Limit of partial sum I am trying to find the limit of this infinite sequence:
$$\lim_{n \rightarrow\infty}  \frac{1}{n}\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+\ldots+1\right)$$ 
I can see that:
$$\left(\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\sqrt{\frac{3}{n}}+\ldots+1\right) \lt n$$
So the whole expression is bounded by $1$, but I am having a hard time finding the limit. Any help pointing me into the right direction will be appreciated.
 A: It is the Riemann Sum that converges to $\displaystyle \int_{0}^1 \sqrt{x}dx$.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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Another interesting approach is the use of Stolz-Ces$\grave{a}$ro Theorem:

\begin{align}
&\color{#f00}{\lim_{n \to \infty}{1 \over n}\pars{\root{1 \over n} +
\root{2 \over n} + \root{3 \over n} + \cdots + 1}} =
\lim_{n \to \infty}{1 \over n}\sum_{k = 1}^{n}\root{k \over n} =
\lim_{n \to \infty}{1 \over n^{3/2}}\sum_{k = 1}^{n}k^{1/2}
\\[5mm] = &\
\lim_{n \to \infty}{\pars{n + 1}^{1/2} \over \pars{n + 1}^{3/2} - n^{3/2}} =
\lim_{n \to \infty}{\pars{n + 1}^{1/2}\bracks{\pars{n + 1}^{3/2} + n^{3/2}} \over
\pars{n + 1}^{3} - n^{3}}
\\[5mm] = &\
\lim_{n \to \infty}{\pars{n + 1}^{1/2}\bracks{\pars{n + 1}^{3/2} + n^{3/2}} \over
3n^{2} + 3n + 1}
=
\color{#f00}{2 \over 3}\lim_{n \to \infty}\bracks{%
{\pars{1 + 1/n}^{1/2} \over
1 + 1/n + 1/\pars{3n^{2}}}\,{\pars{1 + 1/n}^{3/2} + 1 \over 2}} = \color{#f00}{2 \over 3}
\end{align}
A: 
I thought it might be instructive to present a way forward that does not rely on Riemann sums.  To that end, we proceed.

Let the sum $S_n$ be given by
$$S_n=\frac{1}{n^{3/2}}\sum_{k=1}^n \sqrt{k}$$
Now, not that we can bound $S_n$ by the integrals
$$\frac {1}{n^{3/2}}\int_0^n \sqrt{x}\,dx\le \frac{1}{n^{3/2}}\sum_{k=1}^n\le \frac{1}{n^{3/2}}\int_1^{n+1}\sqrt{x}\,dx$$
Application of the squeeze theorem finishes the trick.
