Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$. 
$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$$

$$\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}} =\lim_{n\to \infty} \dfrac1{n}\sum^{n}_{k = 1} \sqrt{\dfrac k n} $$
While searching this question I found,
Turning infinite sum into integral. 
Like in the accepted answer I first compared my series to LRAM,  
$$\int_a^b f(x)\,dx=\lim_{n\to\infty}\frac 1n\sum_{i=1}^n f\left(a+\frac{b-a}n i \right)$$
I got $a = 0$, $b = 1$ and $f(x) =\sqrt{x}$ so,
$$\int_0^1 \sqrt{x}\ dx = \dfrac2 3$$ should be the answer.
Is there any simpler method to do this sum ? I have not learnt this method to do infinite sums so I can't use it.  

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$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; (n+1) \sqrt {n+1}  \; - \; 1 \right)  \; < \;  \frac{2}{3} \left( \; (n+1) ( 1 +\sqrt n)  \; - \; 1 \right) =  \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n  \right)   $$
$$ \frac{2}{3} n \sqrt n < \mbox{SUM} <  \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n  \right)   $$
For anyone worried about the little estimate above,
$$ n + 1 < n + 2 \sqrt n + 1, $$
$$ \sqrt {n+1} \; \;  < \; \; 1 + \sqrt n. $$
