Calculate the limit of a sum Calculate 
$$
\lim_{n\to\infty}S_n
$$
when
$$
S_n= \sum _{i=-n}^{n-1}\frac{2n+i}{2n^2}
$$ 
by treating the sum as an area.
The problem here is that I have no idea how I would start. Can someone give me a hint?
 A: Hint. Write the Riemann sums for the integral
$$
\int_{-1}^1 \left(1 + \frac x 2 \right) dx
$$
Edit. To write the left Riemann sum for the function $f(x) = 1 + \frac x 2$ on the interval $I = [-1, 1]$ divide that interval into $2n$ subintevals
$$
I_i := [x_i, x_{i + 1}] \equiv \left[\frac i n, \frac {i + 1} n\right]
\quad -n \leq i \leq n - 1
$$
Each of them will have length $\Delta x = 1/n$. Now on each subinteval $I_i$ construct a rectangle with height $h_i$ equal to the value of the function $f$ on the left end of the inteval
$$
h_i := f(x_i) = 1 + \frac i {2n}
$$
The area of the union of the rectangles is
$$
S_n = \sum_{i = -n}^{n - 1} \Delta x \cdot h_i = \sum_{i = -n}^{n - 1} \frac 1 n\cdot \left( 1 + \frac i {2n} \right) = 
\sum _{i = -n}^{n - 1}\frac {2n + i} {2n^2}
$$
So the limit of $S_n$ for $n\to \infty$ is the area under the function $f$.
A: Hint: (ignoring that the sum is an area):
\begin{align}
S_n &= \sum_{i = -n}^{n-1} \left( \frac{2n}{2n^2} + \frac{i}{2n^2} \right)
\\
&= \frac{1}{n} \sum_{i = -n}^{n-1} 1 + \frac{1}{2n^2}\sum_{i=-n}^{n-1} i
\end{align}
Now, calculate each sum separately.  For the first sum, notice that we are adding $1$ to itself $2n$ times.  For the second term, notice that a of the terms will cancel.  Once you have an explicit form for $S_n$, then take the limit.
A: $2n^2S_n=\sum _{i=-n}^{n-1}(2n+i)$
$=(2n-n)+\{ 2n-(n-1)\}+\{ 2n-(n-2)\}+\cdots+\{ 2n+(n-2)\}+\{ 2n+(n-1)\}$
$=2n-n+2n-0+4n(n-1)$  (Observe that there are terms $2n-r,2n+r$ for $1\le r\le n-1$, each summing up to $4n$ and $2n-0$ and $2n-n$ do not have any pair)
$=4n^2-n $
So, $S_n=\frac{4n^2-n}{2n^2}=2-\frac 1 2\frac 1n$
$\lim_{n\to\infty}S_n=2$
