fourier transform - Aliasing after downsampling

Let me start with time domain representation of the original signal

$$\begin{equation} x_n=\sum_{k=0}^{2N-1}X_ke^{j\frac{2\pi nk}{2N}} \end{equation}$$ where $2N$ is number of time/frequency samples while $n$ and $k$ are time and frequency indices, respectively. Downsampling by factor of $M$ is essentially multiplying the time domain signal with the comb function which selects every M-th sample while others are set to zero, i.e.,

$$\begin{equation} \check{x}_{n}=x_n\cdot III(n;M) \end{equation}$$

where comb function $III(n;M)=\sum_\limits{{m=0}}^\limits{{\frac{2N}{M}-1}}\delta(n-mM)=\sum_\limits{{m=0}}^\limits{{\frac{2N}{M}-1}}e^{j\frac{2\pi nm}{M}}$. In other words downsampled signal is equal to

$$\begin{equation} \check{x}_{n}=\sum_{k=0}^{2N-1}\sum_\limits{{m=0}}^\limits{{\frac{2N}{M}-1}}X_ke^{j\frac{2\pi nk}{2N}}e^{j\frac{2\pi nm}{M}}=\sum_{k=0}^{2N-1}X_k\sum_\limits{{m=0}}^\limits{{\frac{2N}{M}-1}}e^{j2\pi n(\frac{k}{2N}+\frac{m}{M})} \end{equation}$$

• Are described steps correct up to now?


• I assume that at one point $\sum_{k=0}^{2N-1}$ should become $\sum_{k=0}^{(2N/M)-1}$ since by downsampling I am actually reducing max "visible" frequency.This combined with summation over $m$ should ,I suppose, mimic aliasing but I am having problems mathematically showing that. So if somebody could make this a little bit more clear to me.

p.s. I know there a couple of topics related with downsampling and aliasing but although I've read them really carefully I was not able to grasp it completely.