<s>[INST] determining the limit tending to infinity of n times squared roots I'm having troubles showing that
$$
\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}-\frac23n\sqrt{n}}{\sqrt{n}} = \frac12.
$$
I have tried this but I am not able to solve this.
 [/INST] Applied Stolz-Cesaro Theorem:
$$
\lim_{n\rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}-\frac23n\sqrt{n}}{\sqrt{n}} = \lim_{n\rightarrow \infty}\frac{\sqrt{n+1}-\frac{2}{3}(n+1)\sqrt{n+1}+\frac{2}{3}n\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{2n\sqrt{n}-(2n-1)\sqrt{n+1}}{\sqrt{n+1}-\sqrt{n}}=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{4n^3-(2n-1)^2(n+1)}{2n\sqrt{n}+(2n-1)\sqrt{n+1}}(\sqrt{n+1}+\sqrt{n})=\frac{1}{3}\lim_{n\rightarrow \infty}\frac{(3n-1)(\sqrt{n+1}+\sqrt{n})}{2n\sqrt{n}+(2n-1)\sqrt{n+1}}=\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}.
$$
</s>