Limit of $\sum_{j=1}^n\frac{1}{n}\sqrt{\frac{j}{n}}$ as n approaches infinity So I've been going over some exercies meant as preparation for the final exam of my 2nd semester of real analysis. One of the questions is as follows:
Show that $\lim(\sum_{j=1}^n\frac{1}{n}\sqrt{\frac{j}{n}})=2/3$ for n going to infinity. From other similar questions on mathexchange I'm thinking that the solution is to be found somewhere in the definition of the Riemann Integral, but so far I'm not really seeing the solution here.
I'd deeply appreciate some help and explanation
 A: Consider the Riemann sum for integrals 
$$\int_a^bf\left(x\right)dx=\lim_{n\to\infty}\sum_{r=1}^nf\left(a+j\left(\lfloor\frac{\left(b-a\right)}{n}\rfloor\right)\right)\left(\frac{\lfloor\left(b-a\right)}{n}\rfloor\right).$$
Substitute $b=1$ and $a=0$.
This implies $f(x)=\sqrt{x}$
now evaluate the integral
$$\int _0^1\sqrt{x}dx.$$
Which is indeed equal to 2/3
A: Note that
$$
\sum\limits_{j = 1}^n {\frac{1}{n}\sqrt {\frac{j}{n}} }  = \sum\limits_{j = 1}^n {\left( {\frac{{j + 1}}{n} - \frac{j}{n}} \right)\sqrt {\frac{j}{n}} } .
$$
Can you see now that this is a Riemann sum?
A: The limit is the same as the Riemann sum of $$\sqrt{x}$$ over the interval $[0,1]$. Therefore the limit is $$\int _0 ^1 \sqrt{x} dx.$$
A: 
I thought that it might be instructive to present a way forward without appealing to Riemann sums or integral bounds.  To that end we proceed.


First we evaluate the telescoping series 
$$\sum_{j=1}^n \left(j^{3/2}-(j-1)^{3/2}\right)=n^{3/2}\tag1$$

Next, we expand the general term of the sum in $(1)$ as
$$\begin{align}
j^{3/2}-(j-1)^{3/2}&=j^{3/2}\left(1-\left(1-\frac1j\right)^{3/2}\right)\\\\
&=\frac{3\sqrt j}{2}+O\left(\frac1{\sqrt j}\right)\tag2
\end{align}$$

Then using $(2)$, it is easy to see that
$$\begin{align}\frac23\sum_{j=1}^n \frac{\left(j^{3/2}-(j-1)^{3/2}\right)}{n^{3/2}}=\sum_{j=1}^n \frac{\sqrt j}{n^{3/2}}+O\left(\frac1n\right)\end{align}\tag3$$

Finally, using $(1)$ in $(3)$ reveals
$$\sum_{j=1}^n \frac1n \sqrt{\frac jn}=\frac23+O\left(\frac1n\right)\tag4$$
whereupon taking the limit as $n\to\infty$ yields the coveted result
$$\lim_{n\to\infty}\sum_{j=1}^n \frac1n \sqrt{\frac jn}=\frac23$$
as was to be shown!

TOOLS USED:  Telescoping Series, Taylor's Theorem
A: On the interval $[0,1]$ difine the partition $$\{1/n,  2/n,...,n/n \}$$ 
With the function $f(x)= \sqrt {x}$ consider the right side Reimann Sum. 
Note that $\delta x = 1/n$ so the series in question represents the integral $$\sum_{j=1}^n\frac{1}{n}\sqrt{\frac{j}{n}}=\int _0^1 \sqrt {x} dx = 2/3$$ 
