Evaluate $ \lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n+r}$ by expressing as Riemann sum $$ \lim_{n\to \infty} \left(\frac{1}{n+1} + \frac{1}{n+2} + ..+ \frac{1}{2n}\right)$$
How do i find the limit by expressing it as a definite integral of an appropriate function via Riemann Sums?
I do know that for riemann sums
$\Delta x = \frac{b-a}{n}$ and $ x_i^* = a + \frac{b-a}{n}$
So i would need something like this 
$$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}\cdot f\left(a+i\frac{b-a}{n} \right)$$
But how should i go about solving the question?

Notice That : $$\frac{1}{n+r}=\frac{1}{n}\frac{1}{1+\frac{r}n}$$
So our some can be written as
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r} =
\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}$$
Now by replacing $\frac{1}{n}$ with $dx$,$\frac{r}{n}$ with $x$ and $\sum$ with $\int$ 
$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}=
\lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}$$
Now comparing this with the form you want we have $b=1$ , $ a=0$ and $f\left(x\right)=\frac{1}{1+x}$
$$\lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}=\int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx$$
And resultant integral evaluates to
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r}=\int_0^1\frac{1}{1+x}dx=\ln\left(\frac{2}{1}\right)=\ln(2)$$
