Question

$\underset{n\to \infty }{lim}\sum _{r=1}^{n-1}\frac{1}{\sqrt{n^2-r^2}}$ is

• A. $\pi$
• B. $\pi /2$
• C. $\pi /4$
• D. none of these

Answer 1

The correct option is B $\pi /2$

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