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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

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    @Didier: You seem to be reading$a$condition into the question that I cannot see there. Yes, the construction works for _every_ sum -- but is it somehow my fault that the question as asked is so easy to answer? And I'm still not "advocating" anything. Answering the question being asked does not constitute advocacy. If the OP wanted$a$single integral that had _every_ instance of his question as a Riemann sum, he should have said that.2012-01-09

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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?

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    @yotamoo: For the sum to look like the simplest kind of Riemann sum (equal-sized intervals, function evaluation at endpoints of the intervals) we would like to express our sum as $\sum \frac{L}{n}f(iL/n)$. Summation goes from $i=0$ to $n-1$, or $i=1$ to $n$. For reasonable $f$, the sum will approach $\int_0^L f(x)\,dx$. In this comment, we have $x_i=iL/n$, or maybe $(i-1)L/n$, it doesn't matter.2012-01-07
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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 $\Delta x$.)