<s>[INST] 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*}
 [/INST] 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,
$$2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}}
\quad\implies\quad \lim_{n\to\infty} S_n = 2$$
</s>