limit of the sum $\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $ Prove that : $\displaystyle \lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$

the only thing I could think of is that it can be written like this :
$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{1}{k+n} =\lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{\frac{k}{n}+1}=\int_0^1 \frac{1}{x+1} \ \mathrm{d}x=\ln 2$$
is my answer right ? and are there any other method ?(I'm sure there are) 

The function $f(x)=\frac{1}{1+x}$ is continuous on $[0,1]$, hence the Riemann sum converges to the integral:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac kn}= \int_0^1\frac{\mathrm dx}{1+x}=\ln 2$$
$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n+k}=\ln 2$$
$$\lim_{n\to \infty} \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}=\ln 2$$
