Find $\lim\limits_{n \to \infty} \frac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}$.

By the Stolz-Cesaro Theorem, one has \begin{eqnarray} &&\lim_{n \to \infty} \dfrac{\sqrt 1 + \sqrt 2 + \dots + \sqrt{n}}{n\sqrt{n}}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}}{(n+1)\sqrt{n+1}-n\sqrt n}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}}{(n+1)\sqrt{n+1}-n\sqrt n} \dfrac{(n+1)\sqrt{n+1}+n\sqrt n}{(n+1)\sqrt{n+1}+n\sqrt n}\\ &=&\lim_{n \to \infty} \dfrac{\sqrt{n+1}[(n+1)\sqrt{n+1}+n\sqrt n]}{(n+1)^3-n^3}\\ &=&\frac23. \end{eqnarray}

picture......................................

$$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; (n+1) \sqrt {n+1} \; - \; 1 \right) \; < \; \frac{2}{3} \left( \; (n+1) ( 1 +\sqrt n) \; - \; 1 \right) = \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right) $$ $$ \frac{2}{3} n \sqrt n < \mbox{SUM} < \frac{2}{3} \left( \; n \sqrt n + n + \sqrt n \right) $$

For anyone worried about the little estimate above, $$ n + 1 < n + 2 \sqrt n + 1, $$ $$ \sqrt {n+1} \; \; < \; \; 1 + \sqrt n. $$