Q80 / 160 · WBJEE 2015 · Mathematics

$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to

  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • Correct answer
    C
    $\frac{2}{3}$
  • D
    0

Solution

$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ $=\lim _{n \rightarrow \infty}\left(\frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}+\sqrt{n}}{n \sqrt{n}}-\frac{n}{n \times n}\right)$ $=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n}\left(\frac{r}{n}\right)^{1 / 2}-\lim _{n \rightarrow \infty} \frac{1}{n} \times \frac{n}{n}=\int_{0}^{1} \sqrt{x} d x+0=\frac{2}{3}$

Hence, the correct answer is option: 'C'.

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