Determining Fourier Series coefficient for Discrete time
In the formula displayed for $x[n]$, the exponentials $e^{jk\omega_0 n}$ should be $e^{-jk\omega_0 n}$. I'm assuming that the sum $\sum\limits_{n = \langle N\rangle}$ is extended over all integers $n$ from $0$ to $N-1$. Fix $r\in \{0,\ldots, N-1\}$. Then 

$$\sum_{n = \langle N\rangle}x[n]e^{jr\omega_0 n} = \sum_{n = \langle N\rangle}\sum_{k = \langle N\rangle}a_ke^{-jk\omega_0 n}e^{jr\omega_0 n} = \sum_{k = \langle N\rangle} a_k \sum_{n = \langle N\rangle} e^{j(r-k)\omega_0 n}.\tag{*}$$

If $k = r$, then each of the terms $e^{j(r-k)\omega_0 n}$ equals $1$, in which case 

$$\sum_{n = \langle N\rangle} e^{j(r-k)\omega_0 n} = N.$$

If $k \neq r$, let $z = e^{j(r-k)\omega_0}$. Then $z\neq 1$ and $z^N = e^{2\pi j(r-k)n} = 1$. Thus

$$\sum_{n = \langle N\rangle} e^{j(r-k)\omega_0 n} = \sum_{n = \langle N\rangle} z^n = \frac{1 - z^N}{1 - z} = 0.$$

In summary, 

$$\sum_{n = \langle N\rangle} e^{j(r-k)\omega_0 n} = N\delta_{k,r}.$$

Therefore,

$$\sum_{k = \langle N\rangle} a_k \sum_{n = \langle N\rangle} e^{j(r-k)\omega_0 n} = \sum_{k = \langle N \rangle} a_k (N\delta_{k,r}) = N\sum_{k = \langle N\rangle} a_k \delta_{k,r} = Na_r.\tag{**}$$

Combining (*) and (**), 

$$Na_r = \sum_{n = \langle N\rangle} x[n]e^{jr\omega_0 n} = \sum_{n = \langle N\rangle}x[n]e^{jr(2\pi/N)n}.$$

Hence

$$a_r = \frac{1}{N}\sum_{n = \langle N\rangle} x[n]e^{jr(2\pi/N)n}.$$