What's the limit of this sequence? $\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)$
My attempt:
$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)=\lim_{n \to \infty}\bigg(\frac{\sqrt{1}}{\sqrt{n^3}}+\frac{\sqrt{2}}{\sqrt{n^3}}+\cdots + \frac{\sqrt{n}}{\sqrt{n^3}} \bigg)=0+\cdots+0=0$
 A: Hint: Let $f(x) := \sqrt{x}$. Then
$$\lim_{n \to \infty} \sum_{k=1}^n f\left(\dfrac{k}{n}\right) \dfrac1{n} = \int_0^1 f(x)dx.$$
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$$
\int_{0}^{n}x^{1/2}\,\dd x
<
\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,}
<
\int_{1}^{n + 1}x^{1/2}\,\dd x
\quad\imp\quad
{2 \over 3}\,n^{3/2}
<
\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,}
<
{2 \over 3}\bracks{\pars{n + 1}^{3/2} - 1}
$$
$$
{2 \over 3}\quad
<\quad
{1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\quad
<\quad
{2 \over 3}\bracks{\pars{1 + {1 \over n}}^{3/2} - {1 \over n^{3/2}}}
$$
$$
\vphantom{\Huge A}
$$
$${\large%
\lim_{n \to \infty}
\pars{{1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\,\,}
=
{2 \over 3}}
$$
A: By that same argument, the limit of $$\frac1n(\underbrace{1+\cdots+1}_{\text{$n$ summands}})$$ is zero. Is it?
A: Your limit seems to be 2/3 but I do not know how to prove it except using Njguliyev's hint 
