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 $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}.$$
We can use the following uniform bound
$$a_n = \sum_{k=1}^n \frac{1}{n+k} \leq\sum_{k=1}^n\frac{1}{n+1} = \frac{n}{n+1} < 1$$
Since each $a_n$ is positive, we have that $0\leq a_n < 1$ for all $n$. For monotonicity we use the calculation
$$\begin{align*}a_{n}-a_{n+1} &= \sum_{k=1}^n\frac{1}{n+k}  - \sum_{k=1}^{n+1}\frac{1}{(n+1) + k}\\
&= \frac{1}{n+1} - \frac{1}{2n+1} - \frac{1}{2n+2}\\
&= \frac{(2n+1)(2n+2)- (n+1)(2n+2) - (n+1)(2n+1)}{(n+1)(2n+1)(2n+2)}\\
&= -\frac{(n+1)}{(n+1)(2n+1)(2n+2)}\\
&= -\frac{1}{(2n+1)(2n+2)} < 0\\
\end{align*}$$
We thus have that $\{a_n\}$ is bounded and the statement above shows that $a_n < a_{n+1}$ for any $n$ so that $\{a_n\}$ is also monotonic. Hence the sequence converges.
