Q115 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

$\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots .+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)=$

  • A
    $2 \sqrt{\pi}$
  • B
    $\frac{2}{\sqrt{\pi}}$
  • Correct answer
    C
    $\frac{\pi}{2}$
  • D
    $\frac{3 \pi}{2}$

Solution

$\begin{aligned} & \lim _{\rightarrow \infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right) \\ & =\lim _{n \rightarrow \infty} \frac{1}{n}\left(1+\frac{1}{\sqrt{1-\frac{1}{n^2}}}+\ldots+\frac{1}{\sqrt{1-\frac{(n-1)^2}{n^2}}}\right) \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \frac{1}{\sqrt{1-\left(\frac{r}{n}\right)^2}} \\ & =\int_0^1 \frac{d x}{\sqrt{1-x^2}}=\left[\sin ^{-1} x\right]_0^1=\frac{\pi}{2} .\end{aligned}$

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

Relevant Questions