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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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$\dfrac n{n+i}=\dfrac 1{1+\frac in}$ will help. –  Davide Giraudo Jan 7 '12 at 19:10
    
What are you trying to achieve with the rewriting you show? Simply $\int_0^{n} \frac{1}{n+x} dx$ ought to satisfy the basic condition of "an integral that has your sum as one of its Riemann sums", but perhaps you have further requirements that you have not been explicit about? –  Henning Makholm Jan 7 '12 at 19:12
    
@Davide's hint is probably the right one if your eventual goal is that the sum is at least $\frac 12$ (which is a standard step in proving that the harmonic series diverges). –  Henning Makholm Jan 7 '12 at 19:14
    
    
Your sum should start at $i=0$... –  David Mitra Jan 7 '12 at 19:35
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2 Answers

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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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Can you please explain why doing so (dividing by n)? What do I want my sum to look like before I can integrate over it? Thanks! –  yotamoo Jan 7 '12 at 19:24
    
@yotamoo: What is your definition of a Riemann sum? –  Did Jan 7 '12 at 19:42
    
$\sum_{i=1}^{n}{f(x_i)\Delta x}$ –  yotamoo Jan 7 '12 at 19:46
    
@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. –  André Nicolas Jan 7 '12 at 19:52
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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 $\Delta x$.)

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