Question

The value of $\underset{n\to \infty }{lim}(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\cdots +\frac{1}{\sqrt{n^2-(n-1{)}^2}})$ is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is A $\frac{\pi }{2}$

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