Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$ \begin{align*}S_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)\end{align*}
how to calculate the limit $s_n$?
\begin{align*}\lim_{n\to \infty } \, S_n\end{align*}
 A: Since $$\frac{1}{\sqrt{k}} \ge \frac{2}{\sqrt{k}+\sqrt{k+1}} = 2(\sqrt{k+1}-\sqrt{k}) \ge \frac{1}{\sqrt{k+1}}$$
We find.
$$\begin{align}
& \sum_{k=1}^{n} \frac{1}{\sqrt{k}} \ge 2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k}) = 2(\sqrt{n+1}-1) \ge 2\sqrt{n} - 2\\
\text{and}\quad & \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = 1 + \sum_{k=1}^{n-1} \frac{1}{\sqrt{k+1}} \le 1 + 2\sum_{k=1}^{n-1}(\sqrt{k+1}-\sqrt{k}) = 2\sqrt{n} - 1
\end{align}
$$
As a result,
$$2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} S_n = 2$$
A: \begin{align}
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{k\,}}
&=
{1 \over \sqrt{n}}\,\sum_{k = 1}^{n}{1 \over \sqrt{n\xi_{k}\,}}\,n\Delta\xi
=
\sum_{k = 1}^{n}{1 \over \sqrt{\xi_{k}\,}}\,\Delta\xi
\sim
\int_{1/n}^{1}{{\rm d}\xi \over \xi^{1/2}}
=
\left.\vphantom{\LARGE A}\;2\xi^{1/2}\right\vert_{1/n}^{1}
\\[3mm]&=
2\left(1 - {1 \over \sqrt{n\,}}\right)
=
2 - {2 \over \sqrt{n\,}} \to \color{#ff0000}{\Large 2}
\quad\mbox{when}\quad
n \to \infty
\end{align}
A: Consider the curve $y=\frac{1}{\sqrt{x}}$. We have
$$\int_1^{n+1}\frac{1}{\sqrt{x}}\,dx\lt 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt \int_0^n \frac{1}{\sqrt{x}}\,dx.$$
Evaluate the integrals. We get 
$$2\sqrt{n+1}-2\lt  1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\lt 2\sqrt{n}.$$
Divide everything by $\sqrt{n}$, and use Squeezing to conclude that our limit is $2$.
A: An answer using the Stolz–Cesàro theorem: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$$
