Evaluation of the limit $\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)$

Note that $$ 2(\sqrt{k+1}-\sqrt{k})=\frac{2}{\sqrt{k+1}+\sqrt{k}}\leq\frac{1}{\sqrt{k}}\leq \frac{2}{\sqrt{k}+\sqrt{k-1}}=2(\sqrt{k}-\sqrt{k-1}) $$ Hence $$ 2(\sqrt{n+1}-1)=\sum\limits_{k=1}^n 2(\sqrt{k+1}-\sqrt{k})\leq\sum\limits_{k=1}^n\frac{1}{\sqrt{k}}\leq \sum\limits_{k=1}^n 2(\sqrt{k}-\sqrt{k-1})=2\sqrt{n} $$ so $$ \frac{2\sqrt{1 + n}-2}{\sqrt{n}}\leq\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{k}}\leq2 $$ The rest is clear.

Rewrite as $$\frac 1n \left(\frac 1 {\sqrt{\frac 1n}}+\frac 1 {\sqrt{\frac 2n}}+\dots +\frac 1 {\sqrt{\frac nn}} \right)$$ and interpret this as a Riemann sum for the function $$\frac 1{\sqrt x}.$$

This is a standard Stolz-Cezaro problem

$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_{n \to \infty } \frac{\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)}{\sqrt{n}}=$$ $$=\lim_{n}\frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}$$

rationalize the denominator you get

$$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)=\lim_n \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}}=2$$