DFT of real sinusoids - why sum over -$N/2$ to $N/2-1$ as opposed to $0$ to $N-1$?
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.