Turning infinite sum into integral Could someone explain how one can replace infinite sum with integral?
Examples (or you may use your own, it doesn't matter as I want to understand principles):
$$\frac{1}{n} \sum_{i=1}^{n} \sin \frac{i-1}{n} \pi$$
$$\frac{1}{n} \sum_{i=1}^{n} \sqrt{1 + \frac{i}{n}}$$
I see that $\frac{1}{n}$ denote partition, so my problem is that I don't know how to figure out the function, hence, the integrand.
 A: The basic idea is this:
If $f$ is a positive increasing function
then
$f(n)
\le \int_n^{n+1} f(x) dx
\le f(n+1)
$.
Similarly,
if $f$ is a positive decreasing function
then
$f(n)
\ge \int_n^{n+1} f(x) dx
\ge f(n+1)
$.
For increasing $f$,
$\sum_{i=0}^{n-1} f(i)
\le \int_0^{n} f(x) dx
\le \sum_{i=0}^{n-1} f(i+1)
$
or
$\sum_{i=0}^{n} f(i)-f(n)
\le \int_0^{n} f(x) dx
\le \sum_{i=0}^{n} f(i)-f(0)
$
or
$f(0)
\le \sum_{i=0}^{n} f(i)-\int_0^{n} f(x) dx
\le f(n)
$.
For a decreasing function,
the inequalities are reversed.
From this you can show that
for a bounded piecewise monotonic function,
the sum converges to the integral.
A: The Left-hand Rectangular Approximation Method (LRAM) says that, if $f(x)$ is continuous on $[a,b]$,
$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac {b-a}{n}\sum_{i=1}^n f\left(a+\frac{b-a}n (i-1) \right)$$
Comparing that with your first sum, we see that $\ a=0,\ b=\pi, f(x)=\sin x/\pi$. So, the limit of your sum as $n\to\infty$ is
$$\frac{1}{\pi}\int_0^{\pi} \sin x\,dx$$
Note that this is not your sum, as you asked, not even an infinite sum, but a limit of sums.
The Right-hand Rectangular Approximation Method (RRAM) says that, if $f(x)$ is continuous on $[a,b]$,
$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac {b-a}{n}\sum_{i=1}^n f\left(a+\frac{b-a}n i \right)$$
Comparing that with your second sum, we see that $f(x)=\sqrt x,\ a=1,\ b=2$. So 
the limit of your sum as $n\to\infty$ is
$$\int_1^2 \sqrt x\,dx$$
Note: The main difference in these two methods is that LRAM has the term $f(a)$ but not the term $f(b)$ while RRAM has the term $f(b)$ but not the term $f(a)$ in the sum. This is due to the different indices on the sum. 
