Converting a Riemann sum to an integral Given this sum:
$$ \frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n-1}$$
I am trying to convert (approximate) it to an integral. This is what I have so far:
$$ 
\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n-1} = 
\frac{1}{n} \left( 1 + \frac{n}{n+1} + \cdots + \frac{n}{2n-1} \right) = 
\sum_{i=1}^{n-1}{\frac{1}{n}\frac{n}{n+i}}$$ 
How do I continue from here? Also, how do I set the limits of the integral once I find it? 
What do I want my sum to look like before I can integrate over it? Are there any conditions?
Thanks
 A: Hint: You are close,  a little more manipulation will do it. In the expression $\frac{n}{n+i}$, divide "top" and "bottom" by $n$. We get 
$$\frac{1}{1+\frac{i}{n}},$$
which is the value of $f$ at $\frac{i}{n}$, with $f(x)=\frac{1}{1+x}$.  
We can reach the same conclusion  in one step, by noting that 
$$\frac{1}{n+i}=\frac{1}{n}\frac{1}{1+\frac{i}{n}}.$$
In any case, our sum is equal to
$$\sum_{i=0}^{n-1}\frac{1}{n}f(i/n), \qquad\qquad(\ast)$$
which is a familiar type of Riemann sum. 
The simplest kind of Riemann sum has shape
$$\sum \frac{L}{n}f(iL/n),$$ 
where we sum from $i=0$ to $n-1$ (equal-width intervals, evaluation at left endpoints) or from $i=1$ to $n$ (evaluation at right endpoints). This was the motivation for trying to express our terms as $\frac{1}{n}f(i/n)$. If the function $f$ is well-behaved, the limit as $n \to\infty$ of these Riemann sums is
$$\int_0^L f(x)\,dx.$$
For another way to identify the interval of integration, note that we are evaluating $f$ at the numbers $\frac{0}{n}$, $\frac{1}{n}$, $\frac{2}{n}$, and so on up to $\frac{n-1}{n}$.  What interval are these (equally spaced) division points of?
A: Hint: you want the terms of sum to contain $i/n$ (but keep the ${1\over n}$ as it is, this is your $\Delta x$), so write what you have as $\sum\limits_{i=0}^{n-1} {1\over n} {1\over 1+{i\over n}}$.
To identify the limits of integration: 
Approximately where does the interval start? (Answer: at $i/n$ for $i=0$.)
Approximately where does the interval end?  (Answer: at $i/n$ for $i=n-1$.)
(remember, here, that $n$ is big...)
What is the function?  (Answer: try to recognize $f(i/n)$ in the sum, keeping in mind that the expression $1\over n$ is your $\Delta x$.)
