Using riemann integral, calculate following limit of a sequence $$\lim_{n\to \infty } \frac{1}{n \sqrt{n}}(1+\sqrt3 +\sqrt5 + ...+ \sqrt{2n-1}) $$
Can u give me a hint? I have no idea, I'm trying to convert this as $$\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^n \sqrt\frac{2k-1}n$$But problem is that I don't know how to determine integer function and boundaries of that(almost)integral.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\large Riemann\ Sum\ Approach:}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\pars{{1 \over n}
\sum_{k = 1}^{n}\root{2k - 1 \over n}}}
\\[5mm] = &\
\lim_{n \to \infty}\pars{{1 \over n}
\sum_{k = 1}^{2n - 1}\root{k \over n}} -
\lim_{n \to \infty}\pars{{1 \over n}
\sum_{k = 1}^{n - 1}\root{2k \over n}}
\\[5mm] = &\
\underbrace{\int_{0}^{2}x^{1/2}\,\dd x}
_{\ds{4\root{2} \over 3}}\ -\
\underbrace{\root{2}\int_{0}^{1}x^{1/2}\,\dd x}
_{\ds{2\root{2} \over 3}} =
\bbx{2\root{2} \over 3} \\ &
\end{align}

$\ds{\large Stolz–Cesàro\ Theorem:}$
With the
Stolz–Cesàro theorem:
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\pars{{1 \over n}
\sum_{k = 1}^{n}\root{2k - 1 \over n}}} =
\lim_{n \to \infty}{\sum_{k = 1}^{n}\root{2k - 1} \over n^{3/2}}
\\[5mm] = &\
\lim_{n \to \infty}{\sum_{k = 1}^{n + 1}\root{2k - 1} -
\sum_{k = 1}^{n}\root{2k - 1} \over \pars{n + 1}^{3/2} - n^{3/2}}
\\[5mm] = &\
\lim_{n \to \infty}{\root{2n + 1}
 \over \pars{n + 1}^{3/2} - n^{3/2}}
\\[5mm] = &\
\root{2}\lim_{n \to \infty}{n^{1/2}\root{1 + 1/\pars{2n}}
 \over n^{3/2}\bracks{\pars{1 + 1/n}^{3/2} - 1}}
\\[5mm] = &\
\root{2}\lim_{n \to \infty}{1
 \over n\braces{\bracks{1 + 3/\pars{2n}} - 1}} =
\bbx{2\root{2} \over 3} \\ &\
\end{align}
