Question

$\small\displaystyle \sf \lim_{n \to \infty} \bigg( \frac{1}{n} \sqrt{ \frac{1}{n} } + \frac{1}{n} \sqrt{ \frac{2}{n} } + \frac{1}{n} \sqrt{ \frac{ 3}{n} } + \cdot \cdot \cdot + \frac{1}{n} \sqrt{ \frac{n}{n} } \bigg)$​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \sf \lim_{n \to \infty} \bigg( \frac{1}{n} \sqrt{ \frac{1}{n} } + \frac{1}{n} \sqrt{ \frac{2}{n} } + \frac{1}{n} \sqrt{ \frac{ 3}{n} } + \cdots+ \frac{1}{n} \sqrt{ \frac{n}{n} } \bigg)$

$\displaystyle \sf = \lim_{n \to \infty} \dfrac{1}{n} \bigg( \sqrt{ \frac{1}{n} } + \sqrt{ \frac{2}{n} } + \sqrt{ \frac{ 3}{n} } + \cdots \sqrt{ \frac{n}{n} } \bigg)$

$\displaystyle \sf = \lim_{n \to \infty} \dfrac{1}{n} \bigg( \sum ^{n}_{r = 1} \sqrt{ \frac{r}{n} } \bigg)$

$\displaystyle = \int^{1}_{0} \sqrt{x} \: dx$

$\displaystyle = \dfrac{2}{3} \left[x ^{ \frac{3}{2} } \right]^{1}_{0}$

$\displaystyle = \dfrac{2}{3} \left[1 - 0\right]$

$=\dfrac{2}{3}$

Answer 2

Answer:

$\dag \sf \green{Solution}$

We have,

$\displaystyle \sf \lim_{n \to \infty} \bigg( \frac{1}{n} \sqrt{ \frac{1}{n} } + \frac{1}{n} \sqrt{ \frac{2}{n} } + \frac{1}{n} \sqrt{ \frac{ 3}{n} } + \cdots+ \frac{1}{n} \sqrt{ \frac{n}{n} } \bigg)$

$\displaystyle \sf = \lim_{n \to \infty} \dfrac{1}{n} \bigg( \sqrt{ \frac{1}{n} } + \sqrt{ \frac{2}{n} } + \sqrt{ \frac{ 3}{n} } + \cdots \sqrt{ \frac{n}{n} } \bigg)$

$\displaystyle \sf = \lim_{n \to \infty} \dfrac{1}{n} \bigg( \sum ^{n}_{r = 1} \sqrt{ \frac{r}{n} } \bigg)$

$\displaystyle = \int^{1}_{0} \sqrt{x} \: dx$

$\displaystyle = \dfrac{2}{3} \left[x ^{ \frac{3}{2} } \right]^{1}_{0}$

$\displaystyle = \dfrac{2}{3} \left[1 - 0\right]$

$=\dfrac{2}{3}$

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