How to prove that $\lim\limits_{n \to \infty} \sum\limits_{k = 1}^{n} {\sqrt k \over n^{1.5}} = {2 \over 3}$ I am currently trying to prove: $\lim\limits_{n \to \infty}  \sum\limits_{k = 1}^{n} {\sqrt k \over n^{1.5}} = {2 \over 3}$
I can easily squeeze the series between 0 and 1. 
I don't know many handy tricks/tools one can use to compute those kind of series apart from Euler thought about trying to turn this into an Euler sum. 
I can use a vague argument that the $ \lim  \sum $ sort is $\int\limits_{0}^{1} \sqrt x dx  $, however I don't feel comfortable because I dislike summing up infinitely many infinitely small values to "magically" get the right result.
Thus I would be very happy, if there were actually another way to deal with such kind of problems. 
 A: Hint:
$$\sum_{k =1}^n \frac{\sqrt{k}}{n^{1.5}} = \frac{1}{n} \sum_{k=1}^n \sqrt{\frac{k}{n}}$$
A: If you do not feel comfortable with the Riemann sums magic, just consider that $f(x)=\sqrt{x}$ is a concave function on $[0,1]$, hence:
$$ \frac{1}{n}\left(\frac{1}{2}f(0)+f\left(\frac{1}{n}\right)+\ldots+f\left(\frac{n-1}{n}\right)+\frac{1}{2}f(1)\right)\leq \int_{0}^{1}\sqrt{x}\,dx \tag{1}$$
as well as:
$$\int_{0}^{1}\sqrt{x}\,dx\leq \frac{1}{n}\left(f\left(\frac{1}{n}\right)+\ldots+f(1)\right),\tag{2} $$
since the LHS of $(1)$ gives the trapezoidal approximation of the integral, while the RHS of $(2)$ gives the rectangular approximation. Since the integral equals $\frac{2}{3}$, we have:
$$ \frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}}=\frac{2}{3}+O\left(\frac{1}{n}\right).\tag{3}$$
A: According to the integral test, we have
$$\int_1^{n+1}\frac{\sqrt k}{n^{1.5}}\ dx\le\sum_{k=1}^n\frac{\sqrt k}{n^{1.5}}\le1+\int_1^n\frac{\sqrt k}{n^{1.5}}\ dx$$
Evaluating these, we end up with
$$\frac23\left(1+\frac1n\right)^{1.5}-\frac2{3n^{1.5}}\le\sum_{k=1}^n\frac{\sqrt k}{n^{1.5}}\le\frac1{n^{1.5}}+\frac23$$
Taking the squeeze theorem as $n\to\infty$,
$$\lim_{n\to\infty}\sum_{k=1}^n\frac{\sqrt k}{n^{1.5}}=\frac23$$
