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)
 A: 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, $\int_0^1{(\dfrac{1}{1+x}) \mathrm dx}= \ln 2$ $\blacksquare$
A: 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:  $$\lim_{n\rightarrow\infty}\sum_{k=1}^n{1\over n} {1\over 1+{k\over n}} = \int_0^1 {1\over 1+x}\,dx.$$


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