Let $a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}$ calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals 
Let
$$
a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}}
$$

My goal is to calculate $\displaystyle \lim_{n \to \infty}a_n$ using integrals.
What I have tried so far and got stucked:
Let:
$$f(x) = \sqrt{n + x}$$
Therefore:
$$a_n = \frac{\sqrt{n+1} + \sqrt{n+2} +...+ \sqrt{2n}}{n^{3/2}} = \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{\sqrt{n}}\right)$$
Now, I know that there is a relation between the Riemann sum and integrals generally speaking:
$$
\lim_{n \to \infty} \frac{b-a}{n} \sum_{k = 1}^n f\left(a + \frac{k(b-a)}{n}\right) = \int^{b}_{a}f(x) dx
$$
Therefore, what I think I can conlude so far from $\displaystyle \frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{\sqrt{n}}\right)$ is:
$$
b = 1, a = 0, f(x) = \sqrt{n + x}
$$
But it seems wrong, $f(x)$ has $n$ in it, which is weird as a constant here, so i am not sure i did the whole correlation quite right.

if we have $f(x) > 0$ and $f'(x) > 0,$  then
$$ \int_{a-1}^{b} \; f(x) \;  dx \;  < \; \sum_{j=a}^b \;  f(j) \; < \; \int_{a}^{b+1} \;  f(x) \; dx   $$
Taking $a=n+1,$ $ b = 2n,$
$$ \int_{n}^{2n} \; \sqrt x \;  dx \;  < \; \sum_{j=n+1}^{2n} \;  \sqrt j \; < \; \int_{n+1}^{2n+1} \;  \sqrt x \; dx   $$
and a suitable antiderivative is $$ \frac{2}{3} n^{3/2}  \; .  $$
$$ \frac{2}{3} (2n)^{3/2} -  \frac{2}{3} (n)^{3/2} < \; \sum_{j=n+1}^{2n} \;  \sqrt j \; < \;  \frac{2}{3} (2n+1)^{3/2} -  \frac{2}{3} (n+1)^{3/2} $$
The left hand side is $$ \frac{2n^{3/2}}{3} (2^{3/2} - 1) $$
