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My foreign school book is not clearest at this point, what is really needed with this? What does it mean that "assign definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$"? I am not sure, whether I should assign it like this $\lim_{n\rightarrow \infty} \int_{1}^{n} \frac{1}{n+x} dx$ and noticing that it is continuous then trying to find borders so that $F(a) - F(b)$?

(the book messes up all kind of Leibniz stuff at this point, a bit messy -- and just stating $\int f(x)\,dx= F(a)-F(b)$, but not even paying attention to different borders. As far as I know, it is important to specify whether borders depend on the integration factor)

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    ...good basic video related to this [here](http://www.youtube.com/watch?v=gFpHHTxsDkI) by patrickJMT.2012-02-04

3 Answers 3

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I suspect the following: $ \sum_{k=1}^n {1\over n+k}=\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}} $ Now interpret the right hand sum as a Riemann sum for the function $f(x)={1\over 1+x}$ over $[a,b]=[0,1]$ (for a fixed $n$, the partition of $[0,1]$ is $\{{1\over n}, {2\over n},\ldots, {n\over n} \}$ and the ${1\over n}$ is the common width of the subintervals).

Taking the limit as $n\rightarrow \infty$ gives the corresponding integral: $\lim_{n\rightarrow\infty}\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}} = \int_0^1 {1\over 1+x}\,dx.$


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    Is there still some way to solve it with definite integrals?2012-02-04
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The problem converts into an integral this way:

$\begin{align*}\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{n+k}&=\lim_{n \to \infty} \dfrac{1}{n} \cdot\sum_{k=1}^n \dfrac{n}{n+k}\\&=\lim_{n \to \infty} \dfrac{1}{n} \cdot \sum _{k=1}^n\dfrac{n}{n+k} \\&= \lim_{n\to \infty} \dfrac{1}{n} \cdot \sum_{k=1}^n \dfrac{1}{1+\frac{k}{n}}\end{align*}$

Now interpret the final limit as Riemann sum of the function $f(x)=\dfrac{1}{1+x}$

So, the limit in question equals, $\int_0^1{(\dfrac{1}{1+x}) \mathrm dx}= \ln 2$ $\blacksquare$

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    @hhh Did you try reading the article from Wikipedia, it is a bit Shabby, however. Look at David's Answer below. His picture should give you an idea of what is happening. And, Wolfram math world has an appplet which you could try and fiddle with until you are acquainted.2012-02-04
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Hint: You can write this as $\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}.$ How close is this to the Riemann sum of the integral $\int_0^1 \frac{1}{1+x}dx?$