Limit of $s_n=\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}\right)$

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 $2$.

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, $$2 - \frac{2}{\sqrt{n}} \le S_n \le 2 - \frac{1}{\sqrt{n}} \quad\implies\quad \lim_{n\to\infty} S_n = 2$$

An answer using the Stolz–Cesàro theorem: $$\lim_{n\to\infty} \frac{ \sum_{k=1}^n 1/\sqrt{k} }{\sqrt{n}} = \lim_{n\to\infty} \frac{1/\sqrt{n} }{\sqrt{n} - \sqrt{n-1}} = \lim_{n\to\infty} \frac{\sqrt{n}+\sqrt{n-1} }{\sqrt{n}}=2.$$