DFT of real sinusoids - why sum over -$N/2$ to $N/2-1$ as opposed to $0$ to $N-1$? I'm going through a Coursera course on signal processing, and we're just introduced to DFTs.

We are told that if you have a complex sinusoidal signal 
$x[n]$ where $n=0,1...\ N-1$, its DFT is given as

$$X[k]=\sum_{n=0}^{N-1} x[n] e^{-j2\pi kn/N}$$

And this makes sense to me. Because $x[n]$ has $N$ samples. You take each one, multiply it with the sample of a basis function, essentially taking the dot product of the signal with the basis function, and you get how much of that frequency is present in the input signal.

But we are told that if you have a real sinusoidal signal $y[n]$ where $n=0,1...\ N-1$, then the DFT is given as 

$$X[k]=\sum_{n=-N/2}^{N/2-1} y[n] e^{-j2\pi kn/N}$$

My question: Why do we sum over the different values? What is the intuition and explanation behind this?

I'm not even sure what $y[-N/2]$ would mean.

Is the $-N/2^{th}$ sample the same as the $N/2^{th}$ sample because we assume the input signal is periodic? Even if so, why is taking the DFT summation over such a range beneficial? We could also do this for complex sinusoidal signals since they're periodic too right? What's the advantage?
The DFT (Discrete Fourier Transform) of an $N$-point sequence $\left\{x[n]\right\}$ is given as:
$$X[k]=\sum_{n=0}^{N-1} x[n] e^{-j2\pi kn/N}, \quad k=0, 1, \ldots, N-1\tag{1}$$

If we assume the sequence $\left\{x[n]\right\}$ and its DFT are complex, they can be expressed as follows (with the subscript $r$ and $i$ for real and imaginary part respectively): 

\begin{align}
x[n] &amp;= x_r[n]+jx_i[n], \quad 0\leq n\leq N-1\\
X[k] &amp;= X_r[k]+jX_i[k], \quad 0\leq k\leq N-1
\end{align}

With a bit of algebra and using equation $(1)$, you get:

\begin{align}
X_r[k] &amp;= \sum_{n=0}^{N-1}\left[ x_r[n]\cos\left(\frac{2\pi kn}{N}
\right)+x_i[n]\sin\left(\frac{2\pi kn}{N}
\right)\right]\\
X_i[k] &amp;= -\sum_{n=0}^{N-1}\left[ x_r[n]\sin\left(\frac{2\pi kn}{N}
\right)-x_i[n]\cos\left(\frac{2\pi kn}{N}
\right)\right]
\end{align}
If the sequence $\left\{x[n]\right\}$ is real-valued, this means $ x[n]=x_r[n]$ as $x_i[n]=0$. The DFT will then be expressed as:

$$X[k]=\sum_{n=0}^{N-1} x[n]\left[\cos\left(\frac{2\pi kn}{N}
\right)-j\sin\left(\frac{2\pi kn}{N}
\right)\right]\tag{2}$$

And it can be shown from equation $(1)$ and $(2)$ that for these real-valued signals: 

$$X(N-k)=X^*(k)=X(-k)\tag{3}$$


From the above results, it suffices to compute the DFT for half of the values of $k$ to get the full characteristic of the spectrum. From the $N$ values you compute less (roughly half) $k$ values. The boundaries of the summation are still the same, it still goes from $n=0$ to $N-1$ but your $k$'s shouldn't go up to $N-1$ as it will be a waste of resources. 
$$X[k]=\sum_{n=0}^{N-1}y[n] e^{-j2\pi kn/N}$$

For $N$ even, $ 0\leq k \leq N/2$ suffices. With the results from equation $(3)$ you can get the remaining values from $N/2 + 1$ to $N-1$ or from $-N/2 + 1$ to $-1$. 
You have $ 0\leq n \leq N$. So, $y[-N/2]$ does not exist and $y[N/2]$ is the value of your sequence $\left\{y[n]\right\}$ at $n=N/2$ (if $N$ is even).
However, for a real even sequence $\left\{y[n]\right\}$, the DFT $X[-N/2]$ equals $X[N/2]$ and this is the value at Nyquist.
Also note from equation $(3)$ that if $N$ is even, $X[0]$ and $X[N/2]$ must be real, and if $N$ is odd $X[0]$ is still real but you don't have a frequency bin at exactly $N/2$. Equation $(3)$ applies for real-valued sequences, what has been concluded from that equation doesn't apply for strictly complex-valued sequences. The advantage is that you've saved computations by computing only half of what you need to compute.