Question

The value of the $\underset{n\to \infty }{\lim }\left[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\dots +\frac{1}{\sqrt{n^2-(n-1{)}^2}}\right]$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (C)

$\underset{n\to \infty }{\mathrm{Lim}}\left[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1}}+\frac{1}{\sqrt{n^2-2^2}}+\dots \dots +\frac{1}{\sqrt{n^2-(n-1{)}^2}}\right]$
$=\underset{n\to \infty }{\mathrm{Lim}}\left[\frac{1}{\sqrt{n^2-0^2}}+\frac{1}{\sqrt{n^2-1^2}}+\dots \dots +\frac{1}{\sqrt{n^2-(n-1{)}^2}}\right]$
$=\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=0}^{n-1}\frac{1}{\sqrt{n^2-r^2}}=\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=0}^{n-1}\frac{1}{n}\cdot \frac{1}{\sqrt{1-r^2/n^2}}$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x^2}}={\left[{\sin }^{-1}x\right]}_0^1=\frac{\pi }{2}$