{1}{T_s}\cdot \sum _{n=-\infty }^{\infty }X\left(\omega -\frac{2\pi n}{T_s}\right)$$
This is the reason I am having trouble: $$\frac{1}{T_s}\sum _{n=-\infty }^{\infty }\:\sum _{k=-\infty \:}^{\infty \:}\:\delta \left(k-\frac{2\pi \text{n}}{T_s}\right)X\left(\omega -k\right)$$ Equal to: $$\frac{1}{T_s}\sum _{n=-\infty }^{\infty }\:\sum _{M\:}^{\:}\:\delta \left(M\right)X\left(\omega -M-\frac{2\pi n}{T_s}\right)=\frac{1}{T_s}\sum _{n=-\infty }^{\infty }X\left(\omega -\frac{2\pi n}{T_s}\right)$$
But the problem is, the index of M, since:
$M=k-\frac{2\pi n}{T_s},\:k\in \left(-\infty ,+\infty \right)\:and\:n\in \left(-\infty ,\:+\infty \right)$
So what exactly is the index of $M$? where does it start? where does it end?