Barrows Sum

Note that $\int_1^n \sqrt{x-1}\,\mathrm{d}x \leq \sum_{i=0}^n \sqrt{i} \leq \int_0^n \sqrt{x}\,\mathrm{d}x.$ Then, we can solve both integrals and divide each member of the above inequalities by $(n+1) \sqrt{n}$ , leading us to $\dfrac{2}{3} \dfrac{(n-1)^{3/2}}{(n+1) \sqrt{n}} \leq \dfrac{ \displaystyle \sum_{i=0}^n \sqrt{i}}{(n+1) \sqrt{n}} \leq \dfrac{2}{3} \dfrac{n^{3/2}}{(n+1) \sqrt{n}}.$ We know that when $n \to \infty$ , the lower and upper bounds tend to $\frac{2}{3}$ . This allow us to conclude, by the Squeeze theorem , that $\dfrac{\displaystyle \sum_{i=0}^n \sqrt{i}}{(n+1) \sqrt{n}} = \frac{2}{3}.$