# Help needed with definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$

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)

-
...good basic video related to this here by patrickJMT. – hhh Feb 4 '12 at 14:11

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: $$\lim_{n\rightarrow\infty}\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}} = \int_0^1 {1\over 1+x}\,dx.$$

-
+1 for the illustration with picture! – user21436 Feb 4 '12 at 15:02
+1 thanks! I finally understood the formula after your updated image. – hhh Feb 4 '12 at 18:23
How about if you have a situation so that the height is in different form such as $\frac{1}{n+\frac{k}{n}}$ or $\frac{1}{n+\frac{k}{n^{2}}}$? What about can I have a first term such as $\frac{1}{n^{2}}$ or $\frac{1}{n^{3}}$? – hhh Feb 4 '12 at 21:49
@hhh Generally, for an integral over $[a,b]$, you use something of the form $f(a+[(b-a)i/n])\cdot{1\over n}$. The $1/n$ is the width of the rectangles for the partition points $a+[(b-a)i/n]$ (other widths could be used, like $1/(3n)$). The height is obtained by evaluating $f$ at the partition points. I don't think $1\over n+(i/n)$ would get you one, since there is no way to write that as $f(a+[(b-a)i/n]$ for some $f$. Other expressions would work: e.g if you have $(i/n)^3+{1\over (i/n)^2+1}$, then your $f$ would be $f(x)=x^3+{1\over x^2+1}$ – David Mitra Feb 4 '12 at 22:20
@hhh I don't think that will work. Basically, you chop $[a,b]$ into $n$ pieces so that the maximum length of the pieces is small. The widths can't all be $(1/n)^2$. If you add the lengths of the pieces, you have to get $b-a$. – David Mitra Feb 4 '12 at 22:57

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, $\int_0^1{(\dfrac{1}{1+x}) \mathrm dx}= \ln 2$ $\blacksquare$

-
...thank you, can you suggest some reading about this topic? (My book is not the best one to practise on this, it is introduces Rieman sum after many pages the questions dealing with the rieman sums and the organizaton of material is not most pedagocical-- and I cannot make head-and-tail about what it is really trying to explain, have to look for some English material on this.) – hhh Feb 4 '12 at 14:05
@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. – user21436 Feb 4 '12 at 15:02

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 $$\int_0^1 \frac{1}{1+x}dx?$$

-