5

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

yotamoo
  • 2,753
  • 6
    $\dfrac n{n+i}=\dfrac 1{1+\frac in}$ will help. – Davide Giraudo Jan 07 '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? – hmakholm left over Monica Jan 07 '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). – hmakholm left over Monica Jan 07 '12 at 19:14
  • A closely related earlier question. – Jyrki Lahtonen Jan 07 '12 at 19:29
  • Your sum should start at $i=0$... – David Mitra Jan 07 '12 at 19:35
  • @Henning: Riemann sums are built upon a function $f$ independent on $n$. Your suggestion, which amounts to use $f:x\mapsto \frac1{n+x}[0\leqslant x\leqslant n]$, does not fit this frame. Note that Davide's suggestion leads to $\frac1n\sum\limits_{i=0}^{n-1}f(\frac{i}n)$ with $f:x\mapsto\frac1{1+x}[0\leqslant x\leqslant 1]$ independent on $n$, hence it fits the frame of a classical Riemann sum. – Did Jan 07 '12 at 19:38
  • @Didier, I don't understand. Why can't I integrate a function that depends on $n$ and find Riemann sums for that integral? Is that a special property of the letter $n$, or are you claiming in general that a function that depends on a parameter does not have Riemann sums? Which "frame" is it that you want to "fit" the function into? The question does not specify one. – hmakholm left over Monica Jan 07 '12 at 19:43
  • @Henning: I do not understand what you don't understand. First, EVERY given sum can be written as $\frac1n\sum\limits_{i=0}^{n-1}f_n(\frac{i}n)$ for some $f_n$, hence the notion you advocate is so wide it is empty. Second, to know what everybody calls Riemann sums, see here. – Did Jan 08 '12 at 00:40
  • @Didier, I'm not "advocating" anything here, I'm just going by what the OP appears to ask for: namely an integral such that his sum arises as one of the Riemann sums of that integral. I agree that this is not much to ask for, but I don't see any other condition in the question. I don't know what you're trying to achieve by linking to the Wikipedia article -- do you really think we disagree about what a Riemann sum is? Thus: What I don't understand is how you think my integral is does not satisfy the (very weak) condition being asked for. – hmakholm left over Monica Jan 08 '12 at 02:26
  • @Henning Back to the subject, please... Which sums of $n$ terms, each sum depending on the positive integer valued parameter $n$, are Riemann sums? According to (me and to) WP (and to what you now say you know well), when the $i$th term is $\frac1nf(a+(b-a)\frac{i}n)$ for some $a\lt b$ and some function $f:[a,b]\to\mathbb R$, with $a$, $b$ and $f$ independent on $i$ and $n$. Hence the sums $\sum\limits_{i=0}^{n-1}\frac{1}{n+i}$ are Riemann sums (note the plural), while the sums $\sum\limits_{i=0}^{n-1}\frac{n}{1+in}$ are not Riemann sums. (The WP link answers your request for a "frame".) – Did Jan 08 '12 at 10:09
  • @Didier: Where do you get "with $a$, $b$ and $f$ independent on $n$" from? There is no such requirement in the question as far as I can see. – hmakholm left over Monica Jan 09 '12 at 13:42
  • @Henning: Oh, so now you ARE advocating using a function $f_n$ depending on $n$... Hence we fall back on the fact that EVERY mathematical expression can be written as such (and each one for every integer $n$ at the same time...). What more needs to be said? Sorry but I will take my leave. – Did Jan 09 '12 at 13:59
  • @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. – hmakholm left over Monica Jan 09 '12 at 14:13

2 Answers2

3

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?

André Nicolas
  • 507,029
  • 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 07 '12 at 19:24
  • @yotamoo: What is your definition of a Riemann sum? – Did Jan 07 '12 at 19:42
  • $\sum_{i=1}^{n}{f(x_i)\Delta x}$ – yotamoo Jan 07 '12 at 19:46
  • 1
    @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 07 '12 at 19:52
1

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 $\Delta x$.)

David Mitra
  • 74,748