Sử dụng kết quả ở đây
để có
$\mathop {\lim }\limits_{n \to \infty }\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i}{n}}=\int\limits_{0}^{1}\sqrt x dx$.
Mặt khác
$\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i-1}{n}}=\sum_{i=1}^{n+1}\frac{1}{n}\sqrt{\frac{i-1}{n}}-\frac{1}{n} = \sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i}{n}}-\frac{1}{n} .$
Suy ra
$\mathop {\lim }\limits_{n \to \infty }\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i-1}{n}}=\mathop {\lim }\limits_{n \to \infty }\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i}{n}} -\mathop {\lim }\limits_{n \to \infty }\frac{1}{n}=\mathop {\lim }\limits_{n \to \infty }\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i}{n}}-0$
$=\mathop {\lim }\limits_{n \to \infty }\sum_{i=1}^{n}\frac{1}{n}\sqrt{\frac{i}{n}}=\int\limits_{0}^{1}\sqrt x dx$.