TITLE: DTFT of Impulse train is equal to 0 through my equation.
QUESTION [1 upvotes]: Let me have an impulse train function as below,
$$ x[n] = \sum_{m=-\infty}^{\infty} {\delta[n-f_0 m]} $$
where, $f_0 \in \textbf{Z}$.
Now, I am trying to calculate its DTFT, so I put it into DTFT equation as below,
$$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} {\sum_{m=-\infty}^{\infty} {\delta[n-f_0 m]} e^{-j\omega n}} . $$
I substitute $n-f_0 m = p$, the equation can be re-written as below,
$$ X(e^{j\omega}) = \sum_{m=-\infty}^{\infty} { \left(\sum_{p=-\infty}^{\infty} {\delta[p]} e^{-j\omega p} \right) e^{-j\omega f_0 m}}. $$
In the parenthesis equation goes to 1, $X(e^{j\omega})$ only has the term as below,
$$ X(e^{j\omega}) = \sum_{m=-\infty}^{\infty} {e^{-j\omega f_0 m}}. $$
I divided this equation as two parts.
$$ X(e^{j\omega}) = \sum_{m=0}^{\infty} {e^{-j\omega f_0 m}}+\sum_{m=0}^{\infty} {e^{j\omega f_0 m}}-1. $$
And I calculated this and got the answer which was equal to 0.
$$ X(e^{j\omega}) = \frac{1}{1-e^{-j\omega f_0}} + \frac{1}{1-e^{j\omega f_0}} -1 = 0.$$
At this moment, I was stuck at here, because I could not find where is incorrect.
Please give me 
1) some comments for fixing this issue
2) How can I get the DTFT of impulse train? Is this approach is correct?
Thank you very much in advance!

@Matt L.
The infinite two sums can be considered as infinite series each of them.
So let me consider left part of above equation i.e. $ \sum_{m=0}^{\infty} {e^{-j\omega f_0 m}} $. 
Its common ratio is $e^{-j\omega f_0}$, and its first term is 1.
We can get above fraction, and we also can calculate the 2nd part as the same manner.

From your answer, I tried to calculate the infinite series as below.
If you don't mind, this procedure is correct or not, please let me know.
$$c_k = \frac{1}{N} \sum_{n=0}^{N-1} {\sum_{m=-\infty}^{\infty} {\delta[n-mN] e^{-j \frac{2\pi}{N} nk}} }.$$
Above equation can be written as below, and except the case $m=0$ all $m$'s condition are 0 because the definition of dirac delta sequence.
$$\sum_{n=0}^{N-1} {\sum_{m=-\infty}^{\infty} {\delta[n-mN] e^{-j \frac{2\pi}{N} nk}} } = \sum_{m=-\infty}^{\infty} {\left( \delta[0-mN] e^{-j \frac{2\pi}{N} 0k}+\delta[1-mN] e^{-j \frac{2\pi}{N} 1k} +...+ \delta[(N-1)-mN] e^{-j \frac{2\pi}{N} (N-1)k}\right)}.$$
In other words, $ \delta[1-mN]=\delta[2-mN]=...=\delta[(N-1)-mN]$ equals to 0 whatever $m$ is.
So, $c_k=1/N$.
I understood up to here, and from now on I should understand DTFT of $x[n]$, i.e. $X(e^{j\omega})$.
Thank you for your comment.

@Matt L.
Thank you always for your comment. Now I am struggling with the equation from $x[n]$ to $X(e^{j\omega})$.
What I am doing is as below, I am stuck at the infinite sum.
Because, $x[n]$ is
$$x[n] = \frac{1}{N}\sum_{k=0}^{N-1}{e^{-j\frac{2\pi}{N}nk}},$$
the DTFT can be written as below,
$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} { \sum_{n=-\infty}^{\infty}{e^{-j \left( \frac{2\pi}{N}k - \omega \right)n} } }.$$
I think this can be converted as below, 
$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} {2\pi \delta \left(\omega-\frac{2\pi}{N}k \right)}. $$
Since the DTFT of $e^{j2\pi kn/N}$ is $2\pi\delta(\omega-2\pi k /N)$ as you mentioned previously.
Is the DTFT means that as below? I think the definition is DTFT is as below. If it is, I have no idea why the DTFT of impulse train is $k$ from $-\infty$ to $\infty$.
$$\sum_{n=-\infty}^{\infty}{e^{-j \left( \frac{2\pi}{N}k - \omega \right)n} }.$$
So, the above $X(e^{j\omega})$ could be re-written as
$$ X(e^{j\omega}) = \frac{1}{N} \sum_{k=0}^{N-1} {DTFT(e^{j2\pi kn/N})}.$$
Please give me some comment for fixing this, and understanding the DTFT of impulse train.
Thank you always!

REPLY [6 votes]: A simple approach to show that the DTFT of an impulse train is an impulse train in the frequency domain, is to represent the periodic impulse train by its Fourier series:
$$x[n]=\sum_{m=-\infty}^{\infty}\delta[n-mN]=\sum_{k=0}^{N-1}c_ke^{j2\pi kn/N}\tag{1}$$
where the Fourier coefficients $c_k$ are given by
$$c_k=\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$
For the given $x[n]$, we have $c_k=1/N$ and from (1)
$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}e^{j2\pi kn/N}\tag{2}$$
Since the DTFT of $e^{j2\pi kn/N}$ is $2\pi\delta(\omega-2\pi k /N)$ (periodically continued with period $2\pi$), the DTFT of $x[n]$ is
$$X(e^{j\omega})=\frac{2\pi}{N}\sum_{k=-\infty}^{\infty}\delta(\omega-2\pi k/N)$$
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