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) 

Set $a_n 
= 1/(n+1) + 1/(n+2)+···+ 1/(2n)
=\sum_{k=1}^n \dfrac1{n+k}
$.
$\begin{array}\\
a_{n+1}-a_n 
&=\sum_{k=1}^{n+1} \dfrac1{n+1+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\sum_{k=2}^{n+2} \dfrac1{n+k}-\sum_{k=1}^n \dfrac1{n+k}\\
&=\dfrac1{n+n+2}+\dfrac1{n+n+1}-\dfrac1{n+1}\\
&=\dfrac1{2n+2}+\dfrac1{2n+1}-\dfrac1{n+1}\\
&=-\dfrac1{2 (n + 1) (2 n + 1)}\\
&< 0\\
\text{and}\\
a_{n+1}-a_n 
&=-\dfrac1{2 (n + 1) (2 n + 1)}\\
&>-\dfrac1{2n (n + 1)}\\
\text{so}
&\text{if } m > 0\\
a_{m+n}-a_n
&< 0\\
\text{and}\\
a_{m+n}-a_n 
&=\sum_{k=0}^{m-1}(a_{n+k+1}-a_{n+k})\\ 
&>-\sum_{k=0}^{m-1}(\dfrac1{2(n+k+1) (n + k+2)})\\ 
&=-\dfrac12\sum_{k=0}^{m-1}(\dfrac1{(n+k+1)}-\dfrac1{(n + k+2)})\\ 
&=-\dfrac12(\dfrac1{(n+1)}-\dfrac1{(n +m+1)})\\ 
&>-\dfrac1{2(n+1)}\\ 
\end{array}
$
so,
for $m > 0$,
$|a_{m+n}-a_n|
\lt \dfrac1{2(n+1)}
\to 0
$
as $n \to \infty$
so,
by Cauchy,
$a_n$
converges.
