Answer
$\lim\limits_{n \to \infty}\frac{1}{n}( \sqrt{\frac{1}{n}} +\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+...+\sqrt{\frac{n}{n}}) = \frac{2}{3}$
Work Step by Step
An integral can be expressed as the limit of a Riemann sum:
$\int_{a}^{b}f(x)~dx = \lim\limits_{n \to \infty}\sum_{i=1}^{n}f(x_i^*)\Delta x$
Consider the interval $[0,1]$:
$\Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n}$
$x_i^* = a+i~\Delta x = 0+\frac{i}{n} = \frac{i}{n}$
Note that $x_i^*$ is the right endpoint of each subinterval.
$\lim\limits_{n \to \infty}\frac{1}{n}( \sqrt{\frac{1}{n}} +\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+...+\sqrt{\frac{n}{n}})$
$= \lim\limits_{n \to \infty}\sum_{i=1}^{n}\sqrt{\frac{i}{n}}\cdot \frac{1}{n}$
$= \int_{0}^{1}\sqrt{x}~dx$
$= \frac{2}{3}x^{3/2}~\vert_{0}^{1}$
$= \frac{2}{3}[(1)^{3/2}-(0)^{3/2}]$
$= \frac{2}{3}(1-0)$
$= \frac{2}{3}$
