$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)$
My attempt:
$\lim_{n \to \infty}\frac{1}{n}\bigg(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots + 1 \bigg)=\lim_{n \to \infty}\bigg(\frac{\sqrt{1}}{\sqrt{n^3}}+\frac{\sqrt{2}}{\sqrt{n^3}}+\cdots + \frac{\sqrt{n}}{\sqrt{n^3}} \bigg)=0+\cdots+0=0$
Hint: Let $f(x) := \sqrt{x}$. Then $$\lim_{n \to \infty} \sum_{k=1}^n f\left(\dfrac{k}{n}\right) \dfrac1{n} = \int_0^1 f(x)dx.$$
$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$
$$ \int_{0}^{n}x^{1/2}\,\dd x < \sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,} < \int_{1}^{n + 1}x^{1/2}\,\dd x \quad\imp\quad {2 \over 3}\,n^{3/2} < \sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k\,} < {2 \over 3}\bracks{\pars{n + 1}^{3/2} - 1} $$
$$ {2 \over 3}\quad <\quad {1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\quad <\quad {2 \over 3}\bracks{\pars{1 + {1 \over n}}^{3/2} - {1 \over n^{3/2}}} $$
$$ \vphantom{\Huge A} $$
$${\large% \lim_{n \to \infty} \pars{{1 \over n}\sum_{k = 1}^{n}\sqrt{\vphantom{\large A}k \over n\,}\,\,} = {2 \over 3}} $$
By that same argument, the limit of $$\frac1n(\underbrace{1+\cdots+1}_{\text{$n$ summands}})$$ is zero. Is it?
Your limit seems to be 2/3 but I do not know how to prove it except using Njguliyev's hint
